Cleanup

src/cat/cat/cat-func.tex

@@ -7,6 +7,7 @@
     \[
     \mor{A, B} \times \mor{B, C} \to \mor{A, C}
     \]
+
     that satisfies the following axioms:
     \begin{enumerate}
         \item[(CAT1)] For any $A, B, A', B' \in \obj{\catc}$, $\mor{A, B}$ and $\mor{A', B'}$ are disjoint or equal, where $\mor{A, B} = \mor{A', B'}$ if and only if $A = A'$ and $B = B'$.

src/cat/cat/index.tex

@@ -3,5 +3,5 @@
 
 \textit{I do not know much about categories, however some concepts from it are useful in phrasing certain properties.}
 
-\input{./src/cat/cat/cat-func.tex}
+\input{./cat-func.tex}
-\input{./src/cat/cat/universal.tex}
+\input{./universal.tex}

src/cat/cat/universal.tex

@@ -22,7 +22,7 @@
     \begin{enumerate}
         \item $P \in \obj{\catc}$.
         \item For each $i \in I$, $\pi_i \in \mor{P, A_i}$.
-        \item[\textbf{(U)}] For any pair $(C, \seqi{f})$ satisfying (1) and (2), there exists a unique $f \in \mor{C, P}$ such that the following diagram commutes
+        \item[(U)] For any pair $(C, \seqi{f})$ satisfying (1) and (2), there exists a unique $f \in \mor{C, P}$ such that the following diagram commutes
 
         \[
         \xymatrix{
@@ -31,6 +31,7 @@
         }
         \]
 
+
         for all $i \in I$.
     \end{enumerate}
 \end{definition}
@@ -50,6 +51,7 @@
         }
         \]
 
+
         for all $i \in I$.
     \end{enumerate}
 \end{definition}
@@ -101,6 +103,7 @@
         }
         \]
 
+
         \item[(U)] For any pair $(B, \bracsn{g^i_B}_{i \in I})$ satisfying (1) and (2), there exists a unique $g \in \mor{A, B}$ such that the following diagram commutes
 
         \[
@@ -110,6 +113,7 @@
         }
         \]
 
+
         for all $i \in I$.
     \end{enumerate}
 \end{definition}
@@ -128,6 +132,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
     }
     \]
 
+
     \item[(U)] For any pair $(B, \bracsn{g^B_i}_{i \in I})$, there exists a unique $g \in \mor{B, A}$ such that the following diagram commutes
 
     \[
@@ -137,6 +142,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
     }
     \]
 
+
     for all $i \in I$.
 \end{enumerate}
 \end{definition}
@@ -155,6 +161,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
         }
         \]
 
+
         \item[(U)] For any pair $(B, \bracsn{S^i_B}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom({A, B})$ such that the following diagram commutes
 
         \[
@@ -164,6 +171,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
         }
         \]
 
+
         for all $i \in I$.
     \end{enumerate}
 \end{proposition}
@@ -176,6 +184,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
         0 &k \ne i, j
     \end{cases}
     \]
+
     Let $N \subset M$ be the submodule generated by $\bracs{x_{i, j}|i, j \in I, i \lesssim j, x \in A_i}$, $A = M/N$, and $\pi: M \to M/N$ be the canonical map.
 
     (1): For each $i \in I$, let
@@ -185,6 +194,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
         0 &k \ne i
     \end{cases}
     \]
+
     and $T^i_A = \pi \circ T^i_M$.
 
     (2): Let $i, j \in I$ with $i \lesssim j$, then for any $x \in A_i$, $T^i_Mx - T^j_M T^i_j x \in N$. Hence $T^i_Ax = T^j_A T^i_jx$.
@@ -193,6 +203,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
     \[
     S_0: M \to B \quad x \mapsto \sum_{i \in I}S^i_B \pi_i x
     \]
+
     then $S_0$ is the unique linear map such that $S_0 \circ T^i_M = S^i_B$ for all $i \in I$. For any $i, j \in I$ with $i \lesssim J$, $S^i_B x = S^j_B T^i_j x$, so $\ker S_0 \supset N$. By the first isomorphism theorem, there exists a unique $S \in \hom(A; B)$ such that $S_0 = S \circ \pi$.
 \end{proof}
 
@@ -208,6 +219,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
         A \ar@{->}[u]^{T^A_i} \ar@{->}[ru]_{T^A_j} &
         }
         \]
+
         \item[(U)] For any pair $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom(B; A)$ such that the following diagram commutes
 
         \[
@@ -217,6 +229,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
         }
         \]
 
+
         for all $i \in I$.
     \end{enumerate}
 \end{proposition}
@@ -225,15 +238,18 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
     \[
     A = \bracs{x \in \prod_{i \in I}A_i \bigg | \pi_j(x) = T^i_j\pi_i(x) \forall i, j \in I, i \lesssim j}
     \]
+
     For each $i \in I$, let $T^A_i = \pi_i$, then $(A, \bracsn{T^A_i}_{i \in I})$ satisfies (1) and (2) by definition of $A$.
 
     (U): Let $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2). Let
     \[
     S: B \to \prod_{i \in I}A_i \quad \pi_i(Sx) = S^B_i
     \]
+
     then for any $x \in B$ and $i, j \in I$ with $i \lesssim j$,
     \[
     \pi_j (Sx) = S^B_jx = T^i_j S^B_ix = T^i_j \pi_i(S x)
     \]
+
     so $S \in \hom(B; A)$, and the diagram commutes. Since any map $f: B \to A$ is uniquely determined by its composition with the projections, $S$ is unique.
 \end{proof}

src/cat/gluing/index.tex

@@ -37,5 +37,6 @@
     \[
     T(\lambda x + y) = T_V(\lambda x + y) = \lambda T_Vx + T_Vy = \lambda Tx + Ty
     \]
+
     and $T \in \hom(E; F)$.
 \end{proof}

src/cat/index.tex

@@ -1,6 +1,6 @@
 \part{General Tools}
 \label{part:part-categories}
 
-\input{./src/cat/cat/index.tex}
+\input{./cat/index.tex}
-\input{./src/cat/gluing/index.tex}
+\input{./gluing/index.tex}
-\input{./src/cat/tricks/index.tex}
+\input{./tricks/index.tex}

src/cat/tricks/dyadic.tex

@@ -14,6 +14,7 @@
     \[
     \text{rk}(q) = \min\bracs{n \in \natp|x \in \mathbb{D}_n}
     \]
+
     is the \textbf{dyadic rank} of $q$.
 \end{definition}
 
@@ -32,7 +33,7 @@
 \begin{proof}
     First suppose that $\text{rk}(x) = 1$. In which case, $x \in \bracs{0, 1/2}$, and either $M(x) = \emptyset$ or $M(y) = \bracs{1}$.
 
-    Assume inductively that the proposition holds for all dyadic rationals of rank $n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$. By \ref{lemma:dyadic-decompose}, there exists a unique $y \in \mathbb{D}_n$ such that $x = y + 2^{-n-1}$. Since $x > 0$, $x \ge 1/2^{-n-1}$, so $y \in [0, 1)$. By the inductive assumption, there exists a unique $M(y) \subset \natp \cap [1, n]$ such that $y = \sum_{k \in M(y)}2^{-k}$. In which case, $M(x) = M(y) \cup \bracs{n}$ is the desired set.
+    Assume inductively that the proposition holds for all dyadic rationals of rank $n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$. By \autoref{lemma:dyadic-decompose}, there exists a unique $y \in \mathbb{D}_n$ such that $x = y + 2^{-n-1}$. Since $x > 0$, $x \ge 1/2^{-n-1}$, so $y \in [0, 1)$. By the inductive assumption, there exists a unique $M(y) \subset \natp \cap [1, n]$ such that $y = \sum_{k \in M(y)}2^{-k}$. In which case, $M(x) = M(y) \cup \bracs{n}$ is the desired set.
 \end{proof}
 
 \begin{proposition}
@@ -48,15 +49,18 @@
     \[
     \phi(x) + \phi(y) = g_2 + g_2 \le g_1 = \phi(x + y)
     \]
+
     In the second, $\phi(x) + \phi(y) = \phi(x + y)$.
 
-    Now assume inductively that the proposition holds for all $x, y \in \mathbb{D} \cap (0, 1)$ with $\text{rk}(x) = n$ and $\text{rk}(y) \le n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$ and $y \in \mathbb{D}_{n+1}$. By \ref{lemma:dyadic-decompose}, there exists $x_0 \in \mathbb{D}_n$ such that $x = x_0 + 2^{-n-1}$. If $y \in \mathbb{D}_n$, then
+    Now assume inductively that the proposition holds for all $x, y \in \mathbb{D} \cap (0, 1)$ with $\text{rk}(x) = n$ and $\text{rk}(y) \le n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$ and $y \in \mathbb{D}_{n+1}$. By \autoref{lemma:dyadic-decompose}, there exists $x_0 \in \mathbb{D}_n$ such that $x = x_0 + 2^{-n-1}$. If $y \in \mathbb{D}_n$, then
     \[
     \phi(x) + \phi(y) = \phi(x_0) + \phi(y) + g_{n+1} \le \phi(x_0 + y) + g_{n+1} = \phi(x + y)
     \]
+
     by the inductive assumption. Otherwise, there exists $y_0 \in \mathbb{D}_n$ such that $y = y_0 + 2^{-n-1}$, so
     \[
     \phi(x) + \phi(y) \le \phi(x_0) + \phi(y_0) + g_n = \phi(x_0) + \phi(y_0) + \phi(2^{-n}) \le \phi(x + y)
     \]
+
     by the inductive assumption.
 \end{proof}

src/cat/tricks/index.tex

@@ -1,5 +1,5 @@
 \chapter{Inequalities and Computations}
 \label{chap:tricks}
 
-\input{./src/cat/tricks/dyadic.tex}
+\input{./dyadic.tex}
-\input{./src/cat/tricks/product.tex}
+\input{./product.tex}

src/cat/tricks/product.tex

@@ -7,10 +7,12 @@
     \[
     ab \le \frac{a^p}{p} + \frac{b^q}{q}
     \]
+
     and for any $\eps > 0$,
     \[
     ab \le \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
     \]
+
 \end{lemma}
 \begin{proof}
     Since $x \mapsto \exp(x)$ is convex,
@@ -22,4 +24,5 @@
     \[
     ab = (\eps p)^{1/p}a \cdot \frac{b}{(\eps p)^{1/p}} \le \eps a^p + \frac{b^q}{q}(\eps p)^{-(1/p)q} = \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
     \]
+
 \end{proof}

src/dg/derivative/derivative.tex

@@ -15,6 +15,7 @@
     \[
     f(x_0 + h) = f(x_0) + Th + r(h)
     \]
+
     for all $h \in V$. In which case, $T = D_{\mathcal{HR}}f(x_0)$ is the unique element of $\ch$ satisfying the above, known as the \textbf{$\mathcal{HR}$-derivative} of $f$ at $x_0$.
 \end{definition}
 \begin{proof}
@@ -22,6 +23,7 @@
 	\[
 	f(x_0 + h) - f(x_0) = Sh + r(h) = Th + s(h)
 	\]
+
 	for all $h \in V$, then $(S - T)(h) = (s - r)(h)$. By (T), $S - T = 0$. Hence $S = T$.
 \end{proof}
 
@@ -31,6 +33,7 @@
     \[
     D_{\mathcal{HR}}(\lambda f + g)(x_0) = \lambda D_{\mathcal{HR}}f(x_0) + D_{\mathcal{HR}}g(x_0)
     \]
+
 \end{proposition}
 \begin{proof}
     Let $V \in \cn_E(0)$ and $r, s \in \calr$ such that
@@ -42,4 +45,5 @@
     \[
     (\lambda f + g)(x_0+h) - (\lambda f + g)(x_0) = \underbrace{[\lambda D_{\mathcal{HR}}f(x_0) + D_{\mathcal{HR}}g(x_0)]}_{\in \ch}h + \underbrace{(\lambda r + s)}_{\in \calr}(h)
     \]
+
 \end{proof}

src/dg/derivative/higher.tex

@@ -12,22 +12,24 @@
     \end{enumerate}
     In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) \in L(E; B^{n-1}_\sigma(E; F))$ is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
 
-    The mapping $f: U \to F$ is \textbf{$n$-fold $\sigma$-differentiable on $U$} if it is $n$-fold $\sigma$-differentiable at every point in $U$. Under the identification $B_\sigma(E; B^{n-1}_\sigma(E; F)) = B_\sigma^{n}(E; F)$ given by \ref{proposition:multilinear-identify},
+    The mapping $f: U \to F$ is \textbf{$n$-fold $\sigma$-differentiable on $U$} if it is $n$-fold $\sigma$-differentiable at every point in $U$. Under the identification $B_\sigma(E; B^{n-1}_\sigma(E; F)) = B_\sigma^{n}(E; F)$ given by \autoref{proposition:multilinear-identify},
     \[
     D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
     \]
+
     is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
 \end{definition}
 
-\begin{proposition}[{{\cite[Theorem 5.1.1]{Cartan}}}]
+\begin{theorem}[{{\cite[Theorem 5.1.1]{Cartan}}}]
-\label{proposition:derivative-symmetric}
+\label{theorem:derivative-symmetric-frechet}
-    Let $E, F$ be Banach spaces, $U \subset E$ be open, $n \in \natp$, and $f: U \to F$ be a function $n$-times Fréchet-differentiable at $x_0 \in U$, then $D^nf(x_0) \in L^2(E; F)$ is symmetric.
+    Let $E, F$ be Banach spaces, $U \subset E$ be open, $n \in \natp$, and $f: U \to F$ be a function $n$-times Fréchet-differentiable at $x \in U$, then $D^nf(x) \in L^n(E; F)$ is symmetric.
-\end{proposition}
+\end{theorem}
 \begin{proof}
-    First suppose that $n = 2$. Let $r > 0$ such that $B(x_0, 2r) \subset U$, and define $A: B_E(0, r) \times B_E(0, r) \to F$ by
+    First suppose that $n = 2$. Let $r > 0$ such that $B(x, 2r) \subset U$, and define $A: B_E(0, r) \times B_E(0, r) \to F$ by
     \[
     A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)
     \]
+
     then there exists $r_1 \in \mathcal{R}_{B(E)}$ such that
     \begin{align*}
         A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
@@ -41,23 +43,89 @@
     \[
     B_h(k) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k)
     \]
+
     then
     \[
     B_h(k) - B_h(0) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k) -f(x + h) + f(x)
     \]
+
     Now, there exists $r_2, r_3 \in \mathcal{R}_{B(E)}$ such that for any $k \in B(0, r)$,
     \begin{align*}
     DB_h(k) &= Df(x + h + k) - Df(x + k) - Df(x + h) + Df(x) \\
     &= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) - Df(x) - D^2f(x)(k) + r_2(k) + r_3(h) \\
     &=r_2(k) + r_3(h)
     \end{align*}
-    By the mean value theorem,
+    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem},
     \[
     \norm{B_h(k) - B_h(0)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
     \]
+
     As the above argument is symmetric,
     \[
     \norm{D^2f(x)(h, k) - D^2f(x)(k, h)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
     \]
+
     so $D^2f(x)(h, k) - D^2f(x)(k, h) = 0$.
+
+    Now suppose that the proposition holds for $n$. Identify $L^n(E; F) = L^{2}(E; L^{n-2}(E; F))$, then for any $\seqf[n]{x_j} \subset E$,
+    \[
+    Df(x)(x_1, \cdots, x_n) = Df(x)(x_1, x_2)(x_3, \cdots, x_n) = Df(x)(x_2, x_1)(x_3, \cdots, x_n) = Df(x)(x_2, x_1, x_3, \cdots, x_n)
+    \]
+
+    Since any element $\sigma \in S_n$ that does not fix $x_1$ is the composition of the transposition $(12)$ and an element that fixes $x_1$, $Df(x)$ is symmetric.
+\end{proof}
+
+\begin{theorem}[{{\cite[Proposition 4.5.14]{Bogachev}}}]
+\label{theorem:derivative-symmetric}
+    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed system that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
+\end{theorem}
+\begin{proof}
+    Let $\seqf{h_j} \subset E$, $E_0$ be the subspace generated by $\seqf{h_j}$, and $g = f|_{E_0 \cap U}: E_0 \cap U \to F$. Since $\sigma$ includes all bounded sets contained in finite-dimensional spaces, for any $\phi \in F^*$, the mapping $\phi \circ g: E_0 \cap U \to K$ is $n$-times Fréchet-differentiable, with
+    \[
+    D_{B(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
+    \]
+
+    by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^n(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
+\end{proof}
+
+\begin{proposition}[Power Rule]
+\label{proposition:multilinear-derivative}
+    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a Hausdorff locally convex space, and
+    \[
+    T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
+    \]
+
+    be symmetric. For any $x \in E$ and $1 \le k \le n$, let $x^{(k)}$ denote the tuple of $x$ repeated $k$ times, then:
+    \begin{enumerate}
+        \item The mapping $f: E \to F \quad x \mapsto T(x^{(n)})$ is infinitely $\sigma$-differentiable on $E$.
+        \item For each $1 \le k \le n$ and $x, h \in E$,
+        \[
+        Df(x)(h_1, \cdots, h_k) = \frac{n!}{(n-k)!} T(x^{(n-k)}, h_1, \cdots, h_k)
+        \]
+
+        In particular, $D^kf = n! \cdot T$.
+        \item For each $k > n$ and $x \in E$, $Df(x) = 0$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Suppose inductively that (2) holds for $0 \le k \le n$. Let $G =  B^{k}_\sigma(E; F)$, then $D^k_\sigma f \in B^{n-k}_\sigma(E; G)$ under the identification $B^n_\sigma(E; F) = B^{n-k}_\sigma(E; B^k_\sigma(E; F))$ in \autoref{proposition:multilinear-identify}. By \autoref{theorem:derivative-symmetric}, $D^k_\sigma f$ is also symmetric, so using the Binomial formula,
+    \begin{align*}
+        D^k_\sigma f(x + h) &= \sum_{r = 0}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)}) \\
+        &= f(x) + (n-k)D^k_\sigma f(x^{(n-k-1)}, h) \\
+        &+ \underbrace{\sum_{r = 2}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)})}_{r(h)}
+    \end{align*}
+
+    For each $k \ge 2$, let $A \in \sigma$ and $U \in \cn_F(0)$, then since $D^k_\sigma f \in B^{n-k}_\sigma(E; F)$, there exists $t > 0$ such that
+    \[
+    \frac{D^k_\sigma f(x^{(n-k)}, (sA)^{(k)})}{t} = s^{k-1}D^k_\sigma f(x^{(n-k)}, A^{(k)}) \subset U
+    \]
+
+    for all $s \in (0, t)$. Hence $r \in \mathcal{R}_\sigma(E; G)$, and
+    \[
+    D^{k+1}_\sigma f(x + h) = f(x) + \frac{n!}{(n-k-1)!}T(x^{(n-k-1)}, h_1, \cdots, h_{k+1}) + r(h)
+    \]
+
+    by the inductive hypothesis.
+
+    (3): Since $D^n_\sigma f$ is constant, $D^k_\sigma f = 0$ for all $k > n$.
 \end{proof}

src/dg/derivative/index.tex

@@ -1,7 +1,8 @@
 \chapter{Differential Calculus}
 \label{chap:diff}
 
-\input{./src/dg/derivative/derivative.tex}
+\input{./derivative.tex}
-\input{./src/dg/derivative/sets.tex}
+\input{./sets.tex}
-\input{./src/dg/derivative/mvt.tex}
+\input{./mvt.tex}
-\input{./src/dg/derivative/higher.tex}
+\input{./higher.tex}
+\input{./taylor.tex}

src/dg/derivative/mvt.tex

@@ -7,6 +7,7 @@
     \[
     D^+f(x) = \lim_{t \downto 0} \frac{f(x + t) - f(x)}{t}
     \]
+
     exists.
 \end{definition}
 
@@ -26,12 +27,14 @@
     \[
     S = \bracs{x \in [a, b] \bigg | f(y) - f(a) \le g(y) - g(a) + \eps\sum_{n \in N(x)}2^{-n} \quad \forall y \in [a, x]}
     \]
+
     then by continuity of $f$ and $g$, $S$ is closed.
 
     Let $x \in S$ and suppose that $s < b$. If $x \in [a, b] \setminus N$, then since
     \[
     \lim_{t \downto 0}\frac{f(x + t) - f(x)}{t} < \lim_{t \downto 0}\frac{g(x + t) - f(x)}{t}
     \]
+
     there exists $\delta > 0$ such that $f(x + t) - f(x) < g(x + t) < g(x)$ for all $t \in [0, \delta)$. In which case, $[x, x + \delta) \subset S$.
 
     If $x = x_n \in N$, then by continuity of $f$ and $g$, there exists $\delta > 0$ such that $f(x + t) - f(x) \le g(x + t) - g(x) + 2^{-n}$ for all $t \in (0, \delta)$. Hence $[x, x + \delta) \subset S$ as well.
@@ -53,6 +56,7 @@
     \[
     f(b) - f(a) \in [g(b) - g(a)]B
     \]
+
 \end{theorem}
 \begin{proof}
     Let $\phi \in E^*$ and $x \in [a, b] \setminus N$, then since $g$ is non-decreasing,
@@ -61,15 +65,17 @@
         D^+(\phi \circ f)(x) &\le \sup_{z \in B}\phi(z) \cdot \lim_{t \downto 0}\frac{\phi(g(x + t) - g(x))}{t} \\
         D^+(\phi \circ f)(x) &\le  D^+g(x) \cdot \sup_{z \in B}\phi(z)
     \end{align*}
-    By \ref{lemma:right-differentiable-inequality},
+    By \autoref{lemma:right-differentiable-inequality},
     \[
     \phi(f(b) - f(a)) \le (g(b) - g(a)) \cdot \sup_{z \in B}\phi(z)
     \]
 
-    Suppose that $f(b) - f(a) \not\in [g(b) - g(a)]B$. Given that $B$ is closed and convex, by the Hahn-Banach theorem (\ref{theorem:hahn-banach-geometric-2}), there exists $\phi \in E^*$ such that
+
+    Suppose that $f(b) - f(a) \not\in [g(b) - g(a)]B$. Given that $B$ is closed and convex, by the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in E^*$ such that
     \[
     \phi(f(b) - f(a)) > \sup_{x \in B}\phi[(g(b) - g(a))b] = \phi(g(b) - g(a)) \cdot \sup_{x \in B}\phi(x)
     \]
+
     which is impossible. Therefore $f(b) - f(a) \in [g(b) - g(a)]B$.
 \end{proof}
 
@@ -80,16 +86,19 @@
     \[
     f(b) - f(a) \in \overline{\text{Conv}\bracs{Df(x)(b - a)| x \in (a, b) \setminus N}}
     \]
+
 \end{theorem}
 \begin{proof}
-    By \ref{proposition:derivative-sets-real}, $f$ is right-differentiable on $(a, b) \setminus N$ with
+    By \autoref{proposition:derivative-sets-real}, $f$ is right-differentiable on $(a, b) \setminus N$ with
     \[
     D^+f(x) = Df(x) \in \bracs{Df(y)|y \in (a, b) \setminus N}
     \]
-    for all $x \in (a, b)$. Let $g(x) = x$, then by \ref{theorem:right-differentiable-convex-form},
+
+    for all $x \in (a, b)$. Let $g(x) = x$, then by \autoref{theorem:right-differentiable-convex-form},
     \[
     f(b) - f(a) \in \overline{(b - a)\text{Conv}\bracs{Df(x)|x \in (a, b) \setminus N}}
     \]
+
 \end{proof}
 
 \begin{theorem}[Mean Value Theorem]
@@ -98,15 +107,17 @@
     \[
     f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
+
     where $[x, y] = \bracs{(1 - t)x + ty|y \in [0, 1]}$.
 \end{theorem}
 \begin{proof}
-    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$. Since $f$ is Gateaux-differentiable, $g$ is differentiable by the chain rule \ref{proposition:chain-rule-sets-conditions} with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \ref{proposition:derivative-sets-real}.
+    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$. Since $f$ is Gateaux-differentiable, $g$ is differentiable by the chain rule \autoref{proposition:chain-rule-sets-conditions} with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \autoref{proposition:derivative-sets-real}.
 
-    By the Mean Value Theorem (\ref{theorem:mean-value-theorem-line}),
+    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line},
     \[
     f(y) - f(x) = g(1) - g(0) \in \overline{\text{Conv}\bracs{Dg(t)|t \in [0, 1]}} = \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
+
 \end{proof}
 
 \begin{proposition}
@@ -118,9 +129,10 @@
     \[
     f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}} = \bracs{0}
     \]
-    by the Mean Value Theorem (\ref{theorem:mean-value-theorem-line}).
 
-    Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. For any $y \in W$, by \ref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. Therefore $W \supset W \cup (U + y)$, and $W$ is open by \ref{lemma:openneighbourhood}.
+    by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}.
+
+    Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. For any $y \in W$, by \autoref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. Therefore $W \supset W \cup (U + y)$, and $W$ is open by \autoref{lemma:openneighbourhood}.
 
     For any $y \in \ol W \cap V$, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. As $y \in \ol W \cap V$, $U \cap W \ne \emptyset$. Thus $f(y) = f(x)$, $y \in W$, and $W$ is relatively closed.
 

src/dg/derivative/sets.tex

@@ -29,6 +29,7 @@
     \[
     f(x_0 + h) = f(x_0) + Th + r(h)
     \]
+
     for all $h \in V$.
 
     The linear map $T \in L(E; F)$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}, denoted $D_{\sigma}f(x_0)$.
@@ -56,16 +57,19 @@
     \[
     D_\sigma(g \circ f)(x_0) = D_\tau g(f(x_0)) \circ D_\sigma f(x_0)
     \]
+
 \end{proposition}
 \begin{proof}
     Since $g$ is $\tau$-differentiable at $f(x_0)$, there exists $s \in \mathcal{R}_\tau(F; G)$ such that
     \[
     g(f(x_0) + h) = g \circ f (x_0) + D_\tau g(f(x_0))h + s(h)
     \]
+
     for all $h \in F$ such that $f(x_0) + h \in V$. By differentiability of $f$, there exists $r \in \mathcal{R}_\sigma(E; F)$ such that
     \[
     f(x_0 + h) = f(x_0) + D_\sigma f(x_0)h + r(h)
     \]
+
     for all $h \in E$ such that $x_0 + h \in U$. Therefore for all $h \in E$ with $x_0 + h \in U$,
     \begin{align*}
     g \circ f(x_0 + h) &= g \circ f (x_0) + D_\tau g(f(x_0)) \circ D_\sigma f(x_0)h \\
@@ -87,7 +91,7 @@
         \item For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
         \item For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
     \end{enumerate}
-    and by \ref{proposition:chain-rule-sets}, $\sigma$-derivatives and $\tau$-derivatives satisfy the Chain rule.
+    and by \autoref{proposition:chain-rule-sets}, $\sigma$-derivatives and $\tau$-derivatives satisfy the Chain rule.
 \end{proposition}
 \begin{proof}
     (1): Let $A \in \sigma$ and $U \in \cn_G(0)$. Since $T$ is continuous, there exists $V \in \cn_F(0)$ such that $T(V) \subset U$. Since $r \in \mathcal{R}_\sigma(E; G)$, there exists $t > 0$ such that $r(sA)/s \in V$ for all $s \in (0, t)$. In which case, $T \circ r(sA)/s \in U$ for all $s \in (0, t)$.
@@ -98,21 +102,24 @@
     \[
     \limv{n} \frac{1}{t_n}s \circ (T + r)(t_na_n) = \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
     \]
+
     Since $\bracs{t_n^{-1}r(t_na_n)|n \in \natp}$ is a convergent sequence, it is contained in a compact set. Thus
     \[
     B = \bracs{Ta_n + \frac{r(t_na_n)}{t_n}\bigg | n \in \natp} \subset T(A) + \bracs{\frac{r(t_na_n)}{t_n}|n \in \natp}
     \]
+
     is contained in a compact set if $A$ is compact, and bounded if $A$ is bounded. Given that $s \in \mathcal{R}_\sigma(E; F)$, $t^{-1}s(tx) \to 0$ as $t \downto 0$ uniformly on $B$. Therefore
     \[
     \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
     \]
+
 \end{proof}
 
 
 
 
 \begin{remark}
-    In \ref{definition:differentiation-small}, the system $\sigma$ can be chosen based on the bornology of $E$, and the definition of small-ness depends exclusively on $\sigma$. As such, there is an apparent disconnect between differentiation and the topology of the domain.
+    In \autoref{definition:differentiation-small}, the system $\sigma$ can be chosen based on the bornology of $E$, and the definition of small-ness depends exclusively on $\sigma$. As such, there is an apparent disconnect between differentiation and the topology of the domain.
 
     Consider for example a Hilbert space equipped with its norm and weak topology. The norm itself is differentiable with respect to both topologies, because the bounded sets coincide. Moreover, the data for differentiability needs to only come from a neighbourhood of $0$ in the norm topology. As such, a function may be differentiable even if its domain is too small to have an interior.
 
@@ -130,6 +137,7 @@
         \[
         \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
         \]
+
         exists. In which case, the above limit is identified with the derivative of $f$ at $0$.
         \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, if $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.
     \end{enumerate}
@@ -147,5 +155,6 @@
     \[
     \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
     \]
+
     and $Df(x_0) = T$.
 \end{proof}

src/dg/derivative/taylor.tex

@@ -0,0 +1,123 @@
+\section{Taylor's Formula}
+\label{section:taylor}
+
+\begin{theorem}[Taylor's Formula, Lagrange Remainder {{\cite[Theorem 4.7.1]{Bogachev}}}]
+\label{theorem:taylor-lagrange}
+    Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $S \subset [a, b]$ be at most countable, $n \in \natp$, and $f \in C^{n}([a, b]; E)$ be $(n+1)$-fold differentiable on $[a, b] \setminus N$, then
+    \[
+    f(b) - f(a) - \sum_{k = 1}^n \frac{1}{k!}D^kf(a)(b - a)^k
+    \]
+
+    is contained in the closed convex hull\footnote{It may be possible to sharpen the below claim to include the $1/(n+1)!$ factor. However, I was not able to follow the proof for this.} of
+    \[
+    \bracs{D^{n+1}f(s)(t - a)^{n+1}  | s \in (a, b) \setminus N, t \in [a, b]}
+    \]
+
+\end{theorem}
+\begin{proof}
+    If $n = 0$, then the theorem is the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}.
+
+    Suppose inductively that the theorem holds for $n$. Let
+    \[
+    g: [a, b] \to E \quad t \mapsto f(t) - \sum_{k = 1}^{n+1} \frac{1}{k!}D^kf(a)(t - a)^k
+    \]
+
+    then for any $t \in (0, 1)$,
+    \begin{align*}
+    Dg(t) &= Df(t) - \sum_{k = 1}^{n+1} \frac{1}{(k-1)!}D^{k}f(a)(t - a)^{k-1} \\
+    &= Df(t) - Df(a) - \sum_{k = 1}^{n} \frac{1}{k!}D^{k+1}f(a)(t - a)^{k}
+    \end{align*}
+    by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line},
+    \[
+    g(b) - g(a) = f(b) - f(a) - \sum_{k = 1}^{n+1} \frac{1}{k!}D^kf(a)(t - a)^k
+    \]
+
+    is contained in the closed convex hull of
+    \[
+    \bracs{\braks{Df(t) - Df(a) - \sum_{k = 1}^{n} \frac{1}{k!}D^{k+1}f(a)(t - a)^{k}}(b - a) \bigg | t \in [a, b]}
+    \]
+
+
+    By the inductive hypothesis applied to $Df$, for any $t \in [a, b]$,
+    \[
+    Df(t) - Df(a) - \sum_{k = 1}^n \frac{1}{k!}D^{k+1}f(a)(t - a)^k
+    \]
+
+    is contained in the closed convex hull of
+    \[
+    \bracs{D^{n+2}f(s)(t - a)^{n+1} | s \in (a, t) \setminus N}
+    \]
+
+    Therefore
+    \[
+    f(b) - f(a) - \sum_{k = 1}^{n+1} \frac{1}{k!}D^kf(a)(b - a)^k
+    \]
+
+    is contained in the convex hull of
+    \[
+    \bracs{D^{n+2}f(s)(t - a)^{n+2} | s \in (a, t) \setminus N, t \in [a, b]}
+    \]
+
+\end{proof}
+
+\begin{theorem}[Taylor's Formula, Peano Remainder {{\cite[Theorem 4.7.3]{Bogachev}}}]
+\label{theorem:taylor-peano}
+    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
+    \[
+    g(x_0 + h) = g(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
+    \]
+
+\end{theorem}
+\begin{proof}
+    Let
+    \[
+    r(h) = g(x_0 + h) - g(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})
+    \]
+
+    For any $1 \le k \le n$, $D^k_\sigma(x_0) \in B^k_\sigma(E; F)$ is symmetric by \autoref{theorem:derivative-symmetric}. Let $T_k(h) = \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})$, then by \autoref{proposition:multilinear-derivative}, for any $\bracs{t_j}_1^\ell \in E$,
+    \[
+    D^\ell_\sigma T_k(h)(t_1, \cdots, t_\ell) = \begin{cases}
+        0 &\ell > k \\
+        D^k_\sigma(x_0)(t_1, \cdots, t_\ell) &\ell = k \\
+         \frac{1}{(k-\ell)!}D^k_\sigma(x_0)(h^{(k - \ell)}, t_1, \cdots, t_\ell) & \ell < k
+    \end{cases}
+    \]
+
+    so
+    \[
+    D^k_\sigma r(0) = D^k_\sigma g(x_0) - D^k_\sigma(x_0) = 0
+    \]
+
+
+    If $n = 1$, then the theorem holds by definition of the derivative. Now suppose inductively that the theorem holds for $n$. By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem},
+    \[
+    r(h) = r(h) - r(0) \in \overline{\text{Conv}\bracs{D_\sigma r(s)(h)| s \in [0, h]}}
+    \]
+
+    For any $A \in \sigma$ and $t > 0$,
+    \[
+    \frac{r(tA)}{t^{n+1}} \subset \overline{\text{Conv}\bracs{t^{-n}D_\sigma r(s)(h)|s \in [0, h], h \in tA}}
+    \]
+
+    Let $U \in \cn_F(0)$ be convex and circled, then by the inductive assumption applied to $D_\sigma r$, there exists $t_0 \in (0, 1)$ such that for any $t \in (0, t_0)$.
+    \[
+    \frac{D_\sigma r(tA)}{t^n} \subset \bracs{T \in L(E; F)| T(A) \subset U}
+    \]
+
+    Since $U$ is circled,
+    \[
+    \bracs{T \in L(E; F)| T(A) \subset U} = \bracs{T \in L(E; F)| T(tA) \subset U \forall t \in (0, 1)}
+    \]
+
+    so
+    \[
+    \bracs{t^{-n}D_\sigma r(s)(h)|s \in [0, h], h \in tA} \subset U
+    \]
+
+    and
+    \[
+    \frac{r(tA)}{t^{n+1}} \subset \overline{\text{Conv}\bracs{t^{-n}D_\sigma r(s)(h)|s \in [0, h], h \in tA}} \subset \overline{U}
+    \]
+
+    Therefore $r \in \mathcal{R}_\sigma^{n+1}(E; F)$.
+\end{proof}

src/dg/index.tex

@@ -1,4 +1,4 @@
 \part{Differential Geometry}
 \label{part:diffgeo}
 
-\input{./src/dg/derivative/index.tex}
+\input{./derivative/index.tex}

src/fa/index.tex

@@ -2,8 +2,8 @@
 \label{part:fa}
 
 
-\input{./src/fa/tvs/index.tex}
+\input{./tvs/index.tex}
-\input{./src/fa/lc/index.tex}
+\input{./lc/index.tex}
-\input{./src/fa/norm/index.tex}
+\input{./norm/index.tex}
-\input{./src/fa/rs/index.tex}
+\input{./rs/index.tex}
-\input{./src/fa/lp/index.tex}
+\input{./lp/index.tex}

src/fa/lc/continuous.tex

@@ -13,11 +13,11 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    $(1) \Leftrightarrow (2) \Leftrightarrow (3)$: By \ref{definition:continuous-linear}.
+    $(1) \Leftrightarrow (2) \Leftrightarrow (3)$: By \autoref{definition:continuous-linear}.
 
     $(2) \Rightarrow (4)$: $x \mapsto [Tx]_F$ is a continuous seminorm on $E$.
 
-    $(4) \Rightarrow (3)$: Let $U \in \cn_F(0)$ be convex, circled, and radial, then its gauge $[\cdot]_U$ is a continuous seminorm on $F$ by \ref{definition:locally-convex}. Thus there exists a continuous seminorm $[\cdot]_E$ such that $[Tx]_U \le [x]_E$. In which case, $V = \bracs{x \in E| [x]_E < 1} \in \cn_E(0)$ with $T(V) \subset U$. Therefore $T$ is continuous at $0$, and continuous by \ref{definition:continuous-linear}.
+    $(4) \Rightarrow (3)$: Let $U \in \cn_F(0)$ be convex, circled, and radial, then its gauge $[\cdot]_U$ is a continuous seminorm on $F$ by \autoref{definition:locally-convex}. Thus there exists a continuous seminorm $[\cdot]_E$ such that $[Tx]_U \le [x]_E$. In which case, $V = \bracs{x \in E| [x]_E < 1} \in \cn_E(0)$ with $T(V) \subset U$. Therefore $T$ is continuous at $0$, and continuous by \autoref{definition:continuous-linear}.
 \end{proof}
 
 \begin{proposition}
@@ -29,6 +29,7 @@
         \[
         [Tx]_F \le \prod_{j = 1}^n [x_j]_{E_j}
         \]
+
     \end{enumerate}
 \end{proposition}
 \begin{proof}

src/fa/lc/convex.tex

@@ -13,6 +13,7 @@
     \[
     \bracs{tx + (1 - t)y|t \in (0, 1)} \subset A^o
     \]
+
 \end{lemma}
 \begin{proof}
     Fix $t \in (0, 1)$. Using translation, assume without loss of generality that $tx + (1 - t)y = 0$. In which case, $x = \alpha y$ where $\alpha = (1 - t)/t$. By (TVS2), $\alpha A^o \in \cn^o(y)$. Since $y \in \ol{A}$, $\alpha A^o \cap A \ne \emptyset$, so there exists $z \in A^o$ such that $\alpha z \in A$.
@@ -21,10 +22,12 @@
     \[
     \mu z + (1 - \mu)\alpha z = \frac{\alpha z}{\alpha - 1} + \frac{(\alpha - 1 - \alpha)\alpha z}{\alpha - 1} = 0
     \]
+
     By (TVS1) and (TVS2),
     \[
     U = \underbrace{\bracs{\mu w + (1 - \mu)\alpha z|w \in A^o}}_{\subset A} \in \cn^o(0)
     \]
+
     so $0 \in A^o$.
 \end{proof}
 
@@ -39,12 +42,13 @@
     \end{enumerate}
 \end{lemma}
 \begin{proof}
-    (1): By \ref{lemma:convex-interior}.
+    (1): By \autoref{lemma:convex-interior}.
 
-    (2): Let $x, y \in \ol{A}$. By \ref{definition:closure}, there exists filters $\fF, \mathfrak{G} \subset 2^A$ such that $\fF$ converges to $x$ and $\mathfrak{G}$ converges to $y$. In which case,
+    (2): Let $x, y \in \ol{A}$. By \autoref{definition:closure}, there exists filters $\fF, \mathfrak{G} \subset 2^A$ such that $\fF$ converges to $x$ and $\mathfrak{G}$ converges to $y$. In which case,
     \[
     \fU = \bracs{tE + (1 - t)F|t \in [0, 1], E \in \fF, F \in \mathfrak{G}} \subset 2^A
     \]
+
     converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.
 \end{proof}
 
@@ -57,7 +61,8 @@
     \[
     y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o}
     \]
-    by \ref{lemma:convex-interior}.
+
+    by \autoref{lemma:convex-interior}.
 \end{proof}
 
 
@@ -97,7 +102,7 @@
     \end{enumerate}
 \end{lemma}
 \begin{proof}
-    $(3) \Rightarrow (1)$: Let $\eps > 0$, then there exists $V \in \cn(0)$ such that $[x] < \eps$ for all $x \in V$. In which case, for any $x, y \in E$ with $x - y \in V$, $\abs{[x] - [y]} \le [x - y] < \eps$. By \ref{proposition:tvs-uniform}, $[\cdot]$ is uniformly continuous.
+    $(3) \Rightarrow (1)$: Let $\eps > 0$, then there exists $V \in \cn(0)$ such that $[x] < \eps$ for all $x \in V$. In which case, for any $x, y \in E$ with $x - y \in V$, $\abs{[x] - [y]} \le [x - y] < \eps$. By \autoref{proposition:tvs-uniform}, $[\cdot]$ is uniformly continuous.
 \end{proof}
 
 
@@ -122,6 +127,7 @@
     \[
     [\cdot]_A: E \to [0, \infty) \quad x \mapsto \inf\bracsn{\lambda > 0| \lambda^{-1}x \in A}
     \]
+
     is the \textbf{gauge/Minkowski functional} of $A$, and
     \begin{enumerate}
         \item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
@@ -139,6 +145,7 @@
     \[
     (\lambda + \mu)^{-1}(x + y) = t\lambda^{-1}x + (1 - t)\mu^{-1}y
     \]
+
     then $(\lambda + \mu)^{-1} \in A$, and $\lambda + \mu \ge [x + y]_A$. Thus $[x + y]_A \le [x]_A + [y]_A$.
 \end{proof}
 
@@ -153,7 +160,7 @@
     If the above holds, then $E$ is a \textbf{locally convex} space.
 \end{definition}
 \begin{proof}
-    $(1) \Rightarrow (2)$: Let $U \in \cn(0)$ be convex. By \ref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$. Let $W = \bracs{tx + (1 - t)y|x, y \in V}$ be the convex hull of $V$, then $W \subset U$ is convex and circled.
+    $(1) \Rightarrow (2)$: Let $U \in \cn(0)$ be convex. By \autoref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$. Let $W = \bracs{tx + (1 - t)y|x, y \in V}$ be the convex hull of $V$, then $W \subset U$ is convex and circled.
 
     $(2) \Rightarrow (3)$: For each $V \in \cn(0)$ convex, circled, and radial, let $[\cdot]_V: E \to [0, \infty)$ be its gauge, then $[\cdot]_V$ is a seminorm. For each $x, y \in X$ and $r > 0$, $[x - y]_V < r$ if and only if $x - y \in rV$. Thus the uniformity induced by $[\cdot]_V$ corresponds to the uniformity generated by $\bracs{U_{rV}| r > 0}$, where $U_V = \bracs{(x, y) \in E|x - y \in V}$. Since this holds for all $U \in \cn(0)$, the topology of $E$ and the topology induced by $\bracs{[\cdot]_V| V \in \cn(0), \text{ convex, circled, radial}}$ coincide.
 

src/fa/lc/hahn-banach.tex

@@ -9,6 +9,7 @@
     \[
     \phi_{x_0, \lambda}: F + \real x_0 \quad (y + tx_0) \mapsto \phi(y) + \lambda t
     \]
+
     then there exists $\lambda \in \real$ such that $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_0$.
 \end{lemma}
 \begin{proof}
@@ -22,6 +23,7 @@
     \[
     \lambda \in \braks{\sup_{x \in F}[\phi(x) - \rho(x - x_0)], \inf_{x \in F}[\rho(x + x_0) - \phi(x)]}
     \]
+
     then for any $x \in F$ and $t > 0$,
     \begin{align*}
         \phi(x + tx_0) &= \phi(x) + t\lambda = t\braks{\phi(t^{-1}x) + \lambda} \\
@@ -42,26 +44,29 @@
     \end{enumerate}
 \end{theorem}
 \begin{proof}
-    (1): Let $x_0 \in E \setminus F$, then by \ref{lemma:hahn-banach}, there exists $\lambda \in \real$ such that if
+    (1): Let $x_0 \in E \setminus F$, then by \autoref{lemma:hahn-banach}, there exists $\lambda \in \real$ such that if
     \[
     \phi_{x_0, \lambda}: F + \real x_0 \to \real \quad (x + tx_0) \mapsto x + \lambda t
     \]
+
     then $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_0$. Thus for any $F \subsetneq E$, $\phi$ admits an extension to a subspace that strictly contains $F$.
 
     Let
     \[
     \mathbf{F} = \bracs{\Phi \in \hom(F'; \real)|F' \supset F, \Phi|_F = \phi, \Phi \le \rho|_{F'}}
     \]
-    For any $\Phi, \Phi' \in \mathbf{F}$, define $\Phi \le \Phi'$ if $\Phi'$ is an extension of $\Phi$. Let $\cf \subset \mathbf{F}$ be a chain. For any $\Phi \in \cf$, let $\cd(\Phi)$ be its domain. Since $\bigcup_{\phi \in \cf}\cd(\Phi)$ is a subspace and $\cf$ is a chain, by \ref{lemma:glue-linear}, there exists $\Phi \in \hom\paren{\bigcup_{\Phi' \in \cf}\cd(\Phi'); \real}$ such that $\Phi|_{\cd(\Phi')} = \Phi'$ for all $\Phi' \in \cf$. Thus $\Phi \in \mathbf{F}$ is an upper bound of $\cf$.
+
+    For any $\Phi, \Phi' \in \mathbf{F}$, define $\Phi \le \Phi'$ if $\Phi'$ is an extension of $\Phi$. Let $\cf \subset \mathbf{F}$ be a chain. For any $\Phi \in \cf$, let $\cd(\Phi)$ be its domain. Since $\bigcup_{\phi \in \cf}\cd(\Phi)$ is a subspace and $\cf$ is a chain, by \autoref{lemma:glue-linear}, there exists $\Phi \in \hom\paren{\bigcup_{\Phi' \in \cf}\cd(\Phi'); \real}$ such that $\Phi|_{\cd(\Phi')} = \Phi'$ for all $\Phi' \in \cf$. Thus $\Phi \in \mathbf{F}$ is an upper bound of $\cf$.
 
     By Zorn's lemma, $\mathbf{F}$ admits a maximal element $\Phi$. If $\cd(\Phi) \subsetneq E$, then $\Phi$ is not maximal by the preceding discussion. Therefore $\cd(\Phi) = E$ and $\Phi$ is a desired extension.
 
     (2): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$.
 
-    Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by \ref{proposition:polarisation-linear}. By (1), there exists $U \in \hom(E; \real)$ such that $\abs{U} \le \rho$ and $U|_F = u$. For each $x \in E$, let $\Phi(x) = U(x) - iU(ix)$, then $\Phi \in \hom(E; \complex)$ and $\Phi|_F = \phi$ by \ref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then
+    Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by \autoref{proposition:polarisation-linear}. By (1), there exists $U \in \hom(E; \real)$ such that $\abs{U} \le \rho$ and $U|_F = u$. For each $x \in E$, let $\Phi(x) = U(x) - iU(ix)$, then $\Phi \in \hom(E; \complex)$ and $\Phi|_F = \phi$ by \autoref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then
     \[
     |\Phi(x)| = \alpha\Phi(x) = \Phi(\alpha x) = U(\alpha x) \le \rho(\alpha x) = \rho(x)
     \]
+
     so $\abs{\Phi} \le \rho$.
 \end{proof}
 
@@ -72,18 +77,20 @@
 \begin{proof}
     By translation, assume without loss of generality that $0 \in A$. In which case, $A \in \cn^o(0)$ is convex.
 
-    Let $[\cdot]_A: E \to [0, \infty)$ be the gauge (\ref{definition:gauge}) of $A$, then $[\cdot]_A$ is a sublinear functional on $E$. For any $y, z \in E$ and $t > 0$ with $y, z \in tA$,
+    Let $[\cdot]_A: E \to [0, \infty)$ be the \hyperref[gaugeg]{definition:gauge} of $A$, then $[\cdot]_A$ is a sublinear functional on $E$. For any $y, z \in E$ and $t > 0$ with $y, z \in tA$,
     \[
     \abs{[y]_A - [z]_A} \le [y - z]_A \le t
     \]
+
     Hence $[\cdot]_A$ is continuous on $E$.
 
-    Let $\phi_0: \real x \to \real$ be defined by $\lambda x \mapsto \lambda$. Since $x \not\in A$, $[x]_A > 1$. Hence $\phi_0|_{Kx} \le [\cdot]_A|_{Kx}$. By the Hahn-Banach Theorem (\ref{theorem:hahn-banach}), there exists $\phi \in \hom(E; \real)$ such that $\phi|_{\real x} = \phi_0$ and $\phi(y) \le [y]_A$ for all $y \in E$.
+    Let $\phi_0: \real x \to \real$ be defined by $\lambda x \mapsto \lambda$. Since $x \not\in A$, $[x]_A > 1$. Hence $\phi_0|_{Kx} \le [\cdot]_A|_{Kx}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\phi \in \hom(E; \real)$ such that $\phi|_{\real x} = \phi_0$ and $\phi(y) \le [y]_A$ for all $y \in E$.
 
     For any $y \in A \cap (-A)$,
     \[
     \dpb{y, \phi}{E} \le [y]_A < 1 \quad -\dpb{y, \phi}{E} \le [-y]_A < 1
     \]
+
     so $\phi \in E^*$.
 
     Finally, let $y \in A$, then $\dpb{y, \phi}{E} \le [y]_A < 1 = \dpb{x, \phi}{E}$.
@@ -94,9 +101,9 @@
     Let $E$ be a TVS over $\real$, $A, B \subset E$ be non-empty convex sets, such that $A \cap B = \emptyset$ and $A$ is open, then there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ with $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
 \end{theorem}
 \begin{proof}
-    Let $C = A - B$, then $C \subset E$ is an open convex set by \ref{lemma:convex-gymnastics}. Since $A \cap B = \emptyset$, $0 \not\in C$.
+    Let $C = A - B$, then $C \subset E$ is an open convex set by \autoref{lemma:convex-gymnastics}. Since $A \cap B = \emptyset$, $0 \not\in C$.
 
-    By \ref{lemma:hahn-banach-separation}, there exists $\phi \in E^*$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E} < 0$ implies that $\dpb{a, \phi}{E} < \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
+    By \autoref{lemma:hahn-banach-separation}, there exists $\phi \in E^*$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E} < 0$ implies that $\dpb{a, \phi}{E} < \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
 \end{proof}
 
 \begin{theorem}[Hahn-Banach, Second Geometric Form {{\cite[Theorem 1.7]{Brezis}}}]
@@ -104,9 +111,9 @@
     Let $E$ be a locally convex space over $\real$, $A, B \subset E$ be non-empty convex sets such that $A \cap B = \emptyset$, $A$ is closed, and $B$ is compact, then there exists $\phi \in E^* \setminus \bracs{0}$, $\alpha \in \real$, and $r > 0$ with $A \subset \bracs{\phi \le \alpha - r}$ and $B \subset \bracs{\phi \ge \alpha + r}$.
 \end{theorem}
 \begin{proof}
-    Let $C = A - B$, then $C$ is closed by \ref{proposition:tvs-set-operations} with $0 \not\in C$. Since $E$ is locally convex, there exists $U \in \cn^o(0)$ convex such that $U \cap C \ne \emptyset$.
+    Let $C = A - B$, then $C$ is closed by \autoref{proposition:tvs-set-operations} with $0 \not\in C$. Since $E$ is locally convex, there exists $U \in \cn^o(0)$ convex such that $U \cap C \ne \emptyset$.
 
-    By \ref{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ such that $U \subset \bracs{f \le \alpha}$ and $C \subset \bracs{f \ge \alpha}$. In which case, for any $u \in U$, $a \in A$, and $b \in B$,
+    By \autoref{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ such that $U \subset \bracs{f \le \alpha}$ and $C \subset \bracs{f \ge \alpha}$. In which case, for any $u \in U$, $a \in A$, and $b \in B$,
     \begin{align*}
     \dpb{u, \phi}{E} &\le \dpb{a, \phi}{E} - \dpb{b, \phi}{E} \\
     \dpb{b, \phi}{E} + \dpb{u, \phi}{E} &\le \dpb{a, \phi}{E}
@@ -115,6 +122,7 @@
     \[
     \sup_{b \in B}\dpb{b, \phi}{E} + r \le \inf_{a \in A}\dpb{a, \phi}{E}
     \]
+
 \end{proof}
 
 
@@ -132,7 +140,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \ref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the Hahn-Banach theorem (\ref{theorem:hahn-banach}), there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $\phi \le \rho_M \le \rho$.
+    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $\phi \le \rho_M \le \rho$.
 
     (2): By (1) applied to $M = \bracs{0}$.
 

src/fa/lc/index.tex

@@ -2,10 +2,10 @@
 \label{chap:lc}
 
 
-\input{./src/fa/lc/convex.tex}
+\input{./convex.tex}
-\input{./src/fa/lc/continuous.tex}
+\input{./continuous.tex}
-\input{./src/fa/lc/quotient.tex}
+\input{./quotient.tex}
-\input{./src/fa/lc/projective.tex}
+\input{./projective.tex}
-\input{./src/fa/lc/inductive.tex}
+\input{./inductive.tex}
-\input{./src/fa/lc/hahn-banach.tex}
+\input{./hahn-banach.tex}
-\input{./src/fa/lc/spaces-of-linear.tex}
+\input{./spaces-of-linear.tex}

src/fa/lc/inductive.tex

@@ -13,12 +13,13 @@
         \[
         \mathcal{B} = \bracs{U \subset E|U \text{ convex, radial, circled}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
         \]
+
         is a fundamental system of neighbourhoods for $E$ at $0$.
     \end{enumerate}
     The topology $\topo$ is the \textbf{inductive locally convex topology} on $E$ induced by $\seqi{T}$.
 \end{definition}
 \begin{proof}
-    (1), (5): To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply \ref{proposition:tvs-0-neighbourhood-base}.
+    (1), (5): To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply \autoref{proposition:tvs-0-neighbourhood-base}.
     \begin{enumerate}
         \item[(TVB1)] Every set in $\mathcal{B}$ is radial and circled by definition.
         \item[(TVB2)] For any $U \in \mathcal{B}$, $U$ is circled, so $\frac{1}{2}U + \frac{1}{2}U \subset U$. Since $\frac{1}{2}U$ is also circled and radial, $\frac{1}{2}U \in \mathcal{B}$.
@@ -48,6 +49,7 @@
         }
         \]
 
+
         \item[(U)] For any pair $(F, \bracsn{S^i_F}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique $S \in L({E, F})$ such that the following diagram commutes
 
         \[
@@ -57,22 +59,24 @@
         }
         \]
 
+
         for all $i \in I$.
         \item For any locally convex space $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T^i_E \in L(E_i; F)$ for all $i \in I$.
         \item The family
         \[
         \mathcal{B} = \bracs{U \subset E|U \text{ convex, radial, circled}, (T^i_E)^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
         \]
+
         is a fundamental system of neighbourhoods for $E$ at $0$.
     \end{enumerate}
     The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the \textbf{inductive limit} of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
 \end{definition}
 \begin{proof}
-    Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (\ref{proposition:module-direct-limit}). Equip $E$ with the inductive topology (\ref{definition:lc-inductive}) induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).
+    Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (\autoref{proposition:module-direct-limit}). Equip $E$ with the \hyperref[inductive topology]{definition:lc-inductive} induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).
 
-    (U): By (U) of \ref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \ref{definition:lc-inductive}, $S \in L(E; F)$.
+    (U): By (U) of \autoref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \autoref{definition:lc-inductive}, $S \in L(E; F)$.
 
-    (5): By (5) of \ref{definition:lc-inductive}.
+    (5): By (5) of \autoref{definition:lc-inductive}.
 \end{proof}
 
 \begin{remark}
@@ -99,11 +103,13 @@
         \[
         (1 - \alpha)U + w = (1 - \alpha) U + \alpha x \subset V
         \]
+
         so $V \in \cn(0)$.
         \item For any $\lambda \in K$ with $\abs{\lambda} \le 1$, $u \in U$, $w \in W$, and $t \in [0, 1]$,
         \[
         \lambda (1 - t)u + \lambda tw = (1 - t)\lambda u + t \lambda w \in V
         \]
+
         as $U$ and $W$ are both circled.
     \end{itemize}
     so $V \in \cn_E(0)$ is convex and circled.
@@ -128,9 +134,9 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $U \in \cn_{E_n}(0)$. By \ref{lemma:lc-induct-separate}, there exists $\bracs{U_m| m \in \natp, m \ge n} \subset 2^E$ such that $U_n = U$, $U_m \in \cn_{E_m}(0)$ and $U_{m} = U_{m + 1} \cap E_m$ for all $m \in \natp$. Let $V = \bigcup_{m \ge n}U_m$, then $V \cap E_m = U_m$ for all $m \ge n$. In particular, $V \cap E_n = U_n = U$.
+    (1): Let $U \in \cn_{E_n}(0)$. By \autoref{lemma:lc-induct-separate}, there exists $\bracs{U_m| m \in \natp, m \ge n} \subset 2^E$ such that $U_n = U$, $U_m \in \cn_{E_m}(0)$ and $U_{m} = U_{m + 1} \cap E_m$ for all $m \in \natp$. Let $V = \bigcup_{m \ge n}U_m$, then $V \cap E_m = U_m$ for all $m \ge n$. In particular, $V \cap E_n = U_n = U$.
 
-    (2): Let $x \in E \setminus \bracs{0}$, then there exists $n \in \natp$ such that $x \in E_n$. Since $E_n$ is separated, there exists $U \in \cn_{E_n}(0)$ with $x \not\in U$. By \ref{lemma:lc-induct-separate} and (1), there exists $V \in \cn_E(0)$ such that $V \cap E_n = U$, so $x \not\in V$.
+    (2): Let $x \in E \setminus \bracs{0}$, then there exists $n \in \natp$ such that $x \in E_n$. Since $E_n$ is separated, there exists $U \in \cn_{E_n}(0)$ with $x \not\in U$. By \autoref{lemma:lc-induct-separate} and (1), there exists $V \in \cn_E(0)$ such that $V \cap E_n = U$, so $x \not\in V$.
 
     (3), $\neg (b) \Rightarrow \neg (a)$: If $B \not\subset E_n$ for all $n \in \natp$, then there exists a subsequence $\bracsn{n_k}_0^\infty \subset \natp$ and $\seq{x_k} \subset B$ such that $x_k \in E_{n_{k}} \setminus E_{n_{k - 1}}$ for all $k \in \natp$.
 
@@ -148,6 +154,7 @@
     \[
     \fU = \bracs{F + U|F \in \mathcal{F}, U \in \cn_E(0)}
     \]
+
     then $\fU$ is also a Cauchy filter, which converges if and only if $\fF$ does.
 
     Since each $E_n$ is complete, it is sufficient to show that there exists $n \in \natp$ such that $F + U \cap E_n \ne \emptyset$ for all $F \in \fF$ and $U \in \cn_E(0)$.
@@ -156,6 +163,7 @@
     \[
     U = \text{Conv}\paren{\bigcup_{n \in \natp}(U_n \cap E_n)}
     \]
+
     then since each $U_n$ is circled, so is $U$. Thus $U \cap E_n \supset U_n \cap E_n \in \cn_{E_n}(0)$, and $U \in \cn_E(0)$.
 
     Now, suppose that $(F_n + U) \cap E_n \ne \emptyset$. Let $y \in (F_n + U) \cap E_n$, then there exists $N \in \natp$, $\bracs{x_k}_1^N \subset E$, $\bracs{\lambda_k}_1^N \subset [0, 1]$, and $z \in F_n$ such that
@@ -168,11 +176,13 @@
     \[
     \underbracs{y - \sum_{k = 1}^n \lambda_kx_k}_{\in E_n} = \underbrace{z + \sum_{k = n + 1}^N \lambda_kx_k}_{\in F_n + U_n}
     \]
+
     which is impossible. Therefore $(F_n + U) \cap E_n = \emptyset$ for all $n \in \natp$.
 
     Finally, since $\fF$ is a Cauchy filter, there exists $F \in \fF$ such that $F - F \subset U$. Let $z \in F$, then there exists $n \in \natp$ such that $z \in E_n$. In which case, for any $y \in F \cap F_n$,
     \[
     z = y + (z - y) \in y + (F - F) \subset y + U \subset F_n + U
     \]
+
     which contradicts the fact that $(F_n + U) \cap E_n = \emptyset$.
 \end{proof}

src/fa/lc/projective.tex

@@ -7,10 +7,11 @@
     Let $E$ be a vector space over $K \in \RC$, $\seqi{F}$ be locally convex spaces over $K$, and $\seqi{T}$ where $T_i \in \hom(E; F_i)$ for all $i \in I$, then the projective topology on $E$ is locally convex.
 \end{proposition}
 \begin{proof}
-    By \ref{definition:tvs-initial},
+    By \autoref{definition:tvs-initial},
     \[
     \mathcal{B} = \bracs{\bigcap_{j \in J}T_j^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \cn_{F_j}(0)}
     \]
+
     is a fundamental system of neighbourhoods at $0$. For each $i \in I$, $U_i \in \cn_{F_i}(0)$ convex, $T^{-1}(U_i)$ is also convex. Since each $F_i$ is locally convex, $\mathcal{B}$ contains a fundamental system of neighbourhoods at $0$ consisting of only convex sets.
 \end{proof}
 
@@ -19,7 +20,7 @@
     Let $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$ be a downward-directed system of locally convex spaces over $K \in \RC$, then $E = \lim_{\longleftarrow}E_i$ is locally convex.
 \end{proposition}
 \begin{proof}
-    By (U) of \ref{definition:tvs-projective-limit} and \ref{definition:tvs-initial}, $E$ is equipped with the projective topology generated by the projection maps $E \to E_i$. By \ref{proposition:lc-projective-topology}, $E$ is locally convex.
+    By (U) of \autoref{definition:tvs-projective-limit} and \autoref{definition:tvs-initial}, $E$ is equipped with the projective topology generated by the projection maps $E \to E_i$. By \autoref{proposition:lc-projective-topology}, $E$ is locally convex.
 \end{proof}
 
 \begin{proposition}[{{\cite[II.5.4]{SchaeferWolff}}}]
@@ -32,6 +33,7 @@
         \[
         \pi^U_V: E_U \to E_V \quad x + M_U \mapsto x + M_V
         \]
+
         then $\pi^U_V \in L(E_U; E_V)$.
         \item $(\bracsn{E_U}_{U \in \mathcal{B}}, \bracs{\pi^U_V|U, V \in \mathcal{B}, U \subset V})$ is a downward-directed system of topological vector spaces.
         \item The map $\pi \in L(E, \lim_{\longleftarrow}E_U)$ induced by $\bracs{\pi_U}_{U \in \mathcal{B}}$ is a bijection.
@@ -39,10 +41,11 @@
         \[
         E = \lim_{\longleftarrow}E_U = \lim_{\longleftarrow} \ol E_U
         \]
+
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Since $V \supset U$, $[\cdot]_V \ge [\cdot]_U$, so $M_V \supset M_U$. Thus $\ker(\pi_V) \supset M_U$. By (U) of the quotient (\ref{definition:tvs-quotient}), $\pi_V$ factors through $E_U$ as $\pi^U_V$, so $\pi^U_V \in L(E_U; E_V)$.
+    (1): Since $V \supset U$, $[\cdot]_V \ge [\cdot]_U$, so $M_V \supset M_U$. Thus $\ker(\pi_V) \supset M_U$. By (U) of the \hyperref[quotient]{definition:tvs-quotient}, $\pi_V$ factors through $E_U$ as $\pi^U_V$, so $\pi^U_V \in L(E_U; E_V)$.
 
     (2): Since $\mathcal{B}$ is a fundamental system of neighbourhoods, it is downward-directed under inclusion. For any $U, V, W \in \mathcal{B}$ with $U \subset V \subset W$, $M_U \supset M_V \supset M_W$. Thus $\pi^U_W = \pi^V_W \circ \pi^U_V$.
 
@@ -54,11 +57,12 @@
     \[
     \pi_W(x_U) = \pi_W^U \circ \pi_U(x_U) = \pi_W^U p_U(x)
     \]
+
     Thus for any $U' \in \mathcal{B}$ with $U \subset W$, $[x_U - x_{U'}]_W = 0$, and $x_U - x_{U'} \in W$. Therefore $\bracs{x_U}_{U \in \mathcal{B}}$ is a Cauchy net, and converges to $x_0 \in E$ by completeness of $E$.
 
     For any $U \in \mathcal{B}$, $\pi_U(x_0) = \lim_{V \in \mathcal{B}}\pi_U(x_V) = p_U(x)$, so $\pi(x_0) = x$, and $\pi$ is surjective.
 
-    (4): Since $\mathcal{B} \subset \cn_E(0)$ is a fundamental system of neighbourhoods, the topology on $E$ is the projective topology generated by $\bracs{\pi_U|U \in \mathcal{B}}$. As $\pi_U \circ \pi^{-1} = p_U \in L(\lim E_U; E_U)$ for all $U \in \mathcal{B}$, $\pi^{-1} \in L(\lim E_U; E)$ by (U) of the projective topology (\ref{definition:tvs-initial}).
+    (4): Since $\mathcal{B} \subset \cn_E(0)$ is a fundamental system of neighbourhoods, the topology on $E$ is the projective topology generated by $\bracs{\pi_U|U \in \mathcal{B}}$. As $\pi_U \circ \pi^{-1} = p_U \in L(\lim E_U; E_U)$ for all $U \in \mathcal{B}$, $\pi^{-1} \in L(\lim E_U; E)$ by (U) of the \hyperref[projective topology]{definition:tvs-initial}.
 
     Let $x \in \lim\ol{E}_U$ and $V \in \cn(x)$. Since $\mathcal{B}$ is downward-directed and $\lim\ol{E}_U$ is equipped with the projective topology induced by $\bracs{p_U|U \in \mathcal{B}}$, there exists $U \in \mathcal{B}$ and $W \in \cn_{\ol E_U}(x)$ such that $p_U^{-1}(W) \subset V$. As $E_U$ is dense in $\ol E_U$, there exists $y_U \in W$, and $y \in E$ such that $y_U = \pi_U(y)$. Therefore $\pi(y_U) \in p_U^{-1}(W) \subset V$, and $\lim E_U$ is dense in $\lim \ol{E}_U$.
 

src/fa/lc/quotient.tex

@@ -7,6 +7,7 @@
     \[
     \rho_M: E/M \to [0, \infty) \quad x + M \mapsto \inf_{y \in x + M}\rho(y)
     \]
+
     is the \textbf{quotient} of $\rho$ by $M$.
 \end{definition}
 \begin{proof}
@@ -19,6 +20,7 @@
     \[
     \rho_M(x + x') \le \rho(y + y') \le \rho(y) + \rho(y')
     \]
+
     As this holds for all $y \in x + M$ and $y' \in x' + M$, $\rho_M(x + x') \le \rho_M(x) + \rho_M(x')$.
 \end{proof}
 
@@ -36,23 +38,26 @@
         \widetilde E \ar@{->}[r]_{\tilde f} & F
         }
         \]
+
         If $F$ is a TVS over $K$ and $f \in L(E; F)$, then $\td f \in L(E; F)$.
         \item If $\seqi{\rho}$ is a family of seminorms that induces the topology on $E$, then their quotients by $M$ induces the topology on $\td E$.
     \end{enumerate}
     The space $\td E = E/M$ is the \textbf{quotient} of $E$ by $M$.
 \end{definition}
 \begin{proof}
-    By \ref{definition:tvs-quotient}, (2), (3), (U) holds, and $\td E$ is a TVS over $K$.
+    By \autoref{definition:tvs-quotient}, (2), (3), (U) holds, and $\td E$ is a TVS over $K$.
 
     (1): Let $U \subset E$ be convex, then for any $x + M, y + M \in \pi(U)$ and $t \in [0, 1]$,
     \[
     (tx + M) + ((1 - t)y + M) = (tx + (1-t)y) + M \in U + M = \pi(U)
     \]
+
     so $\pi(U)$ is convex. Let $\fB = \bracs{U|U \in \cn_E(0) \text{ convex}}$, then $\bracs{\pi(U)|U \in \fB}$ is a fundamental system of neighbourhoods for the quotient topology on $E/M$. Therefore $E/M$ is locally convex.
 
     (5): By (U), each quotient seminorm is continuous on $\td E$, so the quotient topology contains the topology induced by the quotient seminorms. On the other hand, let $\pi(U) \in \cn_{\td E}(0)$, then there exists $J \subset I$ finite and $r > 0$ such that
     \[
     \bigcap_{j \in J}B_j(0, r) \subset U
     \]
+
     For each $j \in J$, let $\eta_j$ be the quotient of $\rho_j$ by $M$. Let $x + M \in E/M$ with $\eta_j(x) < r$ for all $j \in J$. For each $j \in J$, there exists $y_j \in x + M$ such that $\rho_j(y_j) < r$, so $y_j + M \in \pi(U)$. Therefore $x \in \pi(U)$ as well, and the quotient norms induce the quotient topology on $E/M$.
 \end{proof}

src/fa/lc/spaces-of-linear.tex

@@ -7,12 +7,14 @@
     \[
     [\cdot]_{S, i}: E^T \to [0, \infty) \quad f \mapsto \sup_{x \in S}[f(x)]_{S, i}
     \]
+
     then the $\mathfrak{S}$-uniform topology on $E^T$ is defined by the seminorms
     \[
     \bracs{[\cdot]_{S, i}|S \in \mathfrak{S}, i \in I}
     \]
+
     and hence locally convex.
 \end{proposition}
 \begin{proof}
-    By \ref{proposition:set-uniform-pseudometric}.
+    By \autoref{proposition:set-uniform-pseudometric}.
 \end{proof}

src/fa/lp/definition.tex

@@ -7,6 +7,7 @@
     \[
     \norm{f}_{L^p(X; E)} = \norm{f}_{L^p(\mu; E)} = \norm{f}_{L^p(X, \cm, \mu; E)} = \braks{\int \norm{f}_E^p d\mu}^{1/p} < \infty
     \]
+
     The set $\mathcal{L}^p(X; E) = \mathcal{L}^p(\mu; E) = \mathcal{L}^p(X, \cm, \mu; E)$ is the space of all $p$-integrable functions on $X$.
 \end{definition}
 
@@ -16,6 +17,7 @@
     \[
     \norm{f}_{L^\infty(X; E)} = \norm{f}_{L^\infty(\mu; E)} = \norm{f}_{L^\infty(X, \cm, \mu; E)} = \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0} < \infty
     \]
+
     In which case, $\norm{f}_{L^\infty(X; E)}$ is the \textbf{essential supremum} of $f$.
 \end{definition}
 
@@ -25,6 +27,7 @@
     \[
     \frac{1}{p} + \frac{1}{q} = 1
     \]
+
 \end{definition}
 
 \begin{lemma}
@@ -43,16 +46,19 @@
     \[
     \int \norm{f}_E \norm{g}_F d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)}
     \]
+
 \end{theorem}
 \begin{proof}
     First suppose that $p = 1$ and $q = \infty$. In this case,
     \[
     \int \norm{f}_E \norm{g}_F d\mu \le \norm{g}_{L^\infty(X; F)}\int \norm{f}_Ed\mu = \norm{f}_{L^1(X; E)}\norm{g}_{L^\infty(X; F)}
     \]
-    Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates. Assume without loss of generality that $\norm{f}_{L^p(X; E)} = \norm{g}_{L^q(X; F)} = 1$. By Young's inequality (\ref{lemma:young-inequality}),
+
+    Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates. Assume without loss of generality that $\norm{f}_{L^p(X; E)} = \norm{g}_{L^q(X; F)} = 1$. By \hyperref[Young's inequality]{lemma:young-inequality},
     \[
     \int \norm{f}_E \norm{g}_F d\mu \le \int \frac{\norm{f}_E^p}{p} + \frac{\norm{g}_F^q}{q} d\mu = \frac{1}{p}\int \norm{f}_E d\mu + \frac{1}{q}\int \norm{g}_F^q d\mu = 1
     \]
+
 \end{proof}
 
 \begin{theorem}[Minkowski's Inequality, {{\cite[6.5]{Folland}}}]
@@ -61,6 +67,7 @@
     \[
     \norm{f + g}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)}
     \]
+
 \end{theorem}
 \begin{proof}
     If $p = 1$, then the theorem holds directly.
@@ -70,6 +77,7 @@
     \norm{f + g}_{L^\infty(X; E)} \le \norm{f}_{L^\infty(X; E)} + \norm{g}_{L^\infty(X; E)}
     \]
 
+
     Now suppose that $p \in (1, \infty)$, then $p = q(p - 1)$, and
     \begin{align*}
         \norm{f + g}_E^p &\le (\norm{f}_E + \norm{g}_E)\norm{f + g}_E^{p - 1} \\
@@ -85,10 +93,11 @@
     \[
     L^p(X, \cm, \mu; E) = \mathcal{L}^p(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}}
     \]
+
     is a normed vector space, known as the $E$-valued \textbf{$L^p$ space} on $(X, \cm, \mu)$.
 \end{definition}
 \begin{proof}
-    By Minkowski's Inequality (\ref{theorem:minkowski}).
+    By \hyperref[Minkowski's Inequality]{theorem:minkowski}.
 \end{proof}
 
 \begin{proposition}
@@ -96,11 +105,12 @@
     Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$.
 \end{proposition}
 \begin{proof}
-    Let $f \in L^p(X; E)$. By \ref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$.
+    Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$.
 
-    For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the Dominated Convergence Theorem (\ref{theorem:dct}),
+    For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct},
     \[
     \limv{n}\int \norm{f_n - f}_E^p d\mu = \int \limv{n}\norm{f_n - f}_E^p d\mu = 0
     \]
+
     Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$.
 \end{proof}

src/fa/lp/index.tex

@@ -1,4 +1,4 @@
 \chapter{$L^p$ Spaces}
 \label{chap:lp}
 
-\input{./src/fa/lp/definition.tex}
+\input{./definition.tex}

src/fa/norm/index.tex

@@ -1,6 +1,6 @@
 \chapter{Normed Spaces}
 \label{chap:normed-spaces}
 
-\input{./src/fa/norm/normed.tex}
+\input{./normed.tex}
-\input{./src/fa/norm/linear.tex}
+\input{./linear.tex}
-\input{./src/fa/norm/multilinear.tex}
+\input{./multilinear.tex}

src/fa/norm/linear.tex

@@ -7,7 +7,8 @@
     \[
     \norm{\cdot}_{L(E; F)}: L(E; F) \to [0, \infty) \quad T \mapsto \sup_{\substack{x \in E \\ \norm{x}_E = 1}}Tx
     \]
+
 \end{proposition}
 \begin{proof}
-    By \ref{proposition:lc-spaces-linear-map}.
+    By \autoref{proposition:lc-spaces-linear-map}.
 \end{proof}

src/fa/norm/multilinear.tex

@@ -16,5 +16,6 @@
     \[
     \sup_{y \in B_F(0, 1)}\norm{T_yx}_G = \sup_{y \in B_F(0, 1)}\norm{T(x, y)}_G < \infty
     \]
-    by continuity of $y \mapsto T(x, y)$. By the Uniform Boundedness Principle (\ref{theorem:uniform-boundedness}), $M = \sup_{y \in B_F(0, 1)}\norm{T_y}_{L(E; G)} < \infty$. Thus for any $x \in E$ and $y \in F$, $\norm{T(x, y)}_G \le M\norm{x}_E\norm{y}_F$.
+
+    by continuity of $y \mapsto T(x, y)$. By the \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness}, $M = \sup_{y \in B_F(0, 1)}\norm{T_y}_{L(E; G)} < \infty$. Thus for any $x \in E$ and $y \in F$, $\norm{T(x, y)}_G \le M\norm{x}_E\norm{y}_F$.
 \end{proof}

src/fa/norm/normed.tex

@@ -21,12 +21,13 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1) $\Rightarrow$ (2): Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is also circled. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$ and $\abs{\lambda}^{-1}U \subset V$.
+    (1) $\Rightarrow$ (2): Using \autoref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is also circled. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$ and $\abs{\lambda}^{-1}U \subset V$.
 
-    Let $\norm{\cdot}_E: E \to [0, \infty)$ be the gauge (\ref{definition:gauge}) of $U$. For each $r > 0$, let $B_E(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
+    Let $\norm{\cdot}_E: E \to [0, \infty)$ be the \hyperref[gauge]{definition:gauge} of $U$. For each $r > 0$, let $B_E(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
     \[
     \bracs{\lambda U|\lambda > 0} = \bracs{B_E(0, r)|r > 0}
     \]
+
     is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_E$ induces the topology on $E$.
 \end{proof}
 
@@ -51,14 +52,17 @@
     \[
     \norm{y_n}_F \le \gamma^{n - 1}\norm{y_1}_F = \gamma^{n - 1}\norm{y}_F
     \]
+
     Since $\norm{x_n}_E \le C\norm{y_n}_F$,
     \[
     \sum_{k \in \natp}\norm{x_k}_E \le C\norm{y}_F\sum_{k \in \nat_0}\gamma^k = \frac{C\norm{y}_F}{1 - \gamma}
     \]
+
     In addition,
     \[
     \norm{y - \sum_{k = 1}^n Tx_k}_F = \norm{y_{n+1}}_F \le \gamma^n \norm{y}_F
     \]
+
     so $\sum_{n = 1}^\infty Tx_n = y$.
 \end{proof}
 
@@ -72,11 +76,12 @@
     then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.
 \end{theorem}
 \begin{proof}
-    For each $n \in \natp$, let $A_n = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_n$ is closed with $\bigcup_{n \in \natp}A_n = E$. By the Baire Category Theorem (\ref{theorem:baire}), there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
+    For each $n \in \natp$, let $A_n = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_n$ is closed with $\bigcup_{n \in \natp}A_n = E$. By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
 
     Let $x \in U$ and $r > 0$ such that $\overline{B(x, r)} \subset U$, then for any $y \in E$ with $\norm{y}_E \le r$ and $T \in \mathcal{T}$,
     \[
     \norm{Ty} = \norm{Ty + Tx - Tx}_E = \normn{T\underbrace{(x + y)}_{\in U}}_E + \norm{Tx}_E \le 2n
     \]
+
     so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
 \end{proof}

src/fa/rs/bv.tex

@@ -8,10 +8,12 @@
     \[
     V_{\rho, p}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
     \]
+
     is the \textbf{variation} of $f$ with respect to $\rho$ and $P$. The supremum over all such partitions
     \[
     [f]_{\var, \rho} = \sup_{P \in \scp([a, b])}V_{\rho, P}(f)
     \]
+
     is the \textbf{total variation} of $f$ on $[a, b]$ with respect to $\rho$.
 
     If $E$ is a normed space, then the variation and total variation of $f$ is taken with respect to its norm.
@@ -45,12 +47,14 @@
     = \lim_{g, \fF}\sum_{j = 1}^n \rho(g(x_j) - g(x_{j - 1}))
     = \lim_{g \in \fF}V_{\rho, P}(g)
     \]
+
     By assumption (b), $[0, M_\rho]$ is in the filter generated by $V_{\rho, P}(\fF)$. Thus $V_{\rho, P}(f) \le M_\rho$. As this holds for all $P \in \scp([a, b])$, $V_{\rho, P}(f) \le M_\rho$, and $f \in BV([a, b]; E)$.
 
     (5): For each $n \in \nat^+$, let
     \[
     D_n = \bracs{x \in [a, b]|\forall \eps > 0, \exists y \in (x - \eps, x + \eps): \norm{f(x) - f(y)}_E \ge 1/n}
     \]
+
     then $D = \bigcup_{n \in \nat^+}D_n$ is the set of discontinuity points of $f$. If $D$ is uncountable, then there exists $N \in \nat^+$ such that $D_n$ is infinite.
 
     Fix $N \in \nat^+$. Let $E_1 = D_n \cap (a, b)$ and $I_1 = (a, b)$, then

src/fa/rs/index.tex

@@ -1,7 +1,7 @@
 \chapter{The Riemann-Stieltjes Integral}
 \label{chap:rs-integral}
 
-\input{./src/fa/rs/partition.tex}
+\input{./partition.tex}
-\input{./src/fa/rs/bv.tex}
+\input{./bv.tex}
-\input{./src/fa/rs/rs.tex}
+\input{./rs.tex}
-\input{./src/fa/rs/rs-bv.tex}
+\input{./rs-bv.tex}

src/fa/rs/partition.tex

@@ -7,6 +7,7 @@
     \[
     P = \seqfz{x_j} = [a = x_0 \le \cdots \le x_n = b]
     \]
+
     The collection $\scp([a, b])$ is the set of all partitions of $[a, b]$.
 \end{definition}
 
@@ -23,6 +24,7 @@
     \[
     \sigma(P) = \max_{1 \le j \le n}(x_j - x_{j - 1})
     \]
+
     is the \textbf{mesh} of $P$.
 \end{definition}
 

src/fa/rs/rs-bv.tex

@@ -9,9 +9,10 @@
     \[
     \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_1 \cdot [g]_{\var, 2}
     \]
+
 \end{proposition}
 \begin{proof}
-    By \ref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_1$ on $E_1$ and $[\cdot]_2$ on $E_2$ such that $[xy]_H \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
+    By \autoref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_1$ on $E_1$ and $[\cdot]_2$ on $E_2$ such that $[xy]_H \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
 
     Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
     \begin{align*}
@@ -28,6 +29,7 @@
     \[
     [f]_{u, \rho} = \sup_{x \in [a, b]}\rho(f(x))
     \]
+
     Let $\net{f} \subset RS([a, b], G)$ such that:
     \begin{enumerate}
         \item[(a)] For each continuous seminorm $\rho$ on $E_1$, $[f_\alpha - f]_{u, \rho} \to 0$.
@@ -61,6 +63,7 @@
     \[
     \rho\paren{S(P, c, f, G) - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps
     \]
+
 \end{proof}
 
 \begin{proposition}
@@ -74,6 +77,7 @@
         \[
         \int_a^b fdG = \limv{n}S(P_n, t_n, f, G)
         \]
+
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -88,6 +92,7 @@
     \[
     \rho(S(P, c, f, G) - S(Q, d, f, G)) \le 2 \cdot \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \max(\sigma(P), \sigma(Q)) \end{array}}[f(x) - f(y)]_1 \cdot [G]_{\var, 2}
     \]
+
     by passing through a common refinement. Since $f \in C([a, b]; E_1)$, this bound tends to $0$ as $\max(\sigma(P), \sigma(Q))$ tends to $0$, so $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is a Cauchy net.
 
     In addition, for any $\seq{(P_n, t_n)}$ as in (2), $\limv{n}S(P_n, t_n, f, G)$ exists by sequential completeness. Since $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is Cauchy, the limit $\lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)$ exists as well and is equal to $\limv{n}S(P_n, t_n, f, G)$.

src/fa/rs/rs.tex

@@ -9,6 +9,7 @@
     \[
     S(P, c, f, G) = \sum_{j = 1}^n f(c_j)[G(x_j) - G(x_{j - 1})]
     \]
+
     is the \textbf{Riemann-Stieltjes sum} of $f$ with respect to $G$ and $(P, c)$.
 \end{definition}
 
@@ -20,6 +21,7 @@
     \[
     \int_a^b f dG = \int_a^b f(t)G(dt) = \lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)
     \]
+
     exists. In which case, $\int_a^b fdG$ is the \textbf{Riemann-Stieltjes integral} of $G$.
 
     The set $RS([a, b], G)$ is the vector space of all \textbf{Riemann-Stieltjes integrable functions} with respect to $G$.
@@ -31,6 +33,7 @@
     \[
     S(P, c, f, G) + S(P', c', G, f) = f(b)G(b) - f(a)G(a)
     \]
+
     where $P' = \seqfz[n+1]{y_j} = [a, c_1, \cdots, c_n, b]$ and $c' = \seqf[n+1]{d_j} = [x_0, \cdots, x_n]$.
 \end{lemma}
 \begin{proof}
@@ -53,16 +56,19 @@
     \[
     \int_a^b f dG + \int_a^b G df = f(b)G(b) - f(a)G(a)
     \]
+
 \end{theorem}
 \begin{proof}
     Suppose that $f \in RS([a, b], G)$. Let $U \in \cn_F(0)$, then there exits $P_0 = \seqfz{x_j} \in \scp([a, b])$ such that $S(P, c, f, G) - \int_a^b fdG \in U$ for all $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$. Let
     \[
     Q_0 = [x_0, x_1, x_1, \cdots, x_n, x_n]
     \]
+
     then for any $(Q = \seqfz[m]{y_j}, d = \seqf[m]{d_j}) \in \scp_t([a, b])$ with $Q \ge Q_0$,
     \[
     f(b)G(b) - f(a)G(a) - \int_a^b fdG - S(Q, d, G, f) =
     \int_a^b fdG - S(Q', d', G, f)
     \]
-    by \ref{lemma:sum-by-parts}, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_0$, and $\int_a^b fdG - S(Q', d', G, f) \in U$.
+
+    by \autoref{lemma:sum-by-parts}, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_0$, and $\int_a^b fdG - S(Q', d', G, f) \in U$.
 \end{proof}

src/fa/tvs/bounded.tex

@@ -20,16 +20,18 @@
 \begin{proof}
     Let $U \in \cn(0)$.
 
-    (2): Using \ref{proposition:uniform-neighbourhoods}, assume without loss of generality that $U$ is closed. Let $0 \ne \lambda \in K$ with $\lambda U \supset B$, then since $\lambda U$ is closed, $\lambda U \supset \ol B$.
+    (2): Using \autoref{proposition:uniform-neighbourhoods}, assume without loss of generality that $U$ is closed. Let $0 \ne \lambda \in K$ with $\lambda U \supset B$, then since $\lambda U$ is closed, $\lambda U \supset \ol B$.
 
-    (4), (5): By \ref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$, and $\lambda, \lambda' \in K$ such that $\lambda V \supset A$ and $\lambda' V \supset B$.
+    (4), (5): By \autoref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$, and $\lambda, \lambda' \in K$ such that $\lambda V \supset A$ and $\lambda' V \supset B$.
 
     Let $\mu > \abs{\lambda}, \abs{\lambda'}$, then
     \[
     \mu U \supset \mu V \supset \lambda V \cup \lambda' V \supset A \cup B
     \]
+
     and
     \[
     \mu U \supset \mu(V + V) \supset \lambda V + \lambda' V \supset A + B
     \]
+
 \end{proof}

src/fa/tvs/complete-metric.tex

@@ -31,6 +31,7 @@
     \[
     T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r)
     \]
+
 \end{theorem}
 \begin{proof}
     Let $y_0 = y$ and $x_0 = 0$. Let $N \in \natz$ and suppose inductively that $\seqf[N]{x_n} \subset E$ has been constructed such that:
@@ -76,6 +77,7 @@
     \[
     x = \limv{N}x_N = \limv{N}\sum_{n = 1}^N(x_n - x_{n-1})
     \]
+
     exists in $E$. In addition, $\rho(x) \le \sum_{n \in \natp} \rho(x_n - x_{n-1}) < \sum_{n \in \natp}s_n = s$, so $x \in B_E(0, s)$. Finally, $\eta\paren{Tx - y} = \limv{N}\rho(Tx_N - y) = 0$ and $Tx = y$.
 \end{proof}
 
@@ -92,10 +94,11 @@
 \begin{proof}
     Let $r > 0$ and $\gamma \in (0, 1)$. For any $y_0 \in B_F(0, r) \cap \overline{T(E)}$, there exists $y \in B_F(0, r)$ such that $\eta(y) \le \eta(y_0)$ and $\eta(y - y_0) \le \gamma \eta(y_0)$. By assumption (a), there exists $x \in T^{-1}(y)$ with $\rho(x) \le C\eta(y) \le C\eta(y_0)$.
 
-    By the method of successive approximations (\ref{theorem:successive-approximations}),
+    By the \hyperref[method of successive approximations]{theorem:successive-approximations},
     \[
     T(E) \supset T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r) \cap \overline{T(E)}
     \]
+
     As this holds for all $r > 0$, $T(E) \supset \overline{T(E)}$.
 \end{proof}
 
@@ -112,11 +115,13 @@
     \[
     E = \bigcup_{n \in \natp}nB_E(0, r) \quad \overline{T(E)} = \bigcup_{n \in \natp}\overline{nT(B_E(0, r))}
     \]
-    By the Baire Category Theorem (\ref{theorem:baire}), there exists $N \in \natp$, $s > 0$, and $y \in nT(B_E(0, r))$ such that $B_F(y, s) \subset \overline{nT(B_E(0, r))}$. In which case,
+
+    By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $N \in \natp$, $s > 0$, and $y \in nT(B_E(0, r))$ such that $B_F(y, s) \subset \overline{nT(B_E(0, r))}$. In which case,
     \[
     B_F(0, s) = B_F(y, s) - y \subset \overline{nT(B_E(0, r)) + nT(B_E(0, r))} \subset \overline{nT(B_E(0, r_0))}
     \]
+
     By (TVS2), there exists $t > 0$ such that $n^{-1}B_F(0, s) \supset B_F(0, t)$, so $\overline{T(B_E(0, r_0))} \supset B_F(0, t)$.
 
-    Thus by \ref{proposition:successive-approximation-all}, $B_F(0, t) \subset T(B_E(0, r)) \in \cn_F(0)$ for all $r > r_0$. As $r_0 > 0$ is arbitrary, $T(U) \in \cn_F(0)$ for all $U \in \cn_E(0)$. Therefore $T$ is open by translation-invariance of the topology on $E$.
+    Thus by \autoref{proposition:successive-approximation-all}, $B_F(0, t) \subset T(B_E(0, r)) \in \cn_F(0)$ for all $r > r_0$. As $r_0 > 0$ is arbitrary, $T(U) \in \cn_F(0)$ for all $U \in \cn_E(0)$. Therefore $T$ is open by translation-invariance of the topology on $E$.
 \end{proof}

src/fa/tvs/completion.tex

@@ -16,9 +16,9 @@
     The pair $(\wh E, \iota)$ is the \textbf{Hausdorff completion} of $E$.
 \end{definition}
 \begin{proof}
-    All claims of (1), (2), (U), and (4), except the linearity of maps and the fact that $\wh E$ is a TVS is proven via the Hausdorff completion (\ref{definition:hausdorff-completion}).
+    All claims of (1), (2), (U), and (4), except the linearity of maps and the fact that $\wh E$ is a TVS is proven via the \hyperref[Hausdorff completion]{definition:hausdorff-completion}.
 
-    Using \ref{proposition:initial-completion}, identify $\wh E \times \wh E$ with $\wh{E \times E}$ and $K \times \wh E$ with $\wh{K \times E}$ as uniform spaces. By \ref{proposition:hausdorff-uniform-factor}, there exists operations $\wh E \times \wh E \to \wh E$ and $K \times \wh E \to \wh E$ such that the following diagrams commute
+    Using \autoref{proposition:initial-completion}, identify $\wh E \times \wh E$ with $\wh{E \times E}$ and $K \times \wh E$ with $\wh{K \times E}$ as uniform spaces. By \autoref{proposition:hausdorff-uniform-factor}, there exists operations $\wh E \times \wh E \to \wh E$ and $K \times \wh E \to \wh E$ such that the following diagrams commute
     \[
     \xymatrix{
     \widehat E \times \widehat E \ar@{->}[r] & \widehat E &  & K \times \widehat E \ar@{->}[r] & \widehat E \\
@@ -26,6 +26,7 @@
     }
     \]
 
+
     By continuity and the density of $\iota(E)$ in $E$, $\wh E$ with these operations forms a TVS, and $T$ is linear.
 \end{proof}
 

src/fa/tvs/continuous.tex

@@ -13,9 +13,9 @@
     If the above holds, then $T$ is a \textbf{continuous linear map}. The set $L(E; F)$ denotes the vector space of all continuous linear maps from $E$ to $F$.
 \end{definition}
 \begin{proof}
-    $(1) \Rightarrow (2) \Rightarrow (3)$: By \ref{proposition:uniform-continuous} and \ref{definition:continuity}.
+    $(1) \Rightarrow (2) \Rightarrow (3)$: By \autoref{proposition:uniform-continuous} and \autoref{definition:continuity}.
 
-    $(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$.  Using \ref{proposition:tvs-uniform} and \ref{lemma:translation-invariant-symmetric}, assume without loss of generality that $U$ and $V$ are symmetric and translation-invariant.
+    $(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$.  Using \autoref{proposition:tvs-uniform} and \autoref{lemma:translation-invariant-symmetric}, assume without loss of generality that $U$ and $V$ are symmetric and translation-invariant.
 
     For any $x, y \in V$, $x - y \in V(0)$, so $Tx - Ty \in U(0)$, $Ty \in U(Tx)$ by symmetry, and $(Tx, Ty) \in U$. Therefore $T$ is uniformly continuous.
 \end{proof}
@@ -47,6 +47,7 @@
         \prod_{i \in I}E_i \ar@{->}[r]_{\pi_i} & E_i
         }
         \]
+
     \end{enumerate}
     The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
 \end{definition}

src/fa/tvs/definition.tex

@@ -31,6 +31,7 @@
     \[
     U = \bracs{(x + z, y + z)|(x, y) \in U}
     \]
+
     and $\fU$ is \textbf{translation-invariant} if there exists a fundamental system of translation-invariant entourages.
 \end{definition}
 
@@ -39,7 +40,7 @@
     Let $E$ be a vector space and $\fU$ be a translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
 \end{lemma}
 \begin{proof}
-    Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \ref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
+    Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \autoref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
 \end{proof}
 
 
@@ -60,13 +61,13 @@
         \item[(UB2)] Let $V \in \fB_0$, then by (TVS1), there exists $W \in \fB_0$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_W$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_W \circ U_W \subset U_V$.
     \end{enumerate}
 
-    By \ref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
+    By \autoref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
 
     (1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$.
 
-    Let $W \in \cn(0)$, then by \ref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
+    Let $W \in \cn(0)$, then by \autoref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
 
-    Let $V \in \mathfrak{V}$. Using \ref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
+    Let $V \in \mathfrak{V}$. Using \autoref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
 \end{proof}
 
 \begin{proposition}
@@ -75,14 +76,16 @@
     \[
     \ol{A} = \bigcap_{U \in \fB}\bracs{A + U| U \in \fB}
     \]
+
 \end{proposition}
 \begin{proof}
     Let $V \in \cn(0)$ be balanced and $U_V = \bracs{(x, y) \in E \times E| x - y \in V}$, then $y \in U_V(A)$ if and only if there exists $x \in A$ such that $(x, y) \in U_V$. This is equivalent to $x - y \in V$ and $y - x \in V$, so $U_V(A) = A + V$.
 
-    Assume without loss of generality that $\fB$ consists of symmetric entourages. By \ref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \ref{proposition:uniformclosure} implies that
+    Assume without loss of generality that $\fB$ consists of symmetric entourages. By \autoref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \autoref{proposition:uniformclosure} implies that
     \[
     \ol{A} = \bigcap_{V \in \fB}\bracs{U_V(A)| U \in \fB} = \bigcap_{V \in \fB}U + A
     \]
+
 \end{proof}
 
 \begin{proposition}[{{\cite[I.1.1]{SchaeferWolff}}}]
@@ -94,10 +97,11 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    $(1)$: For every $x \in B$, $A + x$ is open by \ref{definition:translation-invariant-topology}, so
+    $(1)$: For every $x \in B$, $A + x$ is open by \autoref{definition:translation-invariant-topology}, so
     \[
     A + B = \bigcup_{x \in B}(A + x)
     \]
+
     is open.
 
     $(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
@@ -105,14 +109,17 @@
     y \in \bigcap_{U \in \fF}\overline{U - B} = \bigcap_{U \in \fF}\overline{(-B) + U}
     \]
 
-    By \ref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
+
+    By \autoref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
     \[
     y \in \bigcap_{U \in \fF}\overline{(-B) + U} \subset \bigcap_{U \in \fF}[(-B) + U + U]
     \]
+
     Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus
     \[
     y \in \bigcap_{U \in \fF}[(-B) + U + U] \subset \bigcap_{V \in \cn(0)}[(x-B) + V] = \overline{x - B} = x - B
     \]
+
     so $x \in y + B \subset A + B$.
 \end{proof}
 
@@ -137,12 +144,13 @@
 \begin{proof}
     Firstly, (TVS2) implies that every neighbourhood of $0$ is circled.
 
-    By \ref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.
+    By \autoref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.
 
     Let $U \in \cn^o(0)$ be open. By (TVS2), there exists $r > 0$ such that $\lambda U \subset U$ for all $\lambda \in K$ with $\abs{\lambda} \le r$. Define
     \[
     V = \bigcup_{\substack{\lambda \in K \\ \abs{\lambda} \le r}} \lambda U \subset U
     \]
+
     then for any $x \in V$, there exists $\lambda \in K$ with $\abs{\lambda} \le r$ and $y \in U$ such that $x = \lambda y$. In which case, for any $\mu \in K$ with $\abs{\mu} \le 1$, $\mu(\lambda y) = (\mu \lambda) y \in \mu\lambda U$. Since $\abs{\mu \lambda} \le r$, $\mu \lambda U \subset V$. Thus $V \subset U$ is balanced.
 
     Let $U \in \cn(0)$ be closed, then there exists a balanced neighbourhood $V \in \cn^o(0)$ such that $V \subset U$. In which case, for any $\lambda \in K$ with $0 < \abs{\lambda} \le 1$, $\lambda \overline{V} = \overline{\lambda V} \subset \overline{V}$ by (TVS2). Therefore $\overline{V} \subset U$ is balanced as well.
@@ -166,19 +174,20 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    \textbf{Forward:} By \ref{proposition:tvs-good-neighbourhood-base}, there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).
+    \textbf{Forward:} By \autoref{proposition:tvs-good-neighbourhood-base}, there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).
 
     \textbf{Converse:} For each $V \in \fB$, let $U_V = \bracs{(x, y) \in E|x - y \in V}$, then $U_V$ is symmetric and translation-invariant by (TVB1). Let
     \[
     \mathfrak{V} = \bracs{U_V|V \in \fB}
     \]
+
     then
     \begin{enumerate}
         \item[(FB1)] For any $V, V' \in \fB$, there exists $W \in \fB$ with $W \subset V \cap V'$. In which case, $U_{V} \cap U_{V'} \supset U_W \in \mathfrak{V}$.
         \item[(UB1)] For any $x \in E$ and $V \in \fB$, $x - x = 0 \in V$, so $\Delta \subset U_V$.
         \item[(UB2)] For any $V \in \fB$, by (TVB1), there exists $W \in \fB$ such that $W + W \subset V$. In which case, for any $x, y, z \in E$ with $x - y, y - z \in W$, $x - z \in V$. Therefore $U_W \circ U_W \subset U_V$.
     \end{enumerate}
-    By \ref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.
+    By \autoref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.
 
     (1): Since $\mathfrak{V}$ is translation-invariant, so is $\fU$.
 
@@ -197,10 +206,11 @@
         \lambda x - \lambda' x' \in \lambda V + \eps(\abs \mu + 1)V \subset \abs{\lambda} V + \eps(\abs \mu + 1)V
         \]
 
+
         Let $W \in \fB$ and $\eps > 0$ such that $\eps(\abs{\mu}+1) \le 1$, then by repeated application of (TVB1), there exists $V \in \fB$ such that $\abs{\lambda} V + V \subset W$. Therefore scalar multiplication is jointly continuous.
     \end{enumerate}
 
-    (Uniqueness): Let $\mathcal{S} \subset 2^E$ be a topology on $E$ satisfying (1) and (2), then for each $x \in E$, $\cn_{(E, \mathcal{S})}(x) = \cn_{(E, \mathcal{T})}(x)$. By \ref{proposition:neighbourhoodcharacteristic}, $\mathcal{S} = \mathcal{T}$.
+    (Uniqueness): Let $\mathcal{S} \subset 2^E$ be a topology on $E$ satisfying (1) and (2), then for each $x \in E$, $\cn_{(E, \mathcal{S})}(x) = \cn_{(E, \mathcal{T})}(x)$. By \autoref{proposition:neighbourhoodcharacteristic}, $\mathcal{S} = \mathcal{T}$.
 \end{proof}
 
 
@@ -209,5 +219,5 @@
     Let $E$ be a TVS over $K \in \RC$, then $E$ is locally connected.
 \end{proposition}
 \begin{proof}
-    Let $U \in \cn(0)$ be radial, then for any $y \in U$, the mapping $t \mapsto ty$ is a path from $0$ to $y$ contained in $U$. Thus $U$ is path-connected. By \ref{proposition:tvs-0-neighbourhood-base}, the radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods, so the path-connected neighbourhoods of $0$ forms a fundamental system as well.
+    Let $U \in \cn(0)$ be radial, then for any $y \in U$, the mapping $t \mapsto ty$ is a path from $0$ to $y$ contained in $U$. Thus $U$ is path-connected. By \autoref{proposition:tvs-0-neighbourhood-base}, the radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods, so the path-connected neighbourhoods of $0$ forms a fundamental system as well.
 \end{proof}

src/fa/tvs/dual.tex

@@ -15,12 +15,14 @@
     \[
     \im{\dpb{x, \phi}{E}} = \re{-i \dpb{x, \phi}{E}} = -\dpb{ix, u}{E}
     \]
+
     so (2) holds.
 
     (2): Since $u \in \hom(E; \real)$, so is $\phi$. On the other hand, for any $x \in E$,
     \[
     \dpb{ix, \phi}{E} = \dpb{ix, u}{E} - i\dpb{-x, u}{E} = i(\dpb{x, u}{E} - i \dpb{ix, u}{E}) = i \dpb{x, \phi}{E}
     \]
+
 \end{proof}
 
 \begin{definition}[Topological Dual]

src/fa/tvs/index.tex

@@ -1,14 +1,14 @@
 \chapter{Topological Vector Spaces}
 \label{chap:tvs}
 
-\input{./src/fa/tvs/definition.tex}
+\input{./definition.tex}
-\input{./src/fa/tvs/metric.tex}
+\input{./metric.tex}
-\input{./src/fa/tvs/bounded.tex}
+\input{./bounded.tex}
-\input{./src/fa/tvs/dual.tex}
+\input{./dual.tex}
-\input{./src/fa/tvs/continuous.tex}
+\input{./continuous.tex}
-\input{./src/fa/tvs/quotient.tex}
+\input{./quotient.tex}
-\input{./src/fa/tvs/completion.tex}
+\input{./completion.tex}
-\input{./src/fa/tvs/complete-metric.tex}
+\input{./complete-metric.tex}
-\input{./src/fa/tvs/projective.tex}
+\input{./projective.tex}
-\input{./src/fa/tvs/inductive.tex}
+\input{./inductive.tex}
-\input{./src/fa/tvs/spaces-of-linear.tex}
+\input{./spaces-of-linear.tex}

src/fa/tvs/inductive.tex

@@ -17,7 +17,8 @@
     \[
     \mathcal{B} = \bracs{U \subset E|U \text{ radial, circled}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
     \]
-    To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply \ref{proposition:tvs-0-neighbourhood-base}.
+
+    To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply \autoref{proposition:tvs-0-neighbourhood-base}.
     \begin{enumerate}
         \item[(TVB1)] Every set in $\mathcal{B}$ is radial and circled by definition.
         \item[(TVB2)] For any $U \in \mathcal{B}$, $U$ is circled, so $\frac{1}{2}U + \frac{1}{2}U \subset U$. Since $\frac{1}{2}U$ is also circled and radial, $\frac{1}{2}U \in \mathcal{B}$.
@@ -47,6 +48,7 @@
         }
         \]
 
+
         \item[(U)] For any pair $(F, \bracsn{S^i_F}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique $S \in L({E, F})$ such that the following diagram commutes
 
         \[
@@ -56,17 +58,18 @@
         }
         \]
 
+
         for all $i \in I$.
         \item For any TVS $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T^i_E \in L(E_i; F)$ for all $i \in I$.
     \end{enumerate}
     The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the \textbf{inductive limit} of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
 \end{definition}
 \begin{proof}
-    Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (\ref{proposition:module-direct-limit}). Equip $E$ with the inductive topology (\ref{definition:tvs-inductive}) induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).
+    Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (\autoref{proposition:module-direct-limit}). Equip $E$ with the \hyperref[inductive topology]{definition:tvs-inductive} induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).
 
-    (U): By (U) of \ref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \ref{definition:tvs-inductive}, $S \in L(E; F)$.
+    (U): By (U) of \autoref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \autoref{definition:tvs-inductive}, $S \in L(E; F)$.
 
-    (5): By (5) of \ref{definition:tvs-inductive}.
+    (5): By (5) of \autoref{definition:tvs-inductive}.
 \end{proof}
 
 \begin{definition}[Strict]

src/fa/tvs/metric.tex

@@ -19,6 +19,7 @@
     \[
     d_i: E \times E \quad (x, y) \mapsto \rho_i(x - y)
     \]
+
     then $d_i$ is a pseudometric. The uniform topology on $E$ induced by $\seqi{d}$ is the \textbf{topology induced by $\seqi{\rho}$}, and
     \begin{enumerate}
         \item The topology induced by $\seqi{\rho}$ is a vector space topology.
@@ -27,11 +28,12 @@
         \[
         \bracs{\bigcap_{j \in J}B_j(0, r)|J \subset I \text{ finite}, r > 0}
         \]
+
         is a fundamental system of neighbourhoods at $0$.
     \end{enumerate}
 \end{definition}
 \begin{proof}
-    (3): By \ref{definition:pseudometric-uniformity}.
+    (3): By \autoref{definition:pseudometric-uniformity}.
 
     (2): Each $d_i$ is translation-invariant.
 
@@ -42,16 +44,19 @@
         \[
         \lambda x - \lambda' x' = \lambda(x - x') + (\lambda - \lambda')x'
         \]
+
         Let $i \in I$ and $\eps > 0$. By (PN5) then there exists $\delta > 0$ such that if $\rho_i(x - x') < \delta$, then $\rho_i(\lambda (x - x')) < \eps$.
 
         On the other hand,
         \[
         (\lambda - \lambda')x' = (\lambda - \lambda')x + (\lambda - \lambda')(x' - x)
         \]
+
         By (PN4), there exists $\delta' \in (0, 1]$ such that if $\abs{\lambda - \lambda'} < \delta'$, then $\rho_i((\lambda - \lambda')x) < \eps$. In which case, since $\delta' \le 1$, (PN2) implies that
         \[
         \rho_i((\lambda - \lambda')x') < \eps + \rho_i(x' - x) < 2\eps
         \]
+
         Therefore $\rho_i(\lambda x - \lambda' x') < 3\eps$.
     \end{enumerate}
 \end{proof}
@@ -68,7 +73,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    $(4) \Rightarrow (1)$: By \ref{definition:tvs-pseudonorm-topology}, for each $r > 0$, $\rho^{-1}([0, r)) \in \cn_E(0)$. Thus for any $x, y \in E$, if $x - y \in \rho^{-1}([0, r))$, then $\abs{\rho(x) - \rho(y)} \le r$. Therefore $\rho \in UC(E; [0, \infty))$.
+    $(4) \Rightarrow (1)$: By \autoref{definition:tvs-pseudonorm-topology}, for each $r > 0$, $\rho^{-1}([0, r)) \in \cn_E(0)$. Thus for any $x, y \in E$, if $x - y \in \rho^{-1}([0, r))$, then $\abs{\rho(x) - \rho(y)} \le r$. Therefore $\rho \in UC(E; [0, \infty))$.
 \end{proof}
 
 \begin{lemma}[{{\cite[Theorem I.6.1]{SchaeferWolff}}}]
@@ -82,16 +87,19 @@
     \[
     U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}
     \]
+
 \end{lemma}
 \begin{proof}
     For each $H \subset \natp$ finite, let
     \[
     U_H = \sum_{n \in H}V_n \quad \rho_H = \sum_{n \in H}2^{-n}
     \]
+
     Define
     \[
     \rho: E \to [0, 1] \quad x \mapsto \inf\bracs{\rho_H|x \in U_H}
     \]
+
     then
     \begin{enumerate}
         \item[(PN1)] Since $0 \in \bigcap_{H \subset  \natp \text{ finite}}U_H$, $\rho(0) = 0$.
@@ -99,10 +107,11 @@
         \[
         \lambda x \in \sum_{n \in H}\lambda U_n \subset \sum_{n \in H}U_n
         \]
+
         so $\rho(\lambda x) \le \rho(x)$.
         \item[(PN3)] Let $x, y \in X$ and $M, N \subset \natp$ finite such that $x \in U_M$ and $y \in U_N$. Assume without loss of generality that $\rho_M + \rho_N < 1$, then there exists a unique $P \subset \nat$ finite such that $\rho_P = \rho_M + \rho_N$. In which case, $U_P \supset U_M + U_N$ by assumption (b). Therefore $\rho(x + y) \le \rho(x) + \rho(y)$.
     \end{enumerate}
-    For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1} < 2^n$, so $U_{n+1} \subset \rho^{-1}([0, 2^{-n}))$ by \ref{proposition:dyadic-semigroup-order}. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}} = U_n$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.
+    For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1} < 2^n$, so $U_{n+1} \subset \rho^{-1}([0, 2^{-n}))$ by \autoref{proposition:dyadic-semigroup-order}. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}} = U_n$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.
     \begin{enumerate}
         \item[(PN4)] Let $x \in X$ and $n \in \natp$. By assumption (a), there exists $\alpha > 0$ such that for any $\lambda \in K$ with $\abs{\lambda} \ge \alpha$, $x \in \lambda U_n$. Therefore for any $\lambda \in K$ with $\abs{\lambda} \le \alpha^{-1}$, $\lambda x \in U_n$, and $\rho(x) \le 2^{-n}$.
         \item[(PN5)] Let $\lambda \in K$ and $n \in \natp$. By assumption (b), there exists $m \in \nat$ such that $\lambda U_{n-m} \subset \sum_{j = 1}^m U_{n-m}^j \subset U_n$.
@@ -112,7 +121,7 @@
 \begin{remark}
 	\label{remark:tvs-sequence-pseudonorm}
 
-	As discussed in \ref{remark:uniform-sequence-pseudometric} on the proof of \ref{lemma:uniform-sequence-pseudometric}, constructing a pseudometric from entourages by building its level sets is difficult because composing symmetric entourages does not necessarily lead to symmetric entourages. The topological vector space does not have this shortcoming, and as such allows this construction.
+	As discussed in \autoref{remark:uniform-sequence-pseudometric} on the proof of \autoref{lemma:uniform-sequence-pseudometric}, constructing a pseudometric from entourages by building its level sets is difficult because composing symmetric entourages does not necessarily lead to symmetric entourages. The topological vector space does not have this shortcoming, and as such allows this construction.
 \end{remark}
 
 \begin{theorem}[Metrisability of Topological Vector Spaces]
@@ -127,15 +136,15 @@
     \end{enumerate}
 \end{theorem}
 \begin{proof}
-    (3) $\Rightarrow$ (4): By \ref{theorem:uniform-metrisable}.
+    (3) $\Rightarrow$ (4): By \autoref{theorem:uniform-metrisable}.
 
-    (4) $\Rightarrow$ (1): By \ref{proposition:tvs-good-neighbourhood-base}, there exists $\seq{U_n} \subset \cn_E(0)$ circled and radial such that for each $n \in \natp$, $U_{n+1} + U_{n+1} \subset U_n$. By \ref{lemma:tvs-sequence-pseudonorm}, there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $N \in \natp$, $U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}$. In which case, $\rho$ induces the topology on $E$.
+    (4) $\Rightarrow$ (1): By \autoref{proposition:tvs-good-neighbourhood-base}, there exists $\seq{U_n} \subset \cn_E(0)$ circled and radial such that for each $n \in \natp$, $U_{n+1} + U_{n+1} \subset U_n$. By \autoref{lemma:tvs-sequence-pseudonorm}, there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $N \in \natp$, $U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}$. In which case, $\rho$ induces the topology on $E$.
 \end{proof}
 
 \begin{remark}
     \label{remark:tvs-metrisable}
 
-    Let $E$ be a TVS over $K \in \RC$. Similar to the case of uniform spaces, there exists a family of pseudonorms $\seqi{\rho}$ that induces the topology on $E$. Using Minkowski functionals (\ref{definition:gauge}), $\seqi{\rho}$ can be taken such that $\rho_i(\lambda x) = \abs{\lambda}\rho_i(x)$ for all $x \in E$ and $\lambda \in K$. However, a single pseudonorm with this property cannot always induce the topology even if the space is metrisable, hence the difference between pseudonorms and seminorms.
+    Let $E$ be a TVS over $K \in \RC$. Similar to the case of uniform spaces, there exists a family of pseudonorms $\seqi{\rho}$ that induces the topology on $E$. Using \hyperref[Minkowski functionals]{definition:gauge}, $\seqi{\rho}$ can be taken such that $\rho_i(\lambda x) = \abs{\lambda}\rho_i(x)$ for all $x \in E$ and $\lambda \in K$. However, a single pseudonorm with this property cannot always induce the topology even if the space is metrisable, hence the difference between pseudonorms and seminorms.
 \end{remark}
 
 \begin{definition}[Locally Bounded]
@@ -148,5 +157,5 @@
     Let $E$ be a locally bounded TVS over $K \in \RC$, then there exists a pseudonorm $\rho: E \to [0, \infty)$ that induces the topology on $E$.
 \end{proposition}
 \begin{proof}
-    Let $U \in \cn^o(0)$ be bounded. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is circled. For each $n \in \natp$, let $U_n = n^{-1}U$. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$, $\abs{\lambda}^{-1}U \subset V$. For any $n \in \natp$ with $n^{-1} < \abs{\lambda}^{-1}$, $U_n \subset V$. Thus $E$ admits a countable fundamental system of neighbourhoods at $0$. By \ref{theorem:tvs-metrisable}, the topology on $E$ is induced by a pseudonorm.
+    Let $U \in \cn^o(0)$ be bounded. Using \autoref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is circled. For each $n \in \natp$, let $U_n = n^{-1}U$. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$, $\abs{\lambda}^{-1}U \subset V$. For any $n \in \natp$ with $n^{-1} < \abs{\lambda}^{-1}$, $U_n \subset V$. Thus $E$ admits a countable fundamental system of neighbourhoods at $0$. By \autoref{theorem:tvs-metrisable}, the topology on $E$ is induced by a pseudonorm.
 \end{proof}

src/fa/tvs/projective.tex

@@ -17,31 +17,34 @@
         \[
         \bracs{\bigcap_{j \in J}T_j^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \cn_{F_j}(0)}
         \]
+
         is a fundamental system of neighbourhoods for $E$ at $0$.
     \end{enumerate}
     The uniformity $\fU$ and its topology are the \textbf{projective uniformity/topology} induced by $\seqi{T}$.
 \end{definition}
 \begin{proof}
-    (1), (U): By \ref{definition:initial-uniformity}.
+    (1), (U): By \autoref{definition:initial-uniformity}.
 
     Let $U \in \fU$, then there exists $J \subset I$ finite and translation-invariant entourages $\seqj{U}$ such that
     \[
     U \subset V = \bigcap_{j \in J}(T_j \times T_j)^{-1}(U_j)
     \]
 
+
     (3): For each $j \in J$, $(x, y) \in (T_j \times T_j)^{-1}(U_j)$, and $z \in E$,
     \[
     (T_j \times T_j)(x + z, y + z) = (T_jx + T_jz, T_jy + T_jz) \in U_j
     \]
+
     so $(T_j \times T_j)^{-1}(U_j)$ is translation-invariant, and so is $V$.
 
     (4): By (TVS1) and (TVS2), for each $j \in J$, there exists an entourage $V_j$ of $F_j$ and $\eps_j > 0$ such that for any $(x, x'), (y, y') \in V_j$ and $\lambda, \lambda' \in K$ with $\abs{\lambda - \lambda'} < \eps_j$, $(x + y, x' + y'), (\lambda x, \lambda' x') \in U_j$.
 
     Therefore, for any $(x, x'), (y, y') \in \bigcap_{j \in J} T_j^{-1}(V_j)$ and $\lambda, \lambda' \in K$ with $\abs{\lambda - \lambda'} < \min_{j \in J}\eps$, $(x + y, x' + y'), (\lambda x, \lambda' x') \in V$.
 
-    (5): By \ref{definition:continuous-linear} and (4) of \ref{definition:initial-uniformity}.
+    (5): By \autoref{definition:continuous-linear} and (4) of \autoref{definition:initial-uniformity}.
 
-    (6): By \ref{definition:initial-uniformity}.
+    (6): By \autoref{definition:initial-uniformity}.
 \end{proof}
 
 \begin{definition}[Projective Limit of Topological Vector Spaces]
@@ -57,6 +60,7 @@
         E \ar@{->}[u]^{T^E_i} \ar@{->}[ru]_{T^E_j} &
         }
         \]
+
         \item[(U)] For any pair $(F, \bracsn{S^F_i}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique $S \in L(F; E)$ such that the following diagram commutes
 
         \[
@@ -66,17 +70,18 @@
         }
         \]
 
+
         for all $i \in I$.
         \item For any TVS $F$ over $K$ and $S \in \hom(F; E)$, $S \in L(F; E)$ if and only if $T^E_i \circ S \in L(F; E_i)$ for all $i \in I$.
     \end{enumerate}
     The pair $(E, \bracsn{T^E_i}_{i \in I})$ is the \textbf{projective limit} of $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$.
 \end{definition}
 \begin{proof}
-    Let $(E, \bracsn{T^E_i}_{i \in I})$ be the inverse limit of $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$ as $K$-vector spaces (\ref{proposition:module-inverse-limit}).
+    Let $(E, \bracsn{T^E_i}_{i \in I})$ be the inverse limit of $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$ as $K$-vector spaces (\autoref{proposition:module-inverse-limit}).
 
     Equip $E$ with the projective topology generated by $\bracsn{T^E_i}_{i \in I}$, then $(E, \bracsn{T^E_i}_{i \in I})$ satisfies (1), (2), and (3).
 
-    (5): By (5) of \ref{definition:tvs-initial}.
+    (5): By (5) of \autoref{definition:tvs-initial}.
 
-    (U): By (U) of \ref{proposition:module-inverse-limit}, there exists a unique $S \in \hom(F; E)$ such that the given diagram commutes. By (4), $S \in L(F; E)$.
+    (U): By (U) of \autoref{proposition:module-inverse-limit}, there exists a unique $S \in \hom(F; E)$ such that the given diagram commutes. By (4), $S \in L(F; E)$.
 \end{proof}

src/fa/tvs/quotient.tex

@@ -15,6 +15,7 @@
         \widetilde E \ar@{->}[r]_{\tilde f} & F
         }
         \]
+
         If $F$ is a TVS over $K$ and $f \in L(E; F)$, then $\td f \in L(E; F)$.
     \end{enumerate}
     The space $\td E = E/M$ is the \textbf{quotient} of $E$ by $M$.
@@ -22,18 +23,19 @@
 \begin{proof}
     Let $\td E = E/M$ be the algebraic quotient of $E$ by $M$, and equip it with the quotient topology by $\pi$.
 
-    (1): By \ref{definition:quotient-topology}, for each $\pi(U) \subset E/M$, $\pi(U)$ is open if and only if $U$ is open. Since the topology on $E$ is translation-invariant, so is the quotient topology on $E/M$. Let
+    (1): By \autoref{definition:quotient-topology}, for each $\pi(U) \subset E/M$, $\pi(U)$ is open if and only if $U$ is open. Since the topology on $E$ is translation-invariant, so is the quotient topology on $E/M$. Let
     \[
     \fB = \bracs{\pi(U)| U \in \cn(0) \text{ circled and radial}}
     \]
-    Since the circled and radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods at $0$ by \ref{proposition:tvs-good-neighbourhood-base}, $\fB$ is a fundamental system of neighbourhoods at $0$ for the quotient topology on $E/M$. In addition,
+
+    Since the circled and radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods at $0$ by \autoref{proposition:tvs-good-neighbourhood-base}, $\fB$ is a fundamental system of neighbourhoods at $0$ for the quotient topology on $E/M$. In addition,
     \begin{enumerate}
         \item[(TVB1)] Let $U \in \cn(0)$ be circled and radial. For any $\lambda \in K$ with $\abs{\lambda} \le 1$, $\lambda \pi(U) = \pi(\lambda U) \subset \pi(U)$, so $\pi(U)$ is also circled. For any $x + M \in E/M$, there exists $\lambda \in K$ such that $x \in \lambda U$. In which case, $x \in \lambda U + M = \pi(U)$, so $\pi(U)$ is also radial.
-        \item[(TVB2)] For any $U \in \cn(0)$ circled and radial, by \ref{proposition:tvs-good-neighbourhood-base}, there exists $W \in \cn(0)$ such that $W + W \subset U$. In which case, $\pi(W) + \pi(W) \subset \pi(U)$.
+        \item[(TVB2)] For any $U \in \cn(0)$ circled and radial, by \autoref{proposition:tvs-good-neighbourhood-base}, there exists $W \in \cn(0)$ such that $W + W \subset U$. In which case, $\pi(W) + \pi(W) \subset \pi(U)$.
     \end{enumerate}
-    By \ref{proposition:tvs-0-neighbourhood-base}, there exists a unique translation-invariant topology on $E/M$ such that $\fB$ is a fundamental system of neighbourhoods at $0$, which must be the quotient topology on $E/M$. In which case, the quotient topology is a vector space topology by (3) of \ref{proposition:tvs-0-neighbourhood-base}.
+    By \autoref{proposition:tvs-0-neighbourhood-base}, there exists a unique translation-invariant topology on $E/M$ such that $\fB$ is a fundamental system of neighbourhoods at $0$, which must be the quotient topology on $E/M$. In which case, the quotient topology is a vector space topology by (3) of \autoref{proposition:tvs-0-neighbourhood-base}.
 
-    (2), (3), (U): By \ref{definition:quotient-topology}.
+    (2), (3), (U): By \autoref{definition:quotient-topology}.
 \end{proof}
 
 \begin{proposition}
@@ -45,5 +47,6 @@
     \[
     M = \bigcap_{V \in \cn(0)}M + V
     \]
+
     which is equivalent to $E/M$ being Hausdorff.
 \end{proof}

src/fa/tvs/spaces-of-linear.tex

@@ -5,11 +5,12 @@
 \label{proposition:tvs-set-uniformity}
     Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be an upward-directed system of sets, $F$ be a TVS over $K \in \RC$, then
     \begin{enumerate}
-        \item The $\mathfrak{S}$-uniformity on $F^T$ (\ref{definition:set-uniform}) is translation invariant.
+        \item The $\mathfrak{S}$-uniformity on $F^T$ (\autoref{definition:set-uniform}) is translation invariant.
         \item The composition defined by
         \[
         T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
         \]
+
         is continuous.
     \end{enumerate}
     For any vector subspace $\cf \subset F^T$, the following are equivalent:
@@ -19,7 +20,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. Using \ref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
+    (1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
 
     (2): Let $f, g, f', g' \in \cf$, $S \in \mathfrak{S}$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
 
@@ -30,10 +31,11 @@
     \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
     &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
     \end{align*}
-    Let $U_0 \in \cn_F(0)$. By (TVS1) and \ref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_F(0)$ circled such that $U + U \subset U_0$. By (TVS2), there exists $V \in \cn_F(0)$ such that $\lambda' V \subset U$. Since $f(S)$ is bounded, there exists $\eps > 0$ with $\eps f(S) \subset U$. In which case, if $\abs{\lambda - \lambda'} < \eps$ and $f(x) - g(x) \in V$ for all $x \in S$, then
+    Let $U_0 \in \cn_F(0)$. By (TVS1) and \autoref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_F(0)$ circled such that $U + U \subset U_0$. By (TVS2), there exists $V \in \cn_F(0)$ such that $\lambda' V \subset U$. Since $f(S)$ is bounded, there exists $\eps > 0$ with $\eps f(S) \subset U$. In which case, if $\abs{\lambda - \lambda'} < \eps$ and $f(x) - g(x) \in V$ for all $x \in S$, then
     \[
     \lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0
     \]
+
     for all $x \in S$.
 \end{proof}
 
@@ -50,25 +52,30 @@
         \[
         I: B_{\mathfrak{S}}^k(E; B_{\mathfrak{S}}(E; F)) \to B^{k+1}_{\mathfrak{S}}(E; F)
         \]
+
         defined by
         \[
         (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
         \]
+
         is an isomorphism.
         \item The map
         \[
         I: \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to B^k_{\mathfrak{S}}(E; F)
         \]
+
         defined by
         \[
         IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
         \]
+
         is an isomorphism.
     \end{enumerate}
     which allows the identification
     \[
     \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = B^k_{\mathfrak{S}}(E; F)
     \]
+
     under the map $I$ in (2).
 \end{proposition}
 \begin{proof}
@@ -76,10 +83,12 @@
     \[
     I^{-1}T: E \to B_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot)
     \]
+
     Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
     \[
     T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
     \]
+
     by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\mathfrak{S}}(E; F)$.
 
     In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F))$.
@@ -92,6 +101,7 @@
     \[
     \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F)) = B^{k+1}_{\mathfrak{S}}(E; F)
     \]
+
     Thus (2) holds for all $k \in \natp$.
 \end{proof}
 

src/measure/bochner-integral/index.tex

@@ -1,4 +1,4 @@
 \chapter{The Bochner Integral}
 \label{chap:bochner-integral}
 
-\input{./src/measure/bochner-integral/strongly.tex}
+\input{./strongly.tex}

src/measure/bochner-integral/strongly.tex

@@ -17,11 +17,12 @@
 \begin{proof}
     (1) $\Rightarrow$ (2): TODO
 
-    (2) $\Rightarrow$ (3): By \ref{proposition:measurable-simple-separable-norm}.
+    (2) $\Rightarrow$ (3): By \autoref{proposition:measurable-simple-separable-norm}.
 
-    (3) $\Rightarrow$ (1): For each $\phi \in E^*$, $\phi \circ f = \limv{n}\phi \circ f_n$ is measurable by \ref{proposition:limit-measurable}. Since
+    (3) $\Rightarrow$ (1): For each $\phi \in E^*$, $\phi \circ f = \limv{n}\phi \circ f_n$ is measurable by \autoref{proposition:limit-measurable}. Since
     \[
     f(X) \subset \ol{\bigcup_{n \in \natp}f_n(X)}
     \]
+
     and each $f_n$ is finitely-valued, $f(X)$ is separable.
 \end{proof}

src/measure/index.tex

@@ -1,8 +1,8 @@
 \part{Measure Theory and Integration}
 \label{part:measure}
 
-\input{./src/measure/sets/index.tex}
+\input{./sets/index.tex}
-\input{./src/measure/measure/index.tex}
+\input{./measure/index.tex}
-\input{./src/measure/measurable-maps/index.tex}
+\input{./measurable-maps/index.tex}
-\input{./src/measure/lebesgue-integral/index.tex}
+\input{./lebesgue-integral/index.tex}
-\input{./src/measure/bochner-integral/index.tex}
+\input{./bochner-integral/index.tex}

src/measure/lebesgue-integral/complex.tex

@@ -7,10 +7,12 @@
     \[
     \int \abs{f} d\mu < \infty
     \]
+
     The set
     \[
     \mathcal{L}^1(X, \cm, \mu; \complex) = \mathcal{L}^1(X, \cm, \mu) = \mathcal{L}^1(X; \complex) = \mathcal{L}^1(X) = \mathcal{L}^1(\mu; \complex) = \mathcal{L}^1(\mu)
     \]
+
     is the vector space of \textbf{$\mu$-integrable functions} on $X$.
 \end{definition}
 \begin{proof}
@@ -18,7 +20,8 @@
     \[
     \int \abs{\lambda f + g}d\mu \le \int \abs{\lambda} \abs{f} + \abs{g}d\mu = \lambda \int \abs{f}d\mu + \int \abs{g}d\mu
     \]
-    by \ref{proposition:lebesgue-non-negative-properties}.
+
+    by \autoref{proposition:lebesgue-non-negative-properties}.
 \end{proof}
 
 \begin{definition}[Positive and Negative Parts]
@@ -27,6 +30,7 @@
     \[
     f^+ = \max(f, 0) \quad f^- = -\min(f, 0)
     \]
+
     are the \textbf{positive} and \textbf{negative} parts of $f$, and $f = f^+ - f^-$.
 \end{definition}
 
@@ -37,10 +41,12 @@
     \[
     \int f d\mu = \int f^+ d\mu - \int f^- d\mu
     \]
+
     is the \textbf{integral} of $f$. If $f$ is $\complex$-valued, then
     \[
     \int f d\mu = \int \text{Re}(f)d\mu + i\int \text{Im}(f)d\mu
     \]
+
     is the \textbf{integral} of $f$.
 \end{definition}
 
@@ -58,7 +64,8 @@
         -\lambda\int f^- d\mu + \lambda\int f^+ d\mu &\lambda < 0
     \end{cases}
     \]
-    by \ref{proposition:lebesgue-non-negative-properties}, so $\int \lambda f d\mu = \lambda \int f d\mu$.
+
+    by \autoref{proposition:lebesgue-non-negative-properties}, so $\int \lambda f d\mu = \lambda \int f d\mu$.
 
     Let $h = f + g$, then $h = h^+ - h^- = f^+ + g^+ - f^- - g^-$, so
     \begin{align*}
@@ -67,7 +74,7 @@
         \int h^+ d\mu - \int h^- d\mu &= \int f^+ d\mu - \int f^- d\mu + \int g^+ d\mu - \int g^- d\mu \\
         &= \int f d\mu + \int g d\mu
     \end{align*}
-    by \ref{proposition:lebesgue-non-negative-properties}, so $\int f + g d\mu = \int f d\mu + \int g d\mu$.
+    by \autoref{proposition:lebesgue-non-negative-properties}, so $\int f + g d\mu = \int f d\mu + \int g d\mu$.
 
     Now suppose that $f, g$ are $\complex$-valued and $\lambda = \alpha + \beta i \in \complex$ with $\alpha, \beta \in \real$, then
     \begin{align*}
@@ -88,6 +95,7 @@
     \[
     \int f = \int f^+ d\mu - \int f^-d\mu \le \int f^+ + \int f^-d\mu = \int \abs{f}d\mu
     \]
+
     and if $f$ is $\complex$-valued and $\alpha = \ol{\sgn(\int f d\mu)}$, then
     \begin{align*}
         \abs{\int f d\mu} &= \alpha \int f d\mu = \int \alpha f \\
@@ -113,7 +121,7 @@
     then $\int fd\mu = \limv{n}\int f_n d\mu$.
 \end{theorem}
 \begin{proof}
-    By (1) and (2), $f$ is measurable with $\int \abs{f}d\mu \le \int \abs{g}d\mu < \infty$, so $f \in \mathcal{L}^1(X)$. Now, since $g + f, g - f \ge 0$, by Fatou's lemma (\ref{lemma:fatou}) and \ref{proposition:lebesgue-integral-properties},
+    By (1) and (2), $f$ is measurable with $\int \abs{f}d\mu \le \int \abs{g}d\mu < \infty$, so $f \in \mathcal{L}^1(X)$. Now, since $g + f, g - f \ge 0$, by \hyperref[Fatou's lemma]{lemma:fatou} and \autoref{proposition:lebesgue-integral-properties},
     \begin{align*}
         \int g d\mu + \int f d\mu &\le \liminf_{n \to \infty}\int g + f_n d\mu = \int g d\mu + \liminf_{n \to \infty}f_n d\mu \\
         \int g d\mu - \int f d\mu &\le \liminf_{n \to \infty}\int g - \int f_n d\mu = \int g d\mu - \limsup_{n \to \infty}\int f_n d\mu
@@ -122,12 +130,13 @@
     \[
     \limsup_{n \to \infty}\int f_n d\mu \le \int f d\mu \le \liminf_{n \to \infty}\int f_n d\mu
     \]
+
     and $\int f d\mu = \limv{n}\int f_n d\mu$.
 \end{proof}
 
 \begin{remark}[There is no dominated convergence theorem for nets]
 \label{remark:dct-no-net}
-    In analysis, one frequently encounters places where only sequential continuity is provided or required. It is my opinion that a good portion of this comes from the lack of an extension of the dominated convergence theorem (\ref{theorem:dct}) to nets. This limitation arises from the monotone convergence theorem (\ref{theorem:mct}), where continuity from below is used.
+    In analysis, one frequently encounters places where only sequential continuity is provided or required. It is my opinion that a good portion of this comes from the lack of an extension of the \hyperref[dominated convergence theorem]{theorem:dct} to nets. This limitation arises from the \hyperref[monotone convergence theorem]{theorem:mct}, where continuity from below is used.
 
     For an example, consider the Lebesgue measure on $[0, 1]$. Let $A$ be the net of all finite subsets of $[0, 1]$, directed by inclusion, then $\lim_{\alpha \in A}\one_\alpha = 1$ pointwise. However, $\int \one_\alpha = 0$ for all $\alpha \in A$.
 \end{remark}

src/measure/lebesgue-integral/index.tex

@@ -1,6 +1,6 @@
 \chapter{The Lebesgue Integral}
 \label{chap:lebesgue-integral}
 
-\input{./src/measure/lebesgue-integral/simple.tex}
+\input{./simple.tex}
-\input{./src/measure/lebesgue-integral/non-negative.tex}
+\input{./non-negative.tex}
-\input{./src/measure/lebesgue-integral/complex.tex}
+\input{./complex.tex}

src/measure/lebesgue-integral/non-negative.tex

@@ -7,6 +7,7 @@
     \[
     \mathcal{L}^+(X, \cm) = \bracs{f: X \to \real| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
     \]
+
     is the space of non-negative $\real$-valued measurable functions on $(X, \cm)$.
 \end{definition}
 
@@ -16,6 +17,7 @@
     \[
     \int f d\mu = \int f(x)\mu(dx) = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le f}
     \]
+
     is the \textbf{Lebesgue integral} of $f$.
 \end{definition}
 
@@ -25,12 +27,14 @@
     \[
     \int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}, \phi \le f}
     \]
+
 \end{lemma}
 \begin{proof}
     Let $\phi \in \Sigma^+(X, \cm)$ with $\phi \le f$, then for any $\alpha \in (0, 1)$, $\alpha \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$. Since
     \[
     \int \phi d\mu = \sup_{\alpha \in (0, 1)}\alpha \int \phi d\mu = \sup_{\alpha \in (0, 1)}\int \alpha \phi d\mu
     \]
+
     the two sides are equal.
 \end{proof}
 
@@ -40,15 +44,17 @@
     \[
     \limv{n}\int f_n d\mu = \int f d\mu
     \]
+
 \end{theorem}
 \begin{proof}
-    By \ref{definition:lebesgue-non-negative}, $\int f_n d\mu \le \int f d\mu$ for each $n \in \natp$.
+    By \autoref{definition:lebesgue-non-negative}, $\int f_n d\mu \le \int f d\mu$ for each $n \in \natp$.
 
-    Let $\phi \in \Sigma^+(X, \cm)$ with $\phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $f_n \upto f$, $\bracs{f_n \ge \phi} \upto X$. In which case, since $A \mapsto \int \one_A \phi d\mu$ is a measure (\ref{proposition:lebesgue-simple-properties}),
+    Let $\phi \in \Sigma^+(X, \cm)$ with $\phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $f_n \upto f$, $\bracs{f_n \ge \phi} \upto X$. In which case, since $A \mapsto \int \one_A \phi d\mu$ is a measure (\autoref{proposition:lebesgue-simple-properties}),
     \[
     \limv{n}\int f_n d\mu \ge \limv{n}\int \one_{\bracs{f_n \ge \phi}} \cdot \phi d\mu = \int \phi d\mu
     \]
-    by continuity from below (\ref{proposition:measure-properties}). Therefore $\limv{n}\int f_n d\mu \ge \int f$ by \ref{lemma:lebesgue-non-negative-strict}.
+
+    by continuity from below (\autoref{proposition:measure-properties}). Therefore $\limv{n}\int f_n d\mu \ge \int f$ by \autoref{lemma:lebesgue-non-negative-strict}.
 \end{proof}
 
 \begin{lemma}[Fatou, {{\cite[Lemma 2.18]{Folland}}}]
@@ -57,6 +63,7 @@
     \[
     \int \liminf_{n \to \infty}f_n d\mu \le \liminf_{n \to \infty}\int f_nd\mu
     \]
+
 \end{lemma}
 \begin{proof}
     For each $n \in \natp$, $\inf_{k \ge n}f_k \le f_n$. By the monotone convergence theorem,
@@ -64,6 +71,7 @@
     \int \liminf_{n \to \infty}f_n d\mu = \limv{n}\int \inf_{k \ge n}f_k d\mu
     \le \liminf_{n \to \infty}\int f_n d\mu
     \]
+
 \end{proof}
 
 \begin{lemma}
@@ -71,7 +79,7 @@
     Let $(X, \cm)$ be a measurable space and $f \in \mathcal{L}^+(X, \cm)$, then there exists $\seq{f_n} \subset \Sigma^+(X, \cm)$ such that $f_n \upto f$ pointwise.
 \end{lemma}
 \begin{proof}
-    By \ref{proposition:measurable-simple-separable-norm}, there exists $\seq{g_n} \subset \Sigma^+(X, \cm)$ such that $0 \le g_n \le f$ for each $n \in \natp$, and $g_n \to f$ pointwise. For each $n \in \natp$, let $f_n = \max_{1 \le k \le n}g_k$, then $f_n \in \Sigma^+(X, \cm)$, $0 \le f_n \le f$, and $f_n \upto f$ pointwise.
+    By \autoref{proposition:measurable-simple-separable-norm}, there exists $\seq{g_n} \subset \Sigma^+(X, \cm)$ such that $0 \le g_n \le f$ for each $n \in \natp$, and $g_n \to f$ pointwise. For each $n \in \natp$, let $f_n = \max_{1 \le k \le n}g_k$, then $f_n \in \Sigma^+(X, \cm)$, $0 \le f_n \le f$, and $f_n \upto f$ pointwise.
 \end{proof}
 
 
@@ -88,19 +96,20 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): By \ref{lemma:lebesgue-simple-monotone}, there exists $\seq{f_n}, \seq{g_n} \subset \Sigma^+(X, \cm)$ with $0 \le f_n \le f$ and $0 \le g_n \le g$ for each $n \in \natp$, $f_n \upto f$, and $g_n \upto g$. By \ref{proposition:lebesgue-simple-properties} and the Monotone Convergence Theorem (\ref{theorem:mct}),
+    (1): By \autoref{lemma:lebesgue-simple-monotone}, there exists $\seq{f_n}, \seq{g_n} \subset \Sigma^+(X, \cm)$ with $0 \le f_n \le f$ and $0 \le g_n \le g$ for each $n \in \natp$, $f_n \upto f$, and $g_n \upto g$. By \autoref{proposition:lebesgue-simple-properties} and the \hyperref[Monotone Convergence Theorem]{theorem:mct},
     \begin{align*}
     \int \alpha f + g d\mu = \limv{n}\int \alpha f_n + g_n d\mu = \alpha\limv{n}\int f_n d\mu + \limv{n}\int g_n d\mu \\
     &= \alpha \int f d\mu + \int g d\mu
     \end{align*}
 
 
-    (2): By (1) and the Monotone Convergence Theorem (\ref{theorem:mct}),
+    (2): By (1) and the \hyperref[Monotone Convergence Theorem]{theorem:mct},
     \[
     \int \sum_{n \in \natp}f_n d\mu = \int \limv{N}\sum_{n = 1}^N f_n d\mu = \limv{N}\sum_{n = 1}^N \int f_nd\mu = \sum_{n \in \natp}\int f_n d\mu
     \]
 
-    (3): By \ref{definition:lebesgue-non-negative}.
+
+    (3): By \autoref{definition:lebesgue-non-negative}.
 
     (4): Since $\int f d\mu \ge \int \infty \cdot \one_{\bracs{f = \infty}}d\mu = \infty \cdot \mu(\bracs{f = \infty})$, $\int f d\mu < \infty$ implies that $\mu(\bracs{f = \infty}) = 0$.
 
@@ -108,6 +117,7 @@
     \[
     \int f d\mu \ge \int \eps \cdot \one_{\bracs{f \ge \eps}} = \eps \cdot \mu(\bracs{f \ge \eps})
     \]
+
     so $\int f d\mu < \infty$ implies that $\mu(\bracs{f \ge \eps}) < \infty$. Since $\bracs{f > 0} = \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$, $\bracs{f > 0}$ is $\sigma$-finite.
 
     (6): As in the proof of (5), $\mu(\bracs{f \ge \eps}) = 0$ for all $\eps > 0$. Thus $\bracs{f > 0} = \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$ is a $\mu$-null set, so $f = 0$ $\mu$-almost everywhere.

src/measure/lebesgue-integral/simple.tex

@@ -12,6 +12,7 @@
     \[
     \int f d\mu = \int f(x) \mu(dx) = \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y})
     \]
+
     is the \textbf{Lebesgue integral} of $f$.
 \end{definition}
 
@@ -30,11 +31,13 @@
     \[
     \alpha f = \sum_{y \in f(X)} (\alpha y) \cdot \one_{\bracs{f = y}}
     \]
+
     is the standard form of $\alpha f$. Therefore
     \[
     \int \alpha f d\mu = \sum_{y \in f(X)}(\alpha y) \cdot \mu({\bracs{f = y}}) = \alpha \int f d\mu
     \]
 
+
     (2): Since $\mu$ is finitely additive,
     \begin{align*}
         \int f d\mu + \int g d\mu &= \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y}) + \sum_{y \in g(X)}y \cdot \mu(\bracs{g = y}) \\
@@ -53,12 +56,15 @@
     \[
     \one_A \cdot f = \sum_{y \in f(X)}y \cdot \one_{A \cap \bracs{f = y}}
     \]
+
     so by (1),
     \[
     \int \one_A \cdot f d\mu = \sum_{y \in f(X)}y \cdot \mu(A \cap \bracs{f = y})
     \]
+
     Since $\mu$ is countably additive, for any $\seq{E_n} \subset \cm$ pairwise disjoint with $E = \bigcup_{n \in \natp}E_n$,
     \[
     \int \one_E \cdot f d\mu = \sum_{y \in f(X)}y \cdot \one_{E \cap \bracs{f = y}} = \sum_{y \in f(X)}\sum_{n \in \natp}y \cdot \one_{E_n \cap \bracs{f = y}} = \sum_{n \in \natp}\int \one_{E_n} \cdot f d\mu
     \]
+
 \end{proof}

src/measure/measurable-maps/index.tex

@@ -1,8 +1,8 @@
 \chapter{Measurable Functions}
 \label{chap:measurable-maps}
 
-\input{./src/measure/measurable-maps/measurable-maps.tex}
+\input{./measurable-maps.tex}
-\input{./src/measure/measurable-maps/product.tex}
+\input{./product.tex}
-\input{./src/measure/measurable-maps/real-valued.tex}
+\input{./real-valued.tex}
-\input{./src/measure/measurable-maps/simple.tex}
+\input{./simple.tex}
-\input{./src/measure/measurable-maps/metric.tex}
+\input{./metric.tex}

src/measure/measurable-maps/measurable-maps.tex

@@ -31,5 +31,6 @@
     \[
     \sigma(\bracs{f_i| i \in I}) = \sigma\paren{f_i^{-1}(\cn_i)|i \in I}
     \]
+
     is the smallest $\sigma$-algebra on $X$ such that each $\seqi{f}$ is measurable.
 \end{definition}

src/measure/measurable-maps/metric.tex

@@ -7,13 +7,15 @@
     \[
     X \to Z \quad x \mapsto F(f_1(x), \cdots, f_n(x))
     \]
+
     is $(\cm, \cn)$-measurable.
 \end{proposition}
 \begin{proof}
-    By \ref{proposition:product-sigma-algebra-metric}, $\cb_{\prod_{j = 1}^n Y_j} = \bigotimes_{j = 1}^n \cb_{Y_j}$, so
+    By \autoref{proposition:product-sigma-algebra-metric}, $\cb_{\prod_{j = 1}^n Y_j} = \bigotimes_{j = 1}^n \cb_{Y_j}$, so
     \[
     X \to \prod_{j = 1}^n Y_j \quad x \mapsto (f_1(x), \cdots, f_n(x))
     \]
+
     is $(\cm, \cb_{\prod_{j = 1}^n Y_j})$-measurable. Therefore the composition is $(\cm, \cn)$-measurable.
 \end{proof}
 
@@ -27,7 +29,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    By \ref{proposition:metric-measurable-fibre-product}.
+    By \autoref{proposition:metric-measurable-fibre-product}.
 \end{proof}
 
 \begin{proposition}
@@ -43,7 +45,8 @@
     \[
     d_{U^c}: X \to [0, 1] \quad x \mapsto d(x, U^c) \wedge 1
     \]
-    is continuous by \ref{proposition:set-distance-continuous}. By \ref{proposition:fattening-closure}, $\bracsn{d_{U^c} > 0} = U$. Thus $\bracs{f \in U} = \bracsn{d_{U^c} \circ f > 0}$.
+
+    is continuous by \autoref{proposition:set-distance-continuous}. By \autoref{proposition:fattening-closure}, $\bracsn{d_{U^c} > 0} = U$. Thus $\bracs{f \in U} = \bracsn{d_{U^c} \circ f > 0}$.
 \end{proof}
 
 \begin{proposition}
@@ -59,9 +62,10 @@
     \[
     \bracs{\limv{n}f_n \text{ exists}} = \bigcap_{k \in \natp}\bigcup_{N \in \natp}\bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracsn{d(f_m, f_n) < 1/k}
     \]
-    is measurable by \ref{proposition:metric-measurables}.
 
-    (2): For each $\phi \in C(X; [0, 1])$, $\phi \circ f = \limv{n}\phi \circ f_n$ is $(\cm, \cb_\real)$-measurable by \ref{proposition:limit-measurable}. Thus $f$ is $(\cm, \cb_\real)$-measurable by \ref{proposition:metric-measurable-compose}.
+    is measurable by \autoref{proposition:metric-measurables}.
+
+    (2): For each $\phi \in C(X; [0, 1])$, $\phi \circ f = \limv{n}\phi \circ f_n$ is $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable}. Thus $f$ is $(\cm, \cb_\real)$-measurable by \autoref{proposition:metric-measurable-compose}.
 \end{proof}
 
 
@@ -89,18 +93,22 @@
     \[
     C(N, x) = \bracs{1 \le n \le N|y_n \in N(f(x))}
     \]
+
     By assumption (b), $1 \in C(N, x) \ne \emptyset$. Let
     \[
     k(N, x) = \min\bracs{n \in C(N, x) \bigg | d(f(x), y_n) = \min_{m \in C(N, x)}d(f(x), y_m)}
     \]
+
     then for any $k \in \natp$,
     \[
     \bracs{x \in X|k(n, x) \le k} = \bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
     \]
-    For each $1 \le m \le N$, $y \mapsto d(y, y_m)$ is continuous. Thus $x \mapsto d(f(x), y_m)$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_m)$ are $(\cm, \cb_\real)$-measurable by \ref{proposition:limit-measurable} and assumption (c). By \ref{proposition:metric-measurables},
+
+    For each $1 \le m \le N$, $y \mapsto d(y, y_m)$ is continuous. Thus $x \mapsto d(f(x), y_m)$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_m)$ are $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable} and assumption (c). By \autoref{proposition:metric-measurables},
     \[
     \bracs{x \in X \bigg | d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)} \in \cm
     \]
+
     for each $1 \le j \le k$. Combining this with assumption (c) shows that $\bracs{x \in X|k(n, x) \le k} \in \cm$.
 
     Let $f_N(x) = y_{k(N, x)}$, then for each $N \in \natp$, $f_N(x) \subset \bracs{y_n|1 \le n \le N}$, so $f_N(x)$ is finitely valued. Since $x \mapsto k{(N, x)}$ is $(\cm, 2^\nat)$-measurable, $f_N$ is $(\cm, \cb_Y)$-measurable.
@@ -109,9 +117,10 @@
     \[
     f(x) \in \ol{N^o(f(x))} = \ol{\bracs{y_n| n \in \natp, y_n \in N^o(f(x))}}
     \]
-    by assumption (a) and \ref{definition:dense}. Thus for any $\eps > 0$, there exists $N \in \nat$ such that $y_N \in N^o(f(x))$ and $d(f(x), y_N) < \eps$. In which case, $d(f(x), f_n(x)) < \eps$ for all $n \ge N$. Therefore $f_n \to f$ pointwise.
 
-    (3) $\Rightarrow$ (1): By \ref{proposition:metric-measurable-limit}.
+    by assumption (a) and \autoref{definition:dense}. Thus for any $\eps > 0$, there exists $N \in \nat$ such that $y_N \in N^o(f(x))$ and $d(f(x), y_N) < \eps$. In which case, $d(f(x), f_n(x)) < \eps$ for all $n \ge N$. Therefore $f_n \to f$ pointwise.
+
+    (3) $\Rightarrow$ (1): By \autoref{proposition:metric-measurable-limit}.
 \end{proof}
 
 \begin{proposition}
@@ -127,6 +136,7 @@
     \[
     N: E \to 2^E \quad y \mapsto B_E(0, \norm{y}_E)
     \]
+
     then
     \begin{enumerate}
         \item[(a)] $y \in \ol{B_E(0, \norm{y}_E)}$.
@@ -135,6 +145,7 @@
         \[
         \bracs{y \in E|y_0 \in N(y)} = \bracs{y \in E|\norm{y_0}_E \le \norm{y}_E} \in \cb_E
         \]
+
     \end{enumerate}
-    By \ref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
+    By \autoref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
 \end{proof}

src/measure/measurable-maps/product.tex

@@ -12,6 +12,7 @@
     \[
     \ce = \bracs{\bigcap_{j \in J}\pi_j^{-1}(E_j) \bigg | J \subset I \text{ finite}, E_j \in \cm_j}
     \]
+
     then $\ce$ is an elementary family, and $\sigma(\ce) = \bigotimes_{i \in I}\cm_i$.
 \end{lemma}
 \begin{proof}
@@ -21,15 +22,18 @@
     \[
     \bigcap_{j \in J}\pi_j^{-1}(E_j), \bigcap_{j \in J'}\pi_j^{-1}(F_j) \in \ce
     \]
+
     Assume without loss of generality that $J = J'$, then
     \[
     \bigcap_{j \in J}\pi_j^{-1}(E_j) \cap \bigcap_{j \in J'}\pi_j^{-1}(F_j) = \bigcap_{j \in J}\pi_j^{-1}(E_j \cap F_j) \in \ce
     \]
+
     \item[(E)] For any $j \in I$, $\prod_{i \in I}X_i = \pi_j^{-1}(X_j) \in \ce$. Thus it is sufficient to show that $\ce$ is closed under complements. Let $\bigcap_{j \in J}\pi_j^{-1}(E_j) \in \ce$, then
     \[
     \braks{\bigcap_{j \in J}\pi_j^{-1}(E_j)}^c = \bigcup_{j \in J}\pi_j^{-1}(E_j^c) = \bigsqcup_{\emptyset \subsetneq K \subset J}
     \underbrace{\paren{\bigcap_{k \in K}\pi_j^{-1}(E_k^c)} \cap \paren{\bigcap_{j \in J \setminus K}\pi_j^{-1}(E_k)}}_{\in \ce}
     \]
+
 \end{enumerate}
 \end{proof}
 
@@ -42,11 +46,12 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (2): By \ref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$.
+    (2): By \autoref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$.
 
     Since $\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \in \bigotimes_{j = 1}^n \cb_{X_j}$ and
     \[
     U = \bigcup_{j \in \natp}\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k})
     \]
+
     The open set $U$ is in $\bigotimes_{j = 1}^n \cb_{X_j}$.
 \end{proof}

src/measure/measurable-maps/real-valued.tex

@@ -33,12 +33,14 @@
     \bracs{F > \alpha} = \bigcup_{n \in \natp}\bracs{f_n > \alpha}
     \]
 
+
     (2): Let $\alpha \in \real$, then
     \[
     \bracs{f < \alpha} = \bigcup_{n \in \natp}\bracs{f_n < \alpha}
     \]
 
-    By \ref{proposition:borel-sigma-extended-generators} and \ref{lemma:measurable-function-generating-set}, $F$ and $f$ are both $(\cm, \cb_{\ol \real})$-measurable.
+
+    By \autoref{proposition:borel-sigma-extended-generators} and \autoref{lemma:measurable-function-generating-set}, $F$ and $f$ are both $(\cm, \cb_{\ol \real})$-measurable.
 
     (3): $\limsup_{n \to \infty}f_n = \inf_{n \in \natp}\sup_{k \ge n}f_k$.
 
@@ -46,5 +48,5 @@
 
     (5): If $\limv{n}f_n$ exists, then it is equal to (3) and (4). In which case, it is measurable.
 
-    Finally, if the above functions are $\real$-valued, then they are $(\cm, \cb_\real)$-measurable by \ref{lemma:extended-real-measurable}.
+    Finally, if the above functions are $\real$-valued, then they are $(\cm, \cb_\real)$-measurable by \autoref{lemma:extended-real-measurable}.
 \end{proof}

src/measure/measurable-maps/simple.tex

@@ -10,6 +10,7 @@
     0 &x \not\in E
     \end{cases}
     \]
+
     is the \textbf{characteristic/indicator function} of $E$.
 \end{definition}
 
@@ -28,6 +29,7 @@
     \[
     f = \sum_{y \in f(X)}y \cdot \one_{\bracs{f = y}}
     \]
+
     is the \textbf{standard form} of $f$.
 
     The set $\Sigma(X, \cm; V)$ is the space of $V$-valued simple functions on $(X, \cm)$, which forms a vector space.

src/measure/measure/complete.tex

@@ -12,6 +12,7 @@
     \[
     \ol{\cm} = \bracs{E \cup F| E \in \cm, F \subset N, N \in \cn}
     \]
+
     then
     \begin{enumerate}
         \item For any $A \in \ol{\cm}$, there exists $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $E \cap N = \emptyset$ and $A = E \sqcup F$.
@@ -35,22 +36,27 @@
     \bigcup_{n \in \natp}A_n = \underbrace{\braks{\bigcup_{n \in \natp}E_n}}_{\in \cm} \cup \underbrace{\braks{\bigcup_{n \in \natp}F_n}}_{\subset \bigcup_{n \in \natp}N_n} \in \ol{\cm}
     \]
 
+
     (3): Let $A \in \ol{\cm}$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$. Since $A \setminus N = E \setminus N \in \cm$ and $A \cup N = E \cup N \in \cm$,
     \[
     \mu(A \setminus N) \le \mu(E) \le \mu(A \cup N)
     \]
+
     Let $E' \in \cm$, $N' \in \cn$, and $F' \subset N'$ such that $A = E' \cup F'$, then
     \[
     \mu(A \setminus (N \cup N')) \le \mu(A \setminus N') \le \mu(E') \le \mu(A \cup N') \le \mu(A \cup (N \cup N'))
     \]
+
     Now, since $N$ and $N'$ are null,
     \[
     \mu(A \cup (N \cup N')) = \mu(A \setminus (N \cup N')) + \mu(N \cup N') = \mu(A \setminus (N \cup N'))
     \]
+
     so
     \[
     \mu(A \setminus (N \cup N')) = \mu(E) = \mu(E') = \mu(A \cup (N \cup N'))
     \]
+
     Thus for any decomposition $A = E \cup F$ where $E \in \cm$ and $F$ is contained in a null set, $\ol{\mu}(A) = \mu(E)$ is well-defined.
 
     To see that $\ol{\mu}$ is a measure, let $\seq{A_n} \subset \ol{\cm}$ be pairwise disjoint. For each $n \in \nat$, let $E_n \in \cm$, $N_n \in \cn$, and $F_n \subset N_n$ such that $A_n = E_n \cup F_n$, then
@@ -58,6 +64,7 @@
     \ol{\mu}\paren{\bigsqcup_{n \in \natp}A_n} = \mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu(E_n) = \sum_{n \in \natp}\mu(A_n)
     \]
 
+
     (4): Let $A \in \ol{\cm}$ with $\ol{\mu}(A) = 0$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$, then $\mu(E) = \mu(E \cup N) = 0$. By definition of $\ol{\cm}$, any subset of $E \cup N$ is also in $\ol{\cm}$. Thus $\ol{\mu}$ is complete.
 
     (U): Let $N \in \cn \subset \cm \subset \cf$, then for any $F \subset N$, $\nu(F) \le \nu(N) = \mu(N)$, and $F \in \cf$ by completeness of $(X, \cf, \nu)$. Thus $\cf \supset \ol{\cm}$.
@@ -66,4 +73,5 @@
     \[
     \nu(A) = \nu(E \cup F) = \nu(E) = \mu(E) = \ol{\mu}(A)
     \]
+
 \end{proof}

src/measure/measure/index.tex

@@ -1,14 +1,14 @@
 \chapter{Positive Measures}
 \label{chap:measures}
 
-\input{./src/measure/measure/measure.tex}
+\input{./measure.tex}
-\input{./src/measure/measure/complete.tex}
+\input{./complete.tex}
-\input{./src/measure/measure/semifinite.tex}
+\input{./semifinite.tex}
-\input{./src/measure/measure/sigma-finite.tex}
+\input{./sigma-finite.tex}
-\input{./src/measure/measure/regular.tex}
+\input{./regular.tex}
-\input{./src/measure/measure/radon.tex}
+\input{./radon.tex}
-\input{./src/measure/measure/outer.tex}
+\input{./outer.tex}
-\input{./src/measure/measure/lebesgue-stieltjes.tex}
+\input{./lebesgue-stieltjes.tex}
-\input{./src/measure/measure/radon.tex}
+\input{./radon.tex}
-\input{./src/measure/measure/riesz.tex}
+\input{./riesz.tex}
-\input{./src/measure/measure/kolmogorov.tex}
+\input{./kolmogorov.tex}

src/measure/measure/kolmogorov.tex

@@ -36,10 +36,12 @@
     \[
     K_n = \bigcap_{j = 1}^n \pi_{[n], [j]}^{-1}(C_n)
     \]
+
     Since each $X_j$ is Hausdorff, $K_n \subset \prod_{j = 1}^n X_j$ is compact with $K_n \subset B_n$ and $K_{n+1} \subset K_n \times X_{n+1}$. Moreover,
     \[
     \mu_{[n]}(B_n \setminus K_n) \le \sum_{j = 1}^n\mu_{[n]}\braks{\pi_{[n], [j]}^{-1}(B_j \setminus C_j)} \le \sum_{j = 1}^n\mu_{[j]}(B_j \setminus C_j) \le \eps/2
     \]
+
     Thus $\mu_{[n]}(K_n) \ge \eps/2$.
 \end{proof}
 
@@ -58,11 +60,12 @@
     \[
     \alg = \bracs{\pi_J^{-1}(B)| B \in \bigotimes_{j \in J}\cb_{X_j}, J \subset I \text{ finite}}
     \]
+
     then $\alg$ is an algebra. For any $B \in \bigotimes_{j \in J}\cb_{X_j}$, define $\mu_0(\pi_J^{-1}(B)) = \mu_J(B)$, then $\mu_0: \alg \to [0, 1]$ is well-defined and finitely additive by the consistency of $\bracs{\mu_J}$.
 
     To show that $\mu_0$ is a premeasure, it is sufficient to show that for any $\{\pi_{J_n}^{-1}(B_n)\}_1^\infty$ with $\pi_{J_n}^{-1}(B_n) \downto \emptyset$, $\mu_0(\pi_{J_n}^{-1}(B_n)) \downto 0$. To this end, suppose for contradiction that $\limv{n}\mu_0(\pi_{J_n}^{-1}(B_n)) = \eps > 0$.
 
-    By inserting additional elements into the sequence and relabeling the indices, assume without loss of generality that $J_n = [n]$ for all $n \in \natp$. By \ref{lemma:kolmogorov-compact-sequence}, there exists $\seq{K_n}$ such that:
+    By inserting additional elements into the sequence and relabeling the indices, assume without loss of generality that $J_n = [n]$ for all $n \in \natp$. By \autoref{lemma:kolmogorov-compact-sequence}, there exists $\seq{K_n}$ such that:
     \begin{enumerate}
         \item $K_n \subset \prod_{j = 1}^n X_j$ is compact.
         \item $K_n \subset B_n$.
@@ -73,4 +76,5 @@
     \[
     x \in \bigcap_{n \in \natp}\pi_{[n]}^{-1}(K_n) \subset \bigcap_{n \in \natp}\pi_{[n]}^{-1}(B_n) \ne \emptyset
     \]
+
 \end{proof}

src/measure/measure/lebesgue-stieltjes.tex

@@ -20,6 +20,7 @@
         \[
         \alg = \bracs{\bigsqcup_{j = 1}^n I_j \bigg | \seqf{I_j} \subset \ci, n \in \natp}
         \]
+
         is a ring.
     \end{enumerate}
 \end{lemma}
@@ -30,7 +31,7 @@
         \item[(E)] $(a, b] \setminus (c, d] = (a, \min(b, c)] \sqcup (\max(a, d), b]$.
     \end{enumerate}
 
-    (2): By \ref{proposition:elementary-family-algebra}, $\alg$ is a ring.
+    (2): By \autoref{proposition:elementary-family-algebra}, $\alg$ is a ring.
 \end{proof}
 
 \begin{definition}[Lebesgue-Stieltjes Measure, {{\cite[Theorem 1.16]{Folland}}}]
@@ -41,6 +42,7 @@
         \[
         \mu((a, b]) = F(b) - F(a)
         \]
+
         \item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
     \end{enumerate}
     Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ satisfying (1), and $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
@@ -53,7 +55,8 @@
         -\mu((x, 0]) &x \le 0
     \end{cases}
     \]
-    then $F$ is a Stieltjes function by monotonicity and continuity from above (\ref{proposition:measure-properties}). For any $-\infty < a < b < \infty$,
+
+    then $F$ is a Stieltjes function by monotonicity and continuity from above (\autoref{proposition:measure-properties}). For any $-\infty < a < b < \infty$,
     \[
     \mu((a, b]) = \begin{cases}
         \mu((0, b]) - \mu((0, a]) &a, b \ge 0 \\
@@ -61,6 +64,7 @@
         \mu((a, 0]) - \mu((b, 0]) &a, b \le 0
     \end{cases}
     \]
+
     In all three cases, $\mu((a, b]) = F(b) - F(a)$.
 
     (U): If $G: \real \to \real$ satisfies (1), then $F(x) = G(x) - G(0)$. Hence $F - G$ is constant.
@@ -70,10 +74,12 @@
     \[
     \alg = \bracs{\bigsqcup_{j = 1}^n (a_j, b_j] \bigg | \seqf{(a_j, b_j]} \subset \ci, n \in \natp}
     \]
+
     Define
     \[
     \mu_0: \alg \to [0, \infty) \quad \bigsqcup_{j = 1}^n (a_j, b_j] \mapsto \sum_{j = 1}^n [F(b_j) - F(a_j)]
     \]
+
     Let $(a, b] \in \ci$ with $(a, b] = \bigsqcup_{j = 1}^n (a_j, b_j]$. Assume without loss of generality that $\seqf{a_j}$ is non-decreasing, then $b_j = a_{j+1}$ for each $1 \le j \le n - 1$, $b = b_n$, and $a = a_1$. Thus
     \begin{align*}
     F(b) - F(a) &= F(b_n) + \sum_{j = 1}^{n-1} [F(b_{j}) - F(b_j)] - F(a_1) \\
@@ -85,10 +91,12 @@
     \[
     \sum_{k = 1}^n \mu_0((a_k, b_k]) = \mu_0 \paren{\bigsqcup_{k = 1}^n (a_k, b_k]} \le \mu_0((a, b])
     \]
+
     Let $\eps > 0$. By right-continuity of $F$, there exists $\delta > 0$ such that $F(a + \delta) - F(a) < \eps/2$, and $\seq{\delta_n} \subset \real_{> 0}$ such that $F(b_n + \delta_n) - F(b_n) < \eps/2^{n+1}$ for all $n \in \natp$. The intervals $\seq{(a_n, b_n + \delta_n)}$ forms an open cover for $[a + \delta, b]$. By compactness, there exists $N \in \natp$ such that
     \[
     [a + \delta, b] \subset \bigcup_{n = 1}^N (a_n, b_n + \delta_n) \subset \bigcup_{n = 1}^N (a_n, b_n + \delta_n]
     \]
+
     Thus
     \begin{align*}
         \mu_0((a, b]) &\le F(a + \delta) - F(a) + \mu_0([a + \delta, b]) \\
@@ -100,5 +108,6 @@
     \[
     \mu_0((a, b]) = \sum_{n \in \natp}\mu_0((a_n, b_n])
     \]
-    Therefore $\mu_0$ is a premeasure on $\alg$. By Carathéodory's Extension Theorem (\ref{theorem:caratheodory-extension}), $\mu_0$ extends uniquely to a Borel measure on $\real$, which satisfies (1).
+
+    Therefore $\mu_0$ is a premeasure on $\alg$. By \hyperref[Carathéodory's Extension Theorem]{theorem:caratheodory-extension}, $\mu_0$ extends uniquely to a Borel measure on $\real$, which satisfies (1).
 \end{proof}

src/measure/measure/measure.tex

@@ -46,10 +46,12 @@
         \[
         \mu\paren{\liminf_{n \to \infty}E_n} \le \liminf_{n \to \infty}\mu(E_n)
         \]
+
         \item For any $\seq{E_n} \subset \cm$ with $\mu\paren{\bigcup_{n \in \nat}E_n} < \infty$,
         \[
         \mu\paren{\limsup_{n \to \infty}E_n} \ge \limsup_{n \to \infty}\mu(E_n)
         \]
+
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -60,15 +62,18 @@
     \mu\paren{\bigcup_{n \in \nat}E_n} = \mu\paren{\bigsqcup_{n \in \nat}F_n} = \sum_{n \in \natp}\mu(F_n) \le \sum_{n \in \natp}\mu(E_n)
     \]
 
+
     (3): Denote $E_0 = \emptyset$. For each $n \in \natp$, let $F_n = E_n \setminus E_{n - 1}$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_n = \bigcup_{n \in \natp}E_n$. In which case,
     \[
     \mu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigsqcup_{n \in \natp}F_n} = \limv{n}\sum_{k = 1}^n \mu(F_n) = \limv{n}\mu(E_n)
     \]
 
+
     (4): Since $\mu(E_1) < \infty$,
     \[
     \limv{n}\mu(E_n) = \mu(E_1) - \limv{n}\mu(E_1 \setminus E_n) = \mu(E_1) - \mu\paren{E_1 \setminus \bigcap_{n \in \natp}E_n} = \mu\paren{\bigcap_{n \in \natp}E_n}
     \]
+
     by (3).
 
     (5): Using (1) and (3),
@@ -99,6 +104,7 @@
     \[
     \alg(F) = \bracs{E \in \cm|\mu(E \cap F) = \nu(E \cap F)}
     \]
+
     then $\alg(F) \supset \mathcal{P}$ by (b), and
     \begin{enumerate}
         \item[(L2)] For any $E, E' \in \alg(F)$ with $E \subset E'$ and $F \in \mathcal{P}$,
@@ -110,12 +116,14 @@
         \[
         \mu\paren{\bigcup_{n \in \nat}E_n \cap F} = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu\paren{\bigcup_{n \in \nat}E_n \cap F}
         \]
-        by continuity from below (\ref{proposition:measure-properties}).
-    \end{enumerate}
-    so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin's $\pi$-$\lambda$ theorem (\ref{theorem:pi-lambda}), $\alg(F) = \cm$.
 
-    Let $\seq{E_n}$ as in assumption (c), then $\mu(E_n \cap F) = \mu(E_n \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (\ref{proposition:measure-properties}),
+        by continuity from below (\autoref{proposition:measure-properties}).
+    \end{enumerate}
+    so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin's $\pi$-$\lambda$ theorem (\autoref{theorem:pi-lambda}), $\alg(F) = \cm$.
+
+    Let $\seq{E_n}$ as in assumption (c), then $\mu(E_n \cap F) = \mu(E_n \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (\autoref{proposition:measure-properties}),
     \[
     \mu(F) = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu(F)
     \]
+
 \end{proof}

src/measure/measure/outer.tex

@@ -11,6 +11,7 @@
         \[
         \mu^*\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu^*(E_n)
         \]
+
     \end{enumerate}
 \end{definition}
 
@@ -20,6 +21,7 @@
     \[
     \mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\rho(E_n) \bigg | \seq{E_n} \subset \ce, \bigcup_{n \in \natp}E_n \supset E}
     \]
+
     then $\mu^*$ is an outer measure.
 \end{proposition}
 \begin{proof}
@@ -39,6 +41,7 @@
     \[
     \mu^*(F) = \mu^*(E \cap F) + \mu^*(E \setminus F)
     \]
+
 \end{definition}
 
 \begin{theorem}[Carathéodory, {{\cite[Theorem 1.11]{Folland}}}]
@@ -49,6 +52,7 @@
         \[
         \mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu^*(F \cap E_n)
         \]
+
         \item $\cm$ is a $\sigma$-algebra.
         \item $(X, \cm, \mu^*|_\cm)$ is a complete measure space.
     \end{enumerate}
@@ -58,11 +62,13 @@
     \[
     \mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \mu^*\paren{F \cap \bigsqcup_{n =1}^N E_n} = \sum_{n = 1}^N \mu^*(F \cap E_n)
     \]
+
     As this holds for all $N \in \nat$,
     \[
     \mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \sum_{n \in \natp}\mu^*(F \cap E_n)
     \]
 
+
     (2): Firstly, $\emptyset, X \in \cm$ since $\mu^*(\emptyset) = 0$. For any $A \in \cm$, $A^c \in \cm$ as well because the definition of $\mu^*$-measurability is symmetric.
 
     Let $A, B \in \cm$ and $F \subset X$, then
@@ -72,7 +78,7 @@
         &= \mu^*(F \cap B) + \mu^*(F \cap A \setminus B) + \mu^*(F \setminus (A \cup B)) \\
         &= \mu^*(F \cap (A \cup B)) + \mu^*(F \setminus (A \cup B))
     \end{align*}
-    so $\cm$ is an algebra. Since (1) implies that $\cm$ is closed under countable disjoint unions, $\cm$ is a $\sigma$-algebra by \ref{lemma:sigma-algebra-condition}.
+    so $\cm$ is an algebra. Since (1) implies that $\cm$ is closed under countable disjoint unions, $\cm$ is a $\sigma$-algebra by \autoref{lemma:sigma-algebra-condition}.
 
     (3): By (1), $\mu|_\cm$ is a measure. Let $N \in \cm$ with $\mu^*(N) = 0$, then for any $E \subset N$ and $F \subset X$,
     \begin{align*}
@@ -93,6 +99,7 @@
         \[
         \mu_0\paren{\bigsqcup_{n \in \natp}A_n} = \sum_{n \in\natp}\mu_0(A_n)
         \]
+
     \end{enumerate}
 \end{definition}
 
@@ -116,28 +123,33 @@
     \[
     \mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\mu_0(E_n) \bigg | \seq{E_n} \subset \alg, \bigcup_{n \in \natp}E_n \supset E}
     \]
-    then $\mu^*$ is an outer measure by \ref{proposition:outer-measure-inf}. By Carathéodory's theorem (\ref{theorem:caratheodory}), there exists a measure space $(X, \cm, \mu)$ such that $\mu = \mu^*|_{\cm}$.
+
+    then $\mu^*$ is an outer measure by \autoref{proposition:outer-measure-inf}. By \hyperref[Carathéodory's theorem]{theorem:caratheodory}, there exists a measure space $(X, \cm, \mu)$ such that $\mu = \mu^*|_{\cm}$.
 
     (1): Let $E \in \alg$ and $F \subset X$. If there exists $\seq{F_n} \subset \alg$ with $\bigcup_{n \in \natp}F_n \supset F$, then
     \[
     \sum_{n \in \natp}\mu_0(F_n) = \sum_{n \in \natp}\mu_0(F_n \cap E) + \sum_{n \in \natp}\mu_0(F_n \setminus E) \ge \mu^*(F \cap E) + \mu^*(F \setminus E)
     \]
+
     for all such $\seq{F_n}$, so $\mu^*(F) = \mu^*(F \cap E) + \mu^*(F \setminus E)$. Otherwise, $\mu^*(F) = \infty$ and the result holds directly.
 
     (2): Let $E \in \alg$ and $\seq{E_n} \subset \alg$ such that $\bigcup_{n \in \natp}E_n \supset E$, then
     \[
     \sum_{n \in \natp}\mu_0(E_n) = \sum_{n \in \natp}\mu_0(E_n \cap E) + \sum_{n \in \natp}\mu_0(E_n \setminus E) \ge \sum_{n \in \natp}\mu_0(E_n \cap E) = \mu_0(E)
     \]
+
     As this holds for all such $\seq{E_n}$, $\mu^*(E) = \mu_0(E)$.
 
-    (U): Let $\seq{E_n} \subset \alg$, then by continuity from below (\ref{proposition:measure-properties}),
+    (U): Let $\seq{E_n} \subset \alg$, then by continuity from below (\autoref{proposition:measure-properties}),
     \[
     \nu\paren{\bigcup_{n \in \natp}E_n} = \limv{n}\nu\paren{\bigcup_{k = 1}^nE_k} = \limv{n}\mu\paren{\bigcup_{k = 1}^n E_k} = \mu\paren{\bigcup_{n \in \natp}E_n}
     \]
+
     Thus if $E \in \cm \cap \cn$ with $\bigcup_{n \in \natp}E_n \supset E$, then
     \[
     \nu(E) \le \nu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu_0(E_n)
     \]
+
     As this holds for all such $\seq{E_n}$, $\nu(E) \le \mu^*(E) = \mu(E)$.
 
     Suppose that $\mu(E) < \infty$, then there exists $\seq{E_n} \subset \alg$ such that $\bigcup_{n \in \natp}E_n \supset E$ and $\sum_{n \in \natp}\mu_0(E_n) < \infty$. In which case,
@@ -148,9 +160,10 @@
     \end{align*}
     so $\nu(E) = \mu(E)$ whenever $\mu(E) < \infty$.
 
-    Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^\infty \subset \alg$ such that $E_n \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}} < \infty$. Let $F_n = \bigcup_{j = 1}^n \bigcup_{k \in \natp}E_{j, k}$, then $\seq{F_n} \subset \sigma(\alg) \subset \cm \cap \cn$ with $F_n \upto X$ and $\mu(F_n) < \infty$ for all $n \in \natp$. For any $E \in \cm \cap \cn$, by continuity from below (\ref{proposition:measure-properties}),
+    Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^\infty \subset \alg$ such that $E_n \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}} < \infty$. Let $F_n = \bigcup_{j = 1}^n \bigcup_{k \in \natp}E_{j, k}$, then $\seq{F_n} \subset \sigma(\alg) \subset \cm \cap \cn$ with $F_n \upto X$ and $\mu(F_n) < \infty$ for all $n \in \natp$. For any $E \in \cm \cap \cn$, by continuity from below (\autoref{proposition:measure-properties}),
     \[
     \nu(E) = \limv{n}\nu(E \cap F_n) = \limv{n}\mu(E \cap F_n) = \mu(E)
     \]
+
     so $\nu|_{\cm \cap \cn} = \mu|_{\cm \cap \cn}$.
 \end{proof}

src/measure/measure/radon.tex

@@ -19,30 +19,36 @@
         \[
         \mu(U) = \sup\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), \supp{f} \subset U}
         \]
+
         \item For any $K \subset X$ compact,
         \[
         \mu(K) = \inf\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), f \ge \one_K}
         \]
+
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $U \subset X$ be open. By Urysohn's lemma (\ref{lemma:lch-urysohn}), for any $K \subset U$ compact, there exists $f_K \in C_c(X; [0, 1])$ such that $f_K|_K = 1$ and $\supp{f_K} \subset U$. In which case,
+    (1): Let $U \subset X$ be open. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, for any $K \subset U$ compact, there exists $f_K \in C_c(X; [0, 1])$ such that $f_K|_K = 1$ and $\supp{f_K} \subset U$. In which case,
     \[
     \mu(K) \le \int f_K d\mu \le \mu(U)
     \]
+
     By (R3'),
     \[
     \mu(U) = \sup_{K \subset U \text{ compact}}\mu(K) \le \sup_{K \subset U \text{ compact}}\int f_K d\mu \le \mu(U)
     \]
 
-    (2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f_U \in C_c(X; [0, 1])$ such that $f_U|_K = 1$ and $\supp{f_U} \subset U$. In which case,
+
+    (2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f_U \in C_c(X; [0, 1])$ such that $f_U|_K = 1$ and $\supp{f_U} \subset U$. In which case,
     \[
     \mu(K) \le \int f_U d\mu \le \mu(U)
     \]
+
     By (R2),
     \[
     \mu(K) = \inf_{U \in \cn^o(K)}\mu(U) \ge \inf_{U \in \cn^o(K)}\int f_U d\mu \ge \mu(K)
     \]
+
 \end{proof}
 
 \begin{proposition}[{{\cite[Proposition 7.5]{Folland}}}]
@@ -64,16 +70,19 @@
     \[
     \mu(K \setminus V) \ge \mu(K) - \mu(V) \ge \mu(U) - \eps/2 - \eps/2 \ge \mu(E) - \eps
     \]
+
     so $\mu$ is inner regular on $E$.
 
     Now suppose that $E$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cb_X$ with $\mu(E_n) < \infty$ for each $n \in \natp$ and $E = \bigsqcup_{n \in \natp}E_n = E$. Let $N \in \natp$ and $\seqf{K_n} \subset 2^X$ compact with $K_n \subset E_n$ for each $n \in \natp$, then $\bigcup_{n = 1}^N K_n \subset E$ is compact. Hence
     \[
     \sup_{\substack{K \subset E \\ \text{compact}}}\mu(K) \ge \sum_{n = 1}^N\sup_{\substack{K_n \subset E_n \\ \text{compact}}}\mu(K_n) = \sum_{n = 1}^N \mu(E_n)
     \]
+
     As the above holds for all $N \in \natp$,
     \[
     \sup_{\substack{K \subset E \\ \text{compact}}}\mu(K) \ge \sum_{n \in \natp}\mu(E_n) = \mu(E)
     \]
+
 \end{proof}
 
 \begin{proposition}[{{\cite[Proposition 7.7]{Folland}}}]
@@ -89,12 +98,14 @@
     \[
     \mu\paren{\bigcup_{n \in \natp}U_n \setminus E} \le \sum_{n \in \natp}\mu(U_n \setminus E_n) < \eps
     \]
+
     so $U = \bigcup_{n \in \natp}U_n$ is the desired open set.
 
     Applying the above result to $E^c$, there exists $V \in \cn^o(E^c)$ such that $\mu(V \setminus E^c) < \eps$. Let $F = V^c$, then $F \subset E$ is closed and
     \[
     \mu(E \setminus F) = \mu(E \cap F^c) = \mu(E \cap V) = \mu(V \setminus E^c) < \eps
     \]
+
 \end{proof}
 
 
@@ -108,29 +119,33 @@
     then $\mu$ is a regular measure on $X$.
 \end{proposition}
 \begin{proof}
-    By assumption (b), $f \mapsto \int f d\mu$ is a positive linear functional on $C_c(X; \real)$. By the Riesz Representation Theorem (\ref{theorem:riesz-radon}), there exists a Radon measure $\nu: \cb_X \to [0, \infty]$ such that for any $f \in C_c(X; \real)$, $\int f d\mu = \int f d\mu$.
+    By assumption (b), $f \mapsto \int f d\mu$ is a positive linear functional on $C_c(X; \real)$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, there exists a Radon measure $\nu: \cb_X \to [0, \infty]$ such that for any $f \in C_c(X; \real)$, $\int f d\mu = \int f d\mu$.
 
-    Let $U \subset X$ be open, then by \ref{proposition:radon-measure-cc},
+    Let $U \subset X$ be open, then by \autoref{proposition:radon-measure-cc},
     \[
     \nu(U) = \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}} \int f d\nu = \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}} \int f d\mu \le \mu(U)
     \]
-    By assumption (a), there exists $\seq{K_n} \subset 2^X$ compact such that $K_n \upto U$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $\seq{f_n} \subset C_c(X; [0, 1])$ such that $\one_{K_n} \le f_n \le \one_U$ for all $n \in \natp$, and $f_n \upto f$ pointwise. Using the Monotone Convergence Theorem (\ref{theorem:mct}),
+
+    By assumption (a), there exists $\seq{K_n} \subset 2^X$ compact such that $K_n \upto U$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seq{f_n} \subset C_c(X; [0, 1])$ such that $\one_{K_n} \le f_n \le \one_U$ for all $n \in \natp$, and $f_n \upto f$ pointwise. Using the \hyperref[Monotone Convergence Theorem]{theorem:mct},
     \[
     \mu(U) = \limv{n} \int f_n d\mu \le \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}}\int f d\mu = \nu(U)
     \]
+
     Therefore $\mu(U) = \nu(U)$.
 
-    Now let $E \in \cb_X$ be arbitrary and $\eps > 0$. By \ref{proposition:radon-measurable-description}, there exists $U \in \cn^o(E)$ and $V \subset E$ closed such that $\nu(U \setminus E), \nu(E \setminus V) < \eps$. In which case, since $U \setminus V$ is open,
+    Now let $E \in \cb_X$ be arbitrary and $\eps > 0$. By \autoref{proposition:radon-measurable-description}, there exists $U \in \cn^o(E)$ and $V \subset E$ closed such that $\nu(U \setminus E), \nu(E \setminus V) < \eps$. In which case, since $U \setminus V$ is open,
     \[
     \mu(U \setminus E) \le \mu(U \setminus V) = \nu(U \setminus V) < \eps
     \]
+
     so $\mu(U) = \mu(U \setminus E) + \mu(E) \le \mu(E) + \eps$. Therefore
     \[
     \mu(E) = \inf_{U \in \cn^o(E)}\mu(U) = \inf_{U \in \cn^o(E)}\nu(U) = \nu(E)
     \]
+
     and $\mu = \nu$.
 
-    Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \ref{proposition:radon-regular-sigma-finite}.
+    Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \autoref{proposition:radon-regular-sigma-finite}.
 \end{proof}
 
 \begin{proposition}
@@ -138,10 +153,11 @@
     Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
 \end{proposition}
 \begin{proof}
-    By \ref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
+    By \autoref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
 
-    Let $A \in \cb_X$ and $\eps > 0$. By \ref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
+    Let $A \in \cb_X$ and $\eps > 0$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
     \[
     \norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
     \]
+
 \end{proof}

src/measure/measure/regular.tex

@@ -7,6 +7,7 @@
     \[
     \mu(E) = \sup\bracs{\mu(K)| K \subset E, K \text{ compact}}
     \]
+
 \end{definition}
 
 \begin{definition}[Outer Regular]
@@ -15,6 +16,7 @@
     \[
     \mu(E) = \sup\bracs{\mu(U)| U \in \cn^o(A)}
     \]
+
 \end{definition}
 
 \begin{definition}[Regular]

src/measure/measure/riesz.tex

@@ -17,10 +17,11 @@
 \begin{proof}
     (1): $\dpb{g - f, I}{C_c(X; \real)} \ge 0$.
 
-    (2): By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $g \in C_c(X; [0, 1])$ such that $g|_K = 1$. In which case,
+    (2): By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $g \in C_c(X; [0, 1])$ such that $g|_K = 1$. In which case,
     \[
     -\norm{f}_u\dpn{g, I}{C_c(X; \real)} \le \dpn{f, I}{C_c(X; \real)} \le \norm{f}_u\dpn{g, I}{C_c(X; \real)}
     \]
+
     so $C_K = \dpn{g, I}{C_c(X; \real)}$ is a desired constant.
 \end{proof}
 
@@ -41,29 +42,34 @@
     \[
     \mu_0: \topo \to [0, \infty] \quad U \mapsto \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}
     \]
+
     and
     \[
     \mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracsn{\mu_0(U)|U \in \cn^o(E)}
     \]
+
     then
     \begin{enumerate}
         \item[(OM1)] Since $\emptyset \in \topo$ and $\mu_0(\emptyset) = 0$, $\mu^*(\emptyset) = 0$.
         \item[(OM2)] Let $E, F \subset X$ with $E \subset F$, then $\cn^o(E) \subset \cn^o(F)$, so $\mu^*(E) \le \mu^*(F)$.
         \item[(OM3)] Let $\seq{E_n} \subset 2^X$, $E = \bigcup_{n \in \natp}E_n$, $U \in \cn^o(E)$, and $\seq{U_n} \subset \topo$ such that $U_n \in \cn^o(E_n)$ for each $n \in \natp$.
 
-        Let $f \prec U$, then by compactness of $\supp{f}$, there exists $N \in \natp$ such that $\supp{f} \subset \bigcup_{n = 1}^N U_n$. By \ref{proposition:lch-partition-of-unity}, there exists a partition of unity $\seqf{\phi_j} \subset C_c(X; [0, 1])$ on $K$ subordinate to $\seqf[N]{U_n}$. In which case,
+        Let $f \prec U$, then by compactness of $\supp{f}$, there exists $N \in \natp$ such that $\supp{f} \subset \bigcup_{n = 1}^N U_n$. By \autoref{proposition:lch-partition-of-unity}, there exists a partition of unity $\seqf{\phi_j} \subset C_c(X; [0, 1])$ on $K$ subordinate to $\seqf[N]{U_n}$. In which case,
         \[
         \dpb{f, I}{C_c(X; \real)} = \sum_{n = 1}^N \dpb{\phi_n f, I}{C_c(X; \real)} \le \sum_{n = 1}^N \mu_0(U_n) \le \sum_{n \in \natp}\mu_0(U_n)
         \]
+
         Since this holds for all $f \prec U$,
         \[
         \mu^*(E) \le \mu_0(U) \le \sum_{n \in \natp}\mu_0(U_n)
         \]
 
+
         Let $\eps > 0$, then there exists $\seq{U_n} \subset \topo$ such that $U_n \supset E_n$ and $\mu_0(U_n) \le \mu^*(E_n) + \eps/2^n$ for each $n \in \natp$. In which case,
         \[
         \mu^*(E) \le \sum_{n \in \natp}\mu_0(U_n) \le \eps + \sum_{n \in \natp}\mu^*(E_n)
         \]
+
         As this holds for all $\eps > 0$, $\mu^*(E) \le \sum_{n \in \natp}\mu^*(E_n)$.
     \end{enumerate}
     Therefore $\mu^*: 2^E \to [0, \infty]$ is an outer measure.
@@ -72,50 +78,61 @@
     \[
     \dpb{f, I}{C_c(X; \real)} + \dpb{g, I}{C_c(X; \real)} \le \mu_0(E)
     \]
+
     Since this holds for all $g \prec E \setminus \supp{f}$,
     \[
     \dpb{f, I}{C_c(X; \real)} + \mu_0(E \setminus \supp{f}) \le \mu_0(E)
     \]
+
     As $E \setminus \supp{f} \supset E \setminus U$,
     \[
     \dpb{f, I}{C_c(X; \real)} + \mu_0(E \setminus U) \le \mu_0(E)
     \]
+
     Finally, the above holds for all $f \prec E \cap U$,
     \[
     \mu_0(E \cap U) + \mu_0(E \setminus U) \le \mu_0(E)
     \]
+
     Now suppose that $E$ is arbitrary. Let $V \in \cn^o(E)$, then
     \[
     \mu^*(E \cap U) + \mu^*(E \setminus U) \le \mu_0(V \cap U) + \mu_0(V \setminus U) \le \mu_0(V)
     \]
+
     As this holds for all such $V$,
     \[
     \mu^*(E) = \mu^*(E \cap U) + \mu^*(E \setminus U)
     \]
 
-    By Carathéodory's Extension Theorem (\ref{theorem:caratheodory-extension}), there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that for all $U \in \topo$, $\mu(U) = \mu_0(U)$.
 
-    (2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f \prec U$ with $f \ge \one_K$. In which case,
+    By \hyperref[Carathéodory's Extension Theorem]{theorem:caratheodory-extension}, there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that for all $U \in \topo$, $\mu(U) = \mu_0(U)$.
+
+    (2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \prec U$ with $f \ge \one_K$. In which case,
     \[
     \inf_{\substack{f \in C_c(X; \real) \\ f \ge \one_K}}\dpb{f, I}{C(X; \real)} \le \inf_{U \in \cn^o(K)}\mu(U) = \mu(K)
     \]
+
     On the other hand, let $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. For any $r \in (0, 1)$, let $g \prec \bracs{f > r}$, then $r^{-1}f \ge g$ and
     \[
     \dpb{g, I}{C_c(X; \real)} \le r^{-1}\dpb{f, I}{C_c(X; \real)}
     \]
+
     As this holds for all $g \prec \bracs{f > r}$,
     \[
     \mu(K) \le \mu(\bracs{f > r}) \le r^{-1}\dpb{f, I}{C_c(X; \real)}
     \]
+
     Since the above holds for all $r \in (0, 1)$,
     \[
     \mu(K) \le \inf_{\substack{f \in C_c(X; \real) \\ f \ge \one_K}}\dpb{f, I}{C_c(X; \real)}
     \]
 
+
     (3): Let $f \in C_c(X; \real)$. Using linearity, assume without loss of generality that $f \in C_c(X; [0, 1])$. Let $N \in \natp$. For each $1 \le n \le N$, let
     \[
     f_n = \min\paren{\max\paren{f - \frac{n - 1}{N}, 0}, \frac{1}{N}}
     \]
+
     For any $x \in X$, there exists $1 \le n \le N$ such that $f(x) \in [(n-1)/N, n/N]$. In which case,
     \begin{itemize}
     \item For each $1 \le j < n$, $f_j(x) = 1/N$.
@@ -128,38 +145,45 @@
     \[
     \frac{1}{N}\mu\paren{K_n} \le \int f_n d\mu \le \frac{1}{N}\mu\paren{K_{n-1}}
     \]
+
     Since $\supp{f_n} \subset \bracs{f \ge (n-1)/N}$, for any $U \supset \bracs{f_n \ge (n - 1)/N}$, $f_n \prec U$, so
     \[
     \dpb{f_n, I}{C_c(X; \real)} \le \frac{1}{N}\mu\paren{K_{n-1}}
     \]
+
     On the other hand, $f_n \ge \frac{1}{N}\one_{K_n}$,
     \[
     \mu(K_n) \le \dpb{f_n, I}{C_c(X; \real)}
     \]
+
     so
     \[
     \frac{1}{N}\sum_{n = 1}^N \mu(K_n) \le \dpb{f, I}{C_c(X; \real)}, \int f d\mu \le \frac{1}{N}\sum_{n = 1}^N \mu(K_{n-1})
     \]
 
+
     Therefore
     \[
     \abs{\int f d\mu - \dpb{f, I}{C_c(X; \real)}} \le \frac{\mu(K_{0}) - \mu(K_N)}{N} \le \frac{\mu(\supp{f})}{N}
     \]
+
     As this holds for all $N \in \natp$, $\int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
 
     (4):
     \begin{enumerate}
-        \item[(R1)] For any $K \subset X$ compact, by Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. In which case, $\mu(K) \le \int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
+        \item[(R1)] For any $K \subset X$ compact, by \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. In which case, $\mu(K) \le \int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
         \item[(R2)] By definition of $\mu^*$, $\mu$ is outer regular.
         \item[(R3')] For any $U \in \topo$,
         \[
         \mu(U) = \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)} \le \sup_{f \prec U}\mu(\supp{f}) \le \sup_{\substack{K \subset U \\ \text{compact}}}\mu(K) \le \mu(U)
         \]
+
     \end{enumerate}
 
-    (U): By \ref{proposition:radon-measure-cc}, $\nu$ also satisfies (2). Thus for any $E \in \cb_X$, by (R2) and (R3'),
+    (U): By \autoref{proposition:radon-measure-cc}, $\nu$ also satisfies (2). Thus for any $E \in \cb_X$, by (R2) and (R3'),
     \[
     \nu(E) = \inf_{U \in \cn^o(E)}\sup_{\substack{K \subset U \\ \text{compact}}}\inf_{\substack{f \in C_c(X; [0, 1]) \\ f \ge \one_K}}\dpb{f, I}{C_c(X; \real)}
     \]
+
     so $\nu$ is uniquely determined by $I$.
 \end{proof}

src/measure/measure/semifinite.tex

@@ -10,6 +10,7 @@
         \[
         \mu(E) = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}
         \]
+
     \end{enumerate}
     If the above holds, then $\mu$ is a \textbf{semifinite measure}.
 \end{definition}
@@ -18,6 +19,7 @@
     \[
     M = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty} < \infty
     \]
+
     Let $\seq{F_n} \subset \cm$ such that $F \subset E$, $\mu(F) < \infty$, and $\mu(F_n) \upto M$. Let $F = \bigcup_{n \in \nat}F_n$, then $\mu(F) = M$, and $\mu(E \setminus F) = \infty$. By (1), there exists $G \in \cm$ with $G \subset E \setminus F$ and $0 < \mu(G) < \infty$. Thus $F \cup G \subset E$ with $M < \mu(F) + \mu(G) = \mu(F \cup G) < \infty$. This contradicts the fact that $M$ is the supremum.
 \end{proof}
 
@@ -27,6 +29,7 @@
     \[
     \mu_0: \cm \to [0, \infty] \quad E \mapsto \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}
     \]
+
     then $\mu_0$ is a semifinite measure, and the \textbf{semifinite part} of $\mu$.
 \end{definition}
 \begin{proof}
@@ -36,18 +39,22 @@
     \[
     \mu(F) = \sum_{n \in \natp}\mu(F \cap E_n) \le \sum_{n \in \natp}\mu_0(E_n)
     \]
+
     Thus $\mu_0\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu_0(E_n)$.
 
     On the other hand, let $n \in \natp$ and $\seqf{F_k} \subset \cm$ such that $F_k \subset E_k$ for each $1 \le k \le n$ and $\mu(F_k) < \infty$, then $F = \bigsqcup_{k = 1}^n F_k \subset \bigsqcup_{k \in \natp}E_k$. Thus
     \[
     \sum_{k = 1}^n \mu(F_k) = \mu(F) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
     \]
+
     As this holds for all such $\seqf{F_k}$,
     \[
     \sum_{k = 1}^n \mu_0(E_k) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
     \]
+
     Since this holds for all $n \in \natp$,
     \[
     \sum_{n \in \natp}\mu_0(E_n) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
     \]
+
 \end{proof}

src/measure/sets/algebra.tex

@@ -85,6 +85,7 @@
     \[
     \cb_{\ol{\real}} = \bracsn{E \subset \ol \real| E \cap \real \in \cb_\real}
     \]
+
     is the \textbf{Borel $\sigma$-algebra} on $\ol{\real}$.
 \end{definition}
 
@@ -126,6 +127,7 @@
     \[
     U = \bigcup_{q \in U \cap \rational}(q - r_q, q + r_q)
     \]
+
     so (8) generates all open sets in $\real$. Conversely, every element of (8) is open, so the $\sigma$-algebra generated by (8) is the Borel $\sigma$-algebra on $\real$.
 \end{proof}
 
@@ -152,7 +154,8 @@
     \[
     \bracs{\infty} = \bigcap_{n \in \nat}[n, \infty] \quad \bracs{-\infty} = \bigcap_{n \in \nat}[-\infty, n]
     \]
-    are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_\real$ by \ref{proposition:borel-sigma-real-generators}.
+
+    are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_\real$ by \autoref{proposition:borel-sigma-real-generators}.
 
     In addition, for any $E \in \cb_{\ol{\real}}$, $E = (E \cap \real) \cup (E \setminus \real)$, where $E \cap \real \in \cb_\real$. Since $\bracs{\infty}, \bracs{-\infty} \in \cm$ and $\cb_\real \subset \cm$, $E \in \cm$ and $\cm = \cb_{\ol \real}$.
 \end{proof}

src/measure/sets/elementary.tex

@@ -21,6 +21,7 @@
     \[
     \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \text{ pairwise disjoint}}
     \]
+
     then is a ring. If $X \in \ce$, then $\ce$ is an algebra.
 \end{proposition}
 \begin{proof}
@@ -28,6 +29,7 @@
     \[
     A \sqcup B = \bigsqcup_{j = 1}^n A_j \sqcup  \bigsqcup_{j = 1}^mB_j \in \alg
     \]
+
     so $\alg$ is closed under disjoint unions.
 
     (A1): By (P1) $\emptyset \in \ce$. By (E1), there exists $\seqf{E_j} \subset \ce$ such that $X = \emptyset^c = \bigsqcup_{j = 1}^nE_j \in \alg$.
@@ -36,14 +38,17 @@
     \[
     B \setminus A = \bigcap_{i = 1}^n B \setminus A_i = \bigcap_{i = 1}^n \bigsqcup_{j = 1}^m E_{i, j} = \bigsqcup_{\alpha \in [1, m]^n} \underbrace{\bigcap_{i = 1}^n E_{i, \alpha_i}}_{\in \ce} \in \ce
     \]
+
     Thus if $B \in \alg$ with $B = \bigsqcup_{j = 1}^n B_j$, then
     \[
     B \setminus A = \bigsqcup_{j = 1}^n B_j \setminus A \in \alg
     \]
 
+
     (A3): Let $A = \bigsqcup_{j = 1}^n A_j, B = \bigsqcup_{j = 1}^m B_j \in \alg$, then
     \[
     A \cap B = \braks{\bigsqcup_{j = 1}^n A_j} \cap \braks{\bigsqcup_{j = 1}^m B_j} = \bigsqcup_{i = 1}^n \bigsqcup_{j = 1}^m \underbrace{A_i \cap B_j}_{\in \ce} \in \alg
     \]
+
     so $\alg$ is closed under intersections. Thus using (A2), $A \cup B = A \setminus B \sqcup B \setminus A \sqcup A \cap B$.
 \end{proof}

src/measure/sets/index.tex

@@ -1,6 +1,6 @@
 \chapter{Set Systems}
 \label{chap:set-system}
 
-\input{./src/measure/sets/algebra.tex}
+\input{./algebra.tex}
-\input{./src/measure/sets/lambda.tex}
+\input{./lambda.tex}
-\input{./src/measure/sets/elementary.tex}
+\input{./elementary.tex}

src/measure/sets/lambda.tex

@@ -34,7 +34,7 @@
     \end{enumerate}
 \end{lemma}
 \begin{proof}
-    $(2) \Rightarrow (1)$: Let $A, B \in \alg$, then $A \cup B = (A^c \cap B^c)^c \in \alg$. Thus $\alg$ is an algebra. By \ref{lemma:sigma-algebra-condition}, $\alg$ is a $\sigma$-algebra.
+    $(2) \Rightarrow (1)$: Let $A, B \in \alg$, then $A \cup B = (A^c \cap B^c)^c \in \alg$. Thus $\alg$ is an algebra. By \autoref{lemma:sigma-algebra-condition}, $\alg$ is a $\sigma$-algebra.
 \end{proof}
 
 
@@ -47,18 +47,21 @@
     \[
     \cm(\ce) = \bracs{E \in \lambda(\mathcal{P})| E \cap F \in \lambda(\mathcal{P}) \forall F \in \ce}
     \]
+
     then
     \begin{enumerate}
         \item[(L2)] Let $E, F \in \cm$ with $E \subset F$, then for any $G \in \ce$,
         \[
         (F \setminus E) \cap G = (F \cap G) \setminus (E \cap G) \in \lambda(\mathcal{P})
         \]
+
         \item[(L3)] Let $\seq{E_n} \in \cm$ with $E_n \subset E_{n+1}$ for all $n \in \natp$, and $F \in \ce$, then
         \[
         \braks{\bigcup_{n \in \natp}E_n} \cap F = \bigcup_{n \in \natp}E_n \cap F \in \lambda(\mathcal{P})
         \]
+
     \end{enumerate}
     so $\cm(\ce)$ is a $\lambda$-system.
 
-    Since $\mathcal{P}$ is a $\pi$-system, $\cm(\mathcal{P}) \supset \mathcal{P}$, so $\cm(\mathcal{P}) = \lambda(\mathcal{P})$. Thus for any $E \in \lambda(\mathcal{P})$ and $F \in \lambda(\mathcal{P})$, $E \cap F \in \lambda(\mathcal{P})$. Therefore $\cm(\lambda(\mathcal{P})) \supset \mathcal{P}$, $\cm(\lambda(\mathcal{P})) = \lambda(\mathcal{P})$, and $\lambda(\mathcal{P})$ satisfies (P2). By \ref{lemma:pi-lambda}, $\lambda(\mathcal{P})$ is a $\sigma$-algebra.
+    Since $\mathcal{P}$ is a $\pi$-system, $\cm(\mathcal{P}) \supset \mathcal{P}$, so $\cm(\mathcal{P}) = \lambda(\mathcal{P})$. Thus for any $E \in \lambda(\mathcal{P})$ and $F \in \lambda(\mathcal{P})$, $E \cap F \in \lambda(\mathcal{P})$. Therefore $\cm(\lambda(\mathcal{P})) \supset \mathcal{P}$, $\cm(\lambda(\mathcal{P})) = \lambda(\mathcal{P})$, and $\lambda(\mathcal{P})$ satisfies (P2). By \autoref{lemma:pi-lambda}, $\lambda(\mathcal{P})$ is a $\sigma$-algebra.
 \end{proof}

src/process/index.tex

@@ -0,0 +1,4 @@
+\part{Stochastic Processes}
+\label{part:stochastic-processes}
+
+\input{./markov/index.tex}

src/process/markov/brownian.tex

@@ -0,0 +1,38 @@
+\section{Brownian Motion}
+\label{section:brownian-markov}
+
+\begin{proposition}[{{\cite[Proposition 3.4]{Baudoin}}}]
+\label{proposition:brownian-markov}
+    Let $(\Omega, \cf, \bp)$ be a probability space, $\bracs{B_t|t \ge 0}$ be a $\real$-valued stochastic process with $B_0 = 0$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\bracs{B_t|t \ge 0}$ is a standard Brownian motion.
+        \item $\bracs{B_t|t \ge 0}$ is a Markov process with semigroup
+        \[
+        \bp_0 = \text{Id} \quad
+        (\bp_t f)(x) = \frac{1}{\sqrt{2\pi t}}\int f(y) \exp\paren{-\frac{(x - y)^2}{2t}}dy
+        \]
+
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Let $\bracs{\cf_t|t \ge 0}$ be the natural filtration of $\bracs{B_t|T \ge 0}$, then for any $s, t \ge 0$ and $\xi \in \real$,
+
+    By the Markov property,
+    \begin{align*}
+        \ev\braks{\exp\paren{i \xi B_{t + s}}|\cf_s} &= \frac{1}{\sqrt{2\pi t}}\int \exp\paren{i \xi y}\exp\paren{-\frac{(B_s - y)^2}{2t}}dy \\
+        &= \frac{1}{\sqrt{2\pi t}}\int \exp\paren{i \xi (B_s + y)}\exp\paren{-\frac{y^2}{2t}}dy \\
+        \ev\braks{\exp\paren{i \xi (B_{t + s} - B_s)}|\cf_s} &= \frac{1}{\sqrt{2\pi t}}\int \exp\paren{i \xi y}\exp\paren{-\frac{y^2}{2t}}dy = e^{-t\xi^2/2}
+    \end{align*}
+    So $\bracs{B_t|t \ge 0}$ admits independent and homogeneous increments, and is the standard Brownian motion.
+
+    Now suppose that $\bracs{B_t| t \ge 0}$ is the standard Brownian motion, then for any $s, t \ge 0$ and bounded Borel function $f: \real \to \real$,
+    \[
+    \ev[f(B_{t+s})|B_s = x] = \frac{1}{\sqrt{2\pi t}}\int f(x + y) \exp\paren{-\frac{y^2}{2t}} dy
+    \]
+
+    so
+    \[
+    \ev[f(B_{t + s})|\cf_s] = \int f(B_s + y)\exp\paren{-\frac{y^2}{2t}}dy
+    \]
+
+\end{proof}

src/process/markov/definition.tex

@@ -0,0 +1,77 @@
+\section{Transition Functions}
+\label{section:markov-transition}
+
+\begin{definition}[Transition Function]
+\label{definition:transition-function}
+    Let $(X, \cm)$ be a measurable space. A \textbf{transition function} $\bracs{P_t| t \ge 0}$ on $X$ is a collection of maps
+    \[
+    P_t: X \times \cm \to [0, 1]
+    \]
+
+    such that:
+    \begin{enumerate}
+        \item For each $t \ge 0$ and $x \in X$, $P_t(x, \cdot)$ is a probability measure on $X$.
+        \item For each $t \ge 0$ and $A \in \cm$, $x \mapsto P_t(x, A)$ is $(\cm, \cb_\real)$-measurable.
+        \item For any $s, t \ge 0$, $x \in X$, and $A \in \cm$,
+        \[
+        P_{t + s}(x, A) = \int P_t(y, A)P_s(x, dy)
+        \]
+
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Semigroup of Transition Function]
+\label{definition:transition-function-semigroup}
+    Let $(X, \cm)$ be a measurable space, $\bracs{P_t|t \ge 0}$ be a transition function on $X$. For each $f: X \to \complex$ bounded and $(\cm, \cb_\complex)$-measurable, let
+    \[
+    (\bp_t f)(x) = \int f(y)P_t(x, dy)
+    \]
+
+    then $\bracs{\bp_t|t \ge 0}$ is a semigroup such that:
+    \begin{enumerate}
+        \item For each $t \ge 0$, $\bp_t \one = \one$.
+        \item For each $t \ge 0$, $\bp_t$ is a positive linear functional.
+    \end{enumerate}
+    known as the \textbf{semigroup} of $\bracs{P_t|t \ge 0}$.
+\end{definition}
+\begin{proof}
+    Let $s, t \ge 0$, then for any $A \in \cm$,
+    \[
+    (\bp_{s+t}\one_A)(x) = \int \one_A(y)P_{s+t}(x, dy) = \int P_t(y, A)P_s(x, dy) = \int \int \one_A(z)P_t(y, dz)P_s(x, dy)
+    \]
+
+    Let $f: X \to \complex$ bounded and $(\cm, \cb_\complex)$-measurable. By \autoref{proposition:measurable-simple-separable-norm}, there exists $\seq{f_n} \subset \Sigma(X; \complex)$ such that $\abs{f_n} \le \abs{f}$ for all $n \in\nat$, and $f_n \to f$ pointwise. By the \hyperref[Dominated Convergence Theorem]{theorem:dct},
+    \begin{align*}
+    (\bp_{s+t}f)(x) &= \limv{n}\int f_n(y)P_{s+t}(x, dy) = \limv{n}\iint f_n(z)P_t(y, dz)P_s(x, dy) \\
+    &= \int \int f_n(z)P_t(y, dz)P_s(x, dy) = (\bp_s\bp_t) f(x)
+    \end{align*}
+\end{proof}
+
+\begin{definition}[Markov Process]
+\label{definition:markov-process}
+    Let $(\Omega, \cf, \bp)$ be a probability space, $(Y, \cm)$ be a measurable space, $\bracs{X_t|t \ge 0}$ be a stochastic process, and $\bracs{\cf_t|t \ge 0}$ be its natural filtration.
+
+    The process $\bracs{X_t|t \ge 0}$ is \textbf{Markov} if there exists a transition function $\bracs{P_t|t \ge 0}$ such that for every $f: Y \to \complex$ bounded and $(\cm, \cb_\complex)$-measurable,
+    \[
+    \ev\braks{f(X_{t+s})|\cf_s} = (\bp_tf)(X_s)
+    \]
+
+\end{definition}
+
+
+\begin{theorem}[{{\cite[Theorem 3.14]{Baudoin}}}]
+\label{theorem:markov-existence}
+    Let $(Y, \cm)$ be a measurable space, $\nu: \cm \to [0, 1]$ be a probability measure on $Y$, and $\bracs{P_t|t \ge 0}$ be a transition function on $Y$, then there exists a probability space $(\Omega, \cf, \bp)$ and a stochastic process $\bracs{X_t|t \ge 0}$ such that:
+    \begin{enumerate}
+        \item The distribution of $X_0$ is $\nu$.
+        \item $\bracs{X_t|t \ge 0}$ is a Markov process associated with $\bracs{P_t|t \ge 0}$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    For each $N \in \natp$, $A \in \cm$ and $B \in \bigotimes_{n = 1}^N\cm$, and $0 = t_0 < t_1 < \cdots < t_N$, let
+    \[
+    \mu_{t_0, t_1, \cdots, t_N} = \int_A\int_B P_{t_1}(z, dx_1)P_{t_2 - t_1}(x_1, dx_2) \cdots P_{t_N - t_{N-1}}(x_{N_1}, dx_N) \nu(dz)
+    \]
+
+    then $\mu_{t_0, t_1, \cdots, t_N}$ is a probability measure on $\cm \times $
+\end{proof}

src/process/markov/index.tex

@@ -0,0 +1,6 @@
+\chapter{Markov Processes}
+\label{chap:markov}
+
+
+\input{./definition.tex}
+\input{./brownian.tex}

src/topology/functions/index.tex

@@ -1,5 +1,5 @@
 \chapter{Function Spaces}
 \label{chap:function-spaces}
 
-\input{./src/topology/functions/set-systems.tex}
+\input{./set-systems.tex}
-\input{./src/topology/functions/uniform.tex}
+\input{./uniform.tex}

src/topology/functions/set-systems.tex

@@ -7,10 +7,12 @@
     \[
     M(S, U) = \bracs{f \in X^T| f(S) \subset U}
     \]
+
     and
     \[
     \ce(\mathfrak{S}, \topo) = \bracs{M(S, U)| S \in \mathfrak{S}, U \in \topo}
     \]
+
     then the topology generated by $\ce$ is the \textbf{$\mathfrak{S}$-open topology} on $T^X$.
 
     If $\cb \subset \topo$ generates $\topo$, then $\ce(\mathfrak{S}, \cb)$ generates the $\mathfrak{S}$-open topology.
@@ -23,10 +25,12 @@
     \[
     E(S, U) = \bracs{(f, g) \in X^T|(f(x), g(x)) \in U \forall x \in S}
     \]
+
     and
     \[
     \mathfrak{E}(\mathfrak{S}, \fU) = \bracs{E(S, U)| S \in \mathfrak{S}, U \in \fU}
     \]
+
     then
     \begin{enumerate}
         \item $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity $\fV$ on $X^T$.
@@ -46,13 +50,15 @@
         \[
         E(T, U \cap U') \subset E(S \cup S', U \cap U') \subset E(S, U) \cap E(S', U')
         \]
+
         \item[(UB1)] For any $U \in \fU$, $\Delta \subset U$. Thus the diagonal in $X^T$ is in $E(S, U)$ for any $S \in \mathfrak{S}$.
         \item[(UB2)] For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \mathfrak{S}$,
         \[
         E(S, V) \circ E(S, V) \subset E(S, V \circ V) \subset E(S, U)
         \]
+
     \end{enumerate}
-    By \ref{proposition:fundamental-entourage-criterion}, $\mathfrak{E}$ is a fundamental system of entourages for the uniformity that it generates.
+    By \autoref{proposition:fundamental-entourage-criterion}, $\mathfrak{E}$ is a fundamental system of entourages for the uniformity that it generates.
 \end{proof}
 
 \begin{proposition}
@@ -61,6 +67,7 @@
     \[
     d_{i, S}: X^T \times X^T \quad (f, g) \mapsto \sup_{x \in S}d_i(f(x), g(x))
     \]
+
     then
     \begin{enumerate}
         \item $\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ is a family of pseudometrics induces the $\mathfrak{S}$-uniformity on $X^T$.
@@ -68,6 +75,7 @@
         \[
         \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \mathfrak{S}}
         \]
+
         is a fundamental system of entourages for the $\mathfrak{S}$-uniformity on $X^T$.
     \end{enumerate}
 \end{proposition}
@@ -76,18 +84,21 @@
     \[
     \bigcap_{j \in J}E(d_{j, S}, r) \subset E\paren{S, \bigcap_{j \in J}E(d_j, r)} \subset E(S, U)
     \]
+
     and the uniformity induced by $\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ contains the $\mathfrak{S}$-uniformity.
 
-    On the other hand, for any $i \in I$ and $r > 0$, $E(d_j, r/2) \in \fU$ by \ref{definition:pseudometric-uniformity}. Therefore $E(S, E(d_j, r/2)) \subset E(d_{j, S}, r)$, so the $\mathfrak{S}$-uniformity contains the induced uniformity.
+    On the other hand, for any $i \in I$ and $r > 0$, $E(d_j, r/2) \in \fU$ by \autoref{definition:pseudometric-uniformity}. Therefore $E(S, E(d_j, r/2)) \subset E(d_{j, S}, r)$, so the $\mathfrak{S}$-uniformity contains the induced uniformity.
 
-    (2): If $\mathfrak{S}$ is upward-directed with respect to inclusion, then by \ref{definition:set-uniform},
+    (2): If $\mathfrak{S}$ is upward-directed with respect to inclusion, then by \autoref{definition:set-uniform},
     \[
     \bracs{E(S, U)| U \in \fU, S \in \mathfrak{S}}
     \]
+
     Following the same steps in (1),
     \[
     \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \mathfrak{S}}
     \]
+
     is a fundamental system of entourages for the $\mathfrak{S}$-uniformity.
 \end{proof}
 
@@ -108,7 +119,8 @@
     \[
     E(F, U)(f) = \bigcap_{x \in F}\pi_x^{-1}(U(f(x)))
     \]
-    which is open in the product topology. The converse is given by \ref{definition:set-uniform}.
+
+    which is open in the product topology. The converse is given by \autoref{definition:set-uniform}.
 \end{proof}
 
 \begin{proposition}

src/topology/functions/uniform.tex

@@ -7,6 +7,7 @@
     \[
     E(U) = \bracsn{(f, g) \in X^T| (f(x), g(x)) \in U \forall x \in X}
     \]
+
     then the \textbf{uniform topology} on $X^T$ is the topology induced by the uniformity generated by $\bracs{E(U)| U \in \fU}$.
 \end{definition}
 
@@ -27,6 +28,7 @@
     \[
     (f(x), g(x)), (g(x), g(y)), (g(y), f(y)) \in W
     \]
+
     so $(f(x), f(y)) \in W \circ W \circ W \subset V$.
 
     Therefore $f$ is continuous at $x$.
@@ -35,5 +37,6 @@
     \[
     (f(x), g(x)), (g(x), g(y)), (g(y), f(y)) \in W
     \]
+
     so $(f(x), f(y)) \in W \circ W \circ W \subset V$.
 \end{proof}

src/topology/index.tex

@@ -1,7 +1,7 @@
 \part{General Topology}
 \label{part:topology}
 
-\input{./src/topology/main/index.tex}
+\input{./main/index.tex}
-\input{./src/topology/uniform/index.tex}
+\input{./uniform/index.tex}
-\input{./src/topology/functions/index.tex}
+\input{./functions/index.tex}
-\input{./src/topology/metric/index.tex}
+\input{./metric/index.tex}

src/topology/main/baire.tex

@@ -36,6 +36,7 @@
     \[
     U \cap \bigcap_{n \in \natp}\ol V_n \subset U \cap \bigcap_{n \in \natp}U_n
     \]
+
     Since $\bigcap_{n \in \natp}\ol V_n \ne \emptyset$ by assumption (b), $U \cap \bigcap_{n \in \natp}U_n \ne \emptyset$, so $\bigcap_{n \in \natp}U_n$ is dense.
 \end{proof}
 
@@ -48,11 +49,11 @@
     \end{enumerate}
 \end{theorem}
 \begin{proof}
-    Let $U \subset X$ be open and $\seq{U_n} \subset 2^X$ be open and dense. Let $V_0 = U$. For each $n \in \natp$, by density of $U_n$, there exists $x \in U_n \cap V_{n - 1}$. Since $X$ is regular (\ref{definition:uniform-separated}/\ref{proposition:compact-hausdorff-normal}), there exists $V_{n} \in \cn^o(x)$ such that $x \in V_{n} \subset \ol U_{n} \subset U_n \cap V_{n-1}$. If $X$ is locally compact, choose $V_n$ to be precompact. If $X$ is completely metrisable, choose $V_n$ such that $\text{diam}(V_n) \le 1/n$.
+    Let $U \subset X$ be open and $\seq{U_n} \subset 2^X$ be open and dense. Let $V_0 = U$. For each $n \in \natp$, by density of $U_n$, there exists $x \in U_n \cap V_{n - 1}$. Since $X$ is regular (\autoref{definition:uniform-separated}/\autoref{proposition:compact-hausdorff-normal}), there exists $V_{n} \in \cn^o(x)$ such that $x \in V_{n} \subset \ol U_{n} \subset U_n \cap V_{n-1}$. If $X$ is locally compact, choose $V_n$ to be precompact. If $X$ is completely metrisable, choose $V_n$ such that $\text{diam}(V_n) \le 1/n$.
 
     Now, if $X$ is locally compact, then by the finite intersection property, $\bigcap_{n \in \natp}\ol{V_n} \ne \emptyset$. If $X$ is completely metrisable, then $\seq{V_n}$ is a Cauchy filter base, and converges to at least one point, so $\bigcap_{n \in \natp}\ol{V_n} \ne \emptyset$.
 
-    By \ref{lemma:baire-condition}, $X$ is a Baire space.
+    By \autoref{lemma:baire-condition}, $X$ is a Baire space.
 \end{proof}
 
 \begin{definition}[Meagre]

src/topology/main/compact.tex

@@ -17,26 +17,30 @@
     \[
     U_J = \bigcup_{j \in J}E_j^c
     \]
+
     then $U_J \subset X$ is open. For any $J, J' \subset I$, $U_J \cup U_{J'} = U_{J \cup J'}$.
 
     Suppose for contradiction that $\bigcap_{i \in I}E_i = \emptyset$, then
     \[
     \mathbf{U} = \bracs{U_J|J \subset I \text{ finite}}
     \]
+
     is an open cover for $X$. By assumption, $U_J \subsetneq X$ for all $J \subset I$ finite. Thus $\mathbf{U}$ admits no finite subcover, contradiction.
 
     (2) $\Rightarrow$ (3): Let $\fF \subset 2^X$ be a filter, then $\bracsn{\overline{E}| E \in \fF}$ satisfies the hypothesis of (2).
 
-    (3) $\Leftrightarrow$ (4): By \ref{definition:accumulation-point}, the cluster points and the limit points of an ultrafilter coincide.
+    (3) $\Leftrightarrow$ (4): By \autoref{definition:accumulation-point}, the cluster points and the limit points of an ultrafilter coincide.
 
     (3) $\Rightarrow$ (1): For each $J \subset I$, let
     \[
     E_J = \bigcap_{j \in J}U_j^c
     \]
+
     then for each $J, J' \subset I$, $E_J \cap E_{J'} = E_{J \cup J'}$. Assume for contradiction that $\mathbf{U}$ admits no finite subcover. Let
     \[
     \fB = \bracs{E_H|J \subset I \text{ finite}}
     \]
+
     then $\fB$ is a filter base consisting of closed sets. By assumption, there exists $x \in \bigcap_{i \in I}U_j^c$, so $\mathbf{U}$ is not an open cover, contradiction.
 \end{proof}
 

src/topology/main/connected.tex

@@ -42,5 +42,5 @@
     Let $X$ be a topological space and $A \subset X$ be connected, then there exists a unique connected set $C \supset A$ such that for any $C' \supset A$ connected, $C \supset C'$. The set $C$ is the \textbf{connected component} of $A$.
 \end{definition}
 \begin{proof}
-    Let $C$ be the union of all connected sets that contain $A$, then $C$ is connected by \ref{proposition:connected-union}, and is the maximum connected set containing $A$ by definition.
+    Let $C$ be the union of all connected sets that contain $A$, then $C$ is connected by \autoref{proposition:connected-union}, and is the maximum connected set containing $A$ by definition.
 \end{proof}

src/topology/main/continuity.tex

@@ -26,7 +26,7 @@
 
     Local continuity, $(2) \Rightarrow (1)$: Since $f(\cn(x))$ converges to $f(x)$, for each $U \in \cn(f(x))$, there exists $V \in \cn(x)$ such that $f(V) \subset V$. Thus $V \subset f^{-1}(U) \in \cn(x)$ by (F1).
 
-    Global continuity, $(1) \Rightarrow (2)$: Let $U \subset Y$ be open, then $U \in \cn(f(x))$ for all $x \in f^{-1}(U)$. Hence $f^{-1}(U) \in \cn(x)$ for all $x \in f^{-1}(U)$. Thus $U$ is open by \ref{lemma:openneighbourhood}.
+    Global continuity, $(1) \Rightarrow (2)$: Let $U \subset Y$ be open, then $U \in \cn(f(x))$ for all $x \in f^{-1}(U)$. Hence $f^{-1}(U) \in \cn(x)$ for all $x \in f^{-1}(U)$. Thus $U$ is open by \autoref{lemma:openneighbourhood}.
 
     Global continuity, $(2) \Rightarrow (1)$: Let $x \in X$ and $V \in \cn(f(x))$, then there exists $U \in \cn(f(x))$ open, so $f^{-1}(V) \supset f^{-1}(U)$ contains an open set that contains $x$. Therefore $f^{-1}(V) \in \cn(x)$.
 
@@ -43,7 +43,7 @@
     then there exists a unique $f \in C(X; Y)$ such that $f|_{U_i} = f_i$ for all $i \in I$.
 \end{lemma}
 \begin{proof}
-    By \ref{lemma:glue-function}, there exists a unique $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
+    By \autoref{lemma:glue-function}, there exists a unique $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
 
     Let $x \in X$. By assumption (a), there exists $i \in I$ such that $x \in U_i$. Since $f_i \in C(U_i; Y)$, for any $V \in \cn_Y(f(x))$, there exists $U \in \cn_{U_i}(x)$ such that $f_i(U) \subset V$. As $U_i$ is open in $X$, $U \in \cn_X(x)$. Thus $f(U) = f_i(U) \subset V$, and $f$ is continuous at $x$.
 \end{proof}

src/topology/main/definition.tex

@@ -51,6 +51,8 @@
     \[
     \topo = \topo(\cb) = \bracs{\bigcup_{i \in I}U_i \bigg | \seqi{U} \subset \cb, I \text{ index set}}
     \]
+
+
     Conversely, if $\cb \subset 2^X$ is a family such that:
     \begin{enumerate}
         \item[(TB1)] For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
@@ -65,6 +67,8 @@
     \[
     \paren{\bigcup_{i \in I}U_i} \cap \paren{\bigcup_{j \in J}V_j} = \bigcup_{i \in I}\bigcup_{j \in J}U_i \cap V_j
     \]
+
+
     Thus to show that $\topo(\cb)$ satisfies (O2), it is sufficient to show that $U \cap V \in \topo(\cb)$ for all $U, V \in \cb$. To this end, for every $x \in U \cap V$, there exists $W_x \in \cb$ such that $x \in W_x \subset U \cap V$. Therefore $U \cap V = \bigcup_{x \in U \cap V}W_x \in \topo(\cb)$, and $\topo(\cb)$ satisfies (O2).
 
     By definition, $\topo(\cb)$ satisfies (O3).
@@ -76,14 +80,16 @@
     \[
     \topo(\ce) = \bracs{\bigcup_{i \in I}U_i \bigg | U_i \in \cb(\ce)}
     \]
+
     where
     \[
     \cb(\ce) = \bracs{\bigcap_{j = 1}^n U_j \bigg | \seqf{U_j} \subset \ce, n \in \nat^+}
     \]
+
     is a base for $\topo(\ce)$. The topology $\topo(\ce)$ is known as the topology \textbf{generated by} $\ce$.
 \end{definition}
 \begin{proof}
-    Since $\cb(\ce) \supset \ce$, $\cb(\ce)$ satisfies (TB1). In addition, $\cb(\ce)$ is closed under finite intersections, so it satisfies (TB2). By \ref{definition:base}, $\cb(\ce)$ is a base for $\topo(\ce)$.
+    Since $\cb(\ce) \supset \ce$, $\cb(\ce)$ satisfies (TB1). In addition, $\cb(\ce)$ is closed under finite intersections, so it satisfies (TB2). By \autoref{definition:base}, $\cb(\ce)$ is a base for $\topo(\ce)$.
 \end{proof}
 
 \begin{definition}[Initial Topology]
@@ -98,6 +104,7 @@
         \mathcal{B} = \bracs{\bigcap_{j \in J}f_j^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \topo_j}
         \]
 
+
         is a base for $\topo$.
     \end{enumerate}
     The topology $\topo$ is known a the \textbf{initial/weak topology} generated by the maps $\seqi{f}$.
@@ -107,6 +114,6 @@
     \begin{enumerate}
         \item For each $i \in I$, $\topo \supset \bracs{f_i^{-1}(U)|U \in \topo_i}$, so $f_i \in C(\topo; Y_i)$.
         \item If $\mathcal{S}$ is a topology such that $f_i \in C(X, \mathcal{S}; Y_i)$, then $\bracs{f_i^{-1}(U)|U \in \topo_i} \subset \mathcal{S}$. Thus $\ce \subset \mathcal{S}$ and $\mathcal{S} \supset \topo$.
-        \item By \ref{definition:generated-topology}, $\cb$ is a base for $\topo$.
+        \item By \autoref{definition:generated-topology}, $\cb$ is a base for $\topo$.
     \end{enumerate}
 \end{proof}

src/topology/main/filters.tex

@@ -30,6 +30,7 @@
     \[
     \fF = \bracs{F \subset X| \exists E \in \fB: E \subset F}
     \]
+
 \end{proposition}
 \begin{proof}
     Filter Base $\Rightarrow$ (FB1): Let $E, F \in \fB$, then $E \cap F \in \fF$. Thus there exists $G \in \fB$ such that $G \subset E \cap F$.
@@ -79,15 +80,17 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
 \[
 \fB = \bracs{\bigcap_{j = 1}^n E_j \bigg | \seqf{E_j} \subset \fB_0, n \in \nat^+}
 \]
+
 \end{definition}
 \begin{proof}
     For any $\seqf{E_j}, \bracsn{F_j}_1^m \subset \fB_0$,
     \[
     G = \paren{\bigcap_{j = 1}^n E_j} \cap \paren{\bigcap_{j = 1}^m F_j} \in \fB
     \]
+
     Thus $\fB$ satisfies (FB1). Since $\bigcap_{j = 1}^n E_j \ne \emptyset$, $\emptyset \not\in \fB$, and $\fB$ satisfies (FB2).
 
-    By \ref{proposition:filterbasecriterion}, $\fB$ is a base for the filter $\fF = \bracs{E \subset X| \exists F \in \fB: F \subset E}$.
+    By \autoref{proposition:filterbasecriterion}, $\fB$ is a base for the filter $\fF = \bracs{E \subset X| \exists F \in \fB: F \subset E}$.
 
     If $\fF' \supset \fB_0$ is a filter, then $\fF \supset \fB$ by (F2), and $\fF' \supset \fF$ by (F1). Thus $\fF$ is the smallest filtetr containing $\fB$.
 \end{proof}
@@ -97,7 +100,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
     Let $X$ be a set, $\fF, \mathfrak{G} \subset 2^X$ be filter subbases. If for any $E \in \fF$ and $F \in \mathfrak{G}$, $E \cap F \ne \emptyset$, then there exists a filter $\fU \supset \fF \cup \mathfrak{G}$.
 \end{lemma}
 \begin{proof}
-    Let $\seqf{E_j} \subset \fF$ and $\seqf[m]{F_j} \subset \mathfrak{G}$, then $\bigcap_{j = 1}^n E_j \in \fF$ and $\bigcap_{j = 1}^m F_j \in \mathfrak{G}$. Thus their intersection is non-empty, and $\fF \cup \mathfrak{G}$ is a filter subbase. By \ref{definition:generatedfilter}, there exists a filter containing $\fF \cup \mathfrak{G}$.
+    Let $\seqf{E_j} \subset \fF$ and $\seqf[m]{F_j} \subset \mathfrak{G}$, then $\bigcap_{j = 1}^n E_j \in \fF$ and $\bigcap_{j = 1}^m F_j \in \mathfrak{G}$. Thus their intersection is non-empty, and $\fF \cup \mathfrak{G}$ is a filter subbase. By \autoref{definition:generatedfilter}, there exists a filter containing $\fF \cup \mathfrak{G}$.
 \end{proof}
 
 \begin{definition}[Ultrafilter]
@@ -111,7 +114,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
     If the above holds, then $\fF$ is an \textbf{ultrafilter}.
 \end{definition}
 \begin{proof}
-    (1) $\Rightarrow$ (2): Let $E \subset X$ with $E^c \not\in \fF$, then $E \cap F \ne \emptyset$ for all $F \in \fF$. By \ref{lemma:filter-extend}, there exists a filter $\fU \supset \fF \cup \bracs{E}$. Since $\fF$ is maximal, $\bracs{E} \in \fU \subset \fF$.
+    (1) $\Rightarrow$ (2): Let $E \subset X$ with $E^c \not\in \fF$, then $E \cap F \ne \emptyset$ for all $F \in \fF$. By \autoref{lemma:filter-extend}, there exists a filter $\fU \supset \fF \cup \bracs{E}$. Since $\fF$ is maximal, $\bracs{E} \in \fU \subset \fF$.
 
     (2) $\Rightarrow$ (1): Let $E \subset X$ with $E \not \in \fF$, then $E^c \in \fF$. Since $E \cap E^c = \emptyset$, there exists no filter containing $\fF \cup \bracs{E}$. Thus $\fF$ is maximal.
 
@@ -133,6 +136,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
     \[
     \mathbf{U} = \bracs{\fU \supset \fF| \fU \subset 2^X \text{ is a filter}}
     \]
+
     and order it by inclusion. For any chain $\mathbf{C} \subset \mathbf{U}$, let $\fB = \bigcup_{\fF \in \textbf{C}}\fF$, then $\fB$ is a filter subbase, and the filter $\fU \in \mathbf{U}$ generated by $\fB$ is an upper bound for $\mathbf{C}$. By Zorn's lemma, $\textbf{U}$ has a maximal element.
 
     (2): Let $E \subset X$. If there exists $F \in \fF$ with $F \cap E^c = \emptyset$, then $E \supset F$ and $E \in \fF$. Otherwise, $\fF \cup \bracs{E^c}$ is a filter subbase, and there exists an ultrafilter containing it by (1). In which case, there exists an ultrafilter $\fU \supset \fF$ with $E \not\in \fU$.
@@ -152,7 +156,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
     If $A \subset X$ and $\fB \subset 2^A$, then $\fB$ \textbf{converges} to $x$ if $\fF(\fB) \supset \bracsn{U \cap A| U \in \cn(x)}$.
 \end{definition}
 \begin{proof}
-    (1) $\Leftrightarrow$ (2): By (2) of \ref{lemma:ultrafilter}.
+    (1) $\Leftrightarrow$ (2): By (2) of \autoref{lemma:ultrafilter}.
 
     (1) $\Rightarrow$ (3): $\cn(x)$ is a fundamental system of neighbourhoods at $x$.
 
@@ -173,7 +177,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
 \begin{proof}
     (1) $\Rightarrow$ (3): Let $U \in \cn(x)$, then $U \cap E \ne \emptyset$ for all $E \in \fB$.
 
-    (3) $\Rightarrow$ (4): By \ref{lemma:filter-extend}, there exists a filter $\fU \supset \cb(x) \cup \fB$. Since $\cb(x)$ and $\fB$ are bases for $\cn(x)$ and $\fF$, respectively, $\fU \supset \cn(x) \cup \fF$.
+    (3) $\Rightarrow$ (4): By \autoref{lemma:filter-extend}, there exists a filter $\fU \supset \cb(x) \cup \fB$. Since $\cb(x)$ and $\fB$ are bases for $\cn(x)$ and $\fF$, respectively, $\fU \supset \cn(x) \cup \fF$.
 
     (4) $\Rightarrow$ (1): Let $U \in \cn(x)$, then since $\fU \supset \fB \cup \cn(x)$, $U \cap E \ne \emptyset$ for all $E \in \fB \subset \fF \subset \fU$. Thus $x \in \bigcap_{E \in \fF}\ol{E}$.
 \end{proof}
@@ -187,5 +191,6 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
     \[
     y = \lim_{x \to x_0 \\ x \in A}f(x)
     \]
+
     is a \textbf{limit} of $f$ at $y$ with respect to $A$. If $A = X$, then $x \in A$ may be omitted.
 \end{definition}

src/topology/main/hausdorff.tex

@@ -16,17 +16,18 @@
     If the above holds, then $X$ is a \textbf{T2/Hausdorff} space.
 \end{definition}
 \begin{proof}
-    $(1) \Rightarrow (2)$: Let $y \in X \setminus \bracs{x}$, then there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \cap V = \emptyset$. By (2) of \ref{definition:closure}, $y \not\in \overline{U} \subset \bigcap_{U \in \cn(x)}\overline{U}$.
+    $(1) \Rightarrow (2)$: Let $y \in X \setminus \bracs{x}$, then there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \cap V = \emptyset$. By (2) of \autoref{definition:closure}, $y \not\in \overline{U} \subset \bigcap_{U \in \cn(x)}\overline{U}$.
 
     $(2) \Rightarrow (3)$: Let $\fF \subset 2^X$ be a filter and $x \in X$ such that $\cn(x) \subset \fF$, then
     \[
     \bracs{x} = \bigcap_{U \in \cn(x)}\ol{U} \supset \bigcap_{U \in \cn(x)}\ol{U} \supset \bracs{x}
     \]
+
     so $x$ is the only cluster point of $\fF$.
 
     $(3) \Rightarrow (4)$: Let $\fF \subset 2^X$ be a filter. If $\fF$ converges to $x \in X$, then $x$ is a cluster point of $\fF$. As $\fF$ admits only one cluster point, $x$ is the only limit point of $\fF$.
 
-    $(4) \Rightarrow (5)$: Let $x \in \overline{\Delta}$, then by (4) of \ref{definition:closure}, there exists $\fF \subset 2^\Delta$ converging to $x$. Let $i, j \in I$, then for any $y \in \Delta$, $\pi_i(y) = \pi_j(y)$. Thus $\pi_i(\fF) = \pi_j(\fF)$. By \ref{proposition:productfilterconvergence}, $\pi_i(\fF)$ converges to $\pi_i(x)$ and $\pi_j(\fF)$ converges to $\pi_j(x)$. By assumption, $\pi_i(x) = \pi_j(x)$. Since this holds for all pairs $i, j \in I$, $x \in \Delta$.
+    $(4) \Rightarrow (5)$: Let $x \in \overline{\Delta}$, then by (4) of \autoref{definition:closure}, there exists $\fF \subset 2^\Delta$ converging to $x$. Let $i, j \in I$, then for any $y \in \Delta$, $\pi_i(y) = \pi_j(y)$. Thus $\pi_i(\fF) = \pi_j(\fF)$. By \autoref{proposition:productfilterconvergence}, $\pi_i(\fF)$ converges to $\pi_i(x)$ and $\pi_j(\fF)$ converges to $\pi_j(x)$. By assumption, $\pi_i(x) = \pi_j(x)$. Since this holds for all pairs $i, j \in I$, $x \in \Delta$.
 
     $(5) \Rightarrow (6)$: Take $I = \bracs{1, 2}$.
 
@@ -38,7 +39,7 @@
     Let $X$ be a topological space, $Y$ be a Hausdorff space, $A \subset X$ be a dense subset, and $F, G \in C(X; Y)$. If $F|_A = G|_A$, then $F = G$.
 \end{proposition}
 \begin{proof}
-    Let $x \in X$. By (4) \ref{definition:closure}, there exists a filter base $\fB \subset 2^A$ that converges to $x$. By (2) of local continuity, $F(\fB)$ converges to $F(x)$ and $G(\fB)$ converges to $G(x)$. Since $F(\fB) = G(\fB)$, $F(x) = G(x)$ by (4) of \ref{definition:hausdorff}.
+    Let $x \in X$. By (4) \autoref{definition:closure}, there exists a filter base $\fB \subset 2^A$ that converges to $x$. By (2) of local continuity, $F(\fB)$ converges to $F(x)$ and $G(\fB)$ converges to $G(x)$. Since $F(\fB) = G(\fB)$, $F(x) = G(x)$ by (4) of \autoref{definition:hausdorff}.
 \end{proof}
 
 \begin{proposition}

src/topology/main/index.tex

@@ -1,24 +1,24 @@
 \chapter{Topological Spaces}
 \label{chap:topological-spaces}
 
-\input{./src/topology/main/definition.tex}
+\input{./definition.tex}
-\input{./src/topology/main/filters.tex}
+\input{./filters.tex}
-\input{./src/topology/main/nets.tex}
+\input{./nets.tex}
-\input{./src/topology/main/neighbourhoods.tex}
+\input{./neighbourhoods.tex}
-\input{./src/topology/main/interiorclosureboundary.tex}
+\input{./interiorclosureboundary.tex}
-\input{./src/topology/main/continuity.tex}
+\input{./continuity.tex}
-\input{./src/topology/main/product.tex}
+\input{./product.tex}
-\input{./src/topology/main/hausdorff.tex}
+\input{./hausdorff.tex}
-\input{./src/topology/main/regular.tex}
+\input{./regular.tex}
-\input{./src/topology/main/normal.tex}
+\input{./normal.tex}
-\input{./src/topology/main/quotient.tex}
+\input{./quotient.tex}
-\input{./src/topology/main/connected.tex}
+\input{./connected.tex}
-\input{./src/topology/main/path-connected.tex}
+\input{./path-connected.tex}
-\input{./src/topology/main/local-path-connected.tex}
+\input{./local-path-connected.tex}
-\input{./src/topology/main/unity.tex}
+\input{./unity.tex}
-\input{./src/topology/main/compact.tex}
+\input{./compact.tex}
-\input{./src/topology/main/sigma-compact.tex}
+\input{./sigma-compact.tex}
-\input{./src/topology/main/para.tex}
+\input{./para.tex}
-\input{./src/topology/main/support.tex}
+\input{./support.tex}
-\input{./src/topology/main/lch.tex}
+\input{./lch.tex}
-\input{./src/topology/main/baire.tex}
+\input{./baire.tex}

src/topology/main/interiorclosureboundary.tex

@@ -18,7 +18,7 @@
 
     $(3) \Rightarrow (1)$: By (F1) of $\cn(x)$, $A \in \cn(x)$.
 
-    Let $U \subset A$ be open, then $U \in \cn(x)$ for all $x \in U$ by \ref{lemma:openneighbourhood}. By (3), $U \subset A^o$, so $A^o$ contains every open subset of $A$. On the other hand, for any $x \in A^o$, (2) implies that there exists $U \in \cn^o(x)$ with $U \subset A$. In which case, $U \subset A^o$, and $A^o \in \cn(x)$. Therefore $A^o$ itself is open by \ref{lemma:openneighbourhood}.
+    Let $U \subset A$ be open, then $U \in \cn(x)$ for all $x \in U$ by \autoref{lemma:openneighbourhood}. By (3), $U \subset A^o$, so $A^o$ contains every open subset of $A$. On the other hand, for any $x \in A^o$, (2) implies that there exists $U \in \cn^o(x)$ with $U \subset A$. In which case, $U \subset A^o$, and $A^o \in \cn(x)$. Therefore $A^o$ itself is open by \autoref{lemma:openneighbourhood}.
 \end{proof}
 
 \begin{definition}[Closure]
@@ -72,9 +72,9 @@
     If the above holds, then $A$ is a \textbf{dense} subset of $X$.
 \end{definition}
 \begin{proof}
-    $(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of \ref{definition:closure}, $U \cap A \ne \emptyset$.
+    $(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of \autoref{definition:closure}, $U \cap A \ne \emptyset$.
 
-    $(2) \Rightarrow (3)$: Let $\emptyset \ne U \subset X$ open. For each $x \in U$ and $V \in \cn(x)$, $U \cap V \in \cn(x)$ by (F2). Thus there exists $y \in A \cap U \cap V$. By (3) of \ref{definition:closure}, $x \in \overline{A \cap U}$.
+    $(2) \Rightarrow (3)$: Let $\emptyset \ne U \subset X$ open. For each $x \in U$ and $V \in \cn(x)$, $U \cap V \in \cn(x)$ by (F2). Thus there exists $y \in A \cap U \cap V$. By (3) of \autoref{definition:closure}, $x \in \overline{A \cap U}$.
 
     $(3) \Rightarrow (1)$: $X$ is open.
 \end{proof}
@@ -92,7 +92,7 @@
     Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^n X_j$ is also separable.
 \end{proposition}
 \begin{proof}
-    Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \ref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
+    Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \autoref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
 \end{proof}
 
 
@@ -101,7 +101,7 @@
     Let $X$ be a topological space and $A \subset X$, then $(A^o)^c = \overline{A^c}$.
 \end{lemma}
 \begin{proof}
-    Let $x \in X$. By (3) of \ref{definition:interior}, $x \in (A^o)^c$ if and only if there exists no $U \in \cn(x)$ such that $U \subset A$. Thus $x \in (A^o)^c$ if and only if $U \cap A^c \ne \emptyset$ for all $U \in \cn(x)$. By (2) of \ref{definition:closure}, this is equivalent to $x \in \ol{A^c}$.
+    Let $x \in X$. By (3) of \autoref{definition:interior}, $x \in (A^o)^c$ if and only if there exists no $U \in \cn(x)$ such that $U \subset A$. Thus $x \in (A^o)^c$ if and only if $U \cap A^c \ne \emptyset$ for all $U \in \cn(x)$. By (2) of \autoref{definition:closure}, this is equivalent to $x \in \ol{A^c}$.
 \end{proof}
 
 
@@ -120,9 +120,9 @@
 \begin{proof}
     $(1) \Rightarrow (2)$: $\cn(x) \supset \fB$.
 
-    $(2) \Rightarrow (3)$: By (2) of \ref{definition:closure}, $x \in \overline{A}$. Since $V \cap A^c \ne \emptyset$ for all $V \in \fB$, there exists no open set $U \subset A$ with $x \in A$. By (2) of \ref{definition:interior}, $x \not\in A^o$.
+    $(2) \Rightarrow (3)$: By (2) of \autoref{definition:closure}, $x \in \overline{A}$. Since $V \cap A^c \ne \emptyset$ for all $V \in \fB$, there exists no open set $U \subset A$ with $x \in A$. By (2) of \autoref{definition:interior}, $x \not\in A^o$.
 
-    $(3) \Rightarrow (4)$: By \ref{lemma:closurecomplement}.
+    $(3) \Rightarrow (4)$: By \autoref{lemma:closurecomplement}.
 
-    $(4) \Rightarrow (1)$: By (2) of \ref{definition:closure}.
+    $(4) \Rightarrow (1)$: By (2) of \autoref{definition:closure}.
 \end{proof}

src/topology/main/lch.tex

@@ -12,9 +12,9 @@
     If the above holds, then $X$ is a \textbf{locally compact Hausdorff (LCH)} space.
 \end{definition}
 \begin{proof}
-    (1) $\Rightarrow$ (2): Let $K \in \cn(x)$ be compact and $U \in \cn(x)$, then $\overline{U \cap K}$ is closed. By \ref{proposition:compact-closed}, $K$ itself is closed, so $\overline{U \cap K} \subset K$ is a closed subset of a compact set, and compact by \ref{proposition:compact-extensions}.
+    (1) $\Rightarrow$ (2): Let $K \in \cn(x)$ be compact and $U \in \cn(x)$, then $\overline{U \cap K}$ is closed. By \autoref{proposition:compact-closed}, $K$ itself is closed, so $\overline{U \cap K} \subset K$ is a closed subset of a compact set, and compact by \autoref{proposition:compact-extensions}.
 
-    (2) $\Rightarrow$ (3): Let $U \in \cn(x)$, then there exists $K \in \cn(x)$ with $x \in K \subset U$. By \ref{proposition:compact-closed}, $K$ is closed, so $\overline{K^o} \subset K$ is compact by \ref{proposition:compact-extensions}.
+    (2) $\Rightarrow$ (3): Let $U \in \cn(x)$, then there exists $K \in \cn(x)$ with $x \in K \subset U$. By \autoref{proposition:compact-closed}, $K$ is closed, so $\overline{K^o} \subset K$ is compact by \autoref{proposition:compact-extensions}.
 \end{proof}
 
 \begin{lemma}
@@ -22,14 +22,16 @@
     Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exits $V \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$.
 \end{lemma}
 \begin{proof}
-    For each $x \in K$, there exists $V_x \in \cn^o(x)$ be precompact such that $x \in V_x \subset \overline{V_x} \subset U$ by (3) of \ref{definition:lch}. Since $K$ is compact, there exists $\seqf{x_j} \subset K$ such that
+    For each $x \in K$, there exists $V_x \in \cn^o(x)$ be precompact such that $x \in V_x \subset \overline{V_x} \subset U$ by (3) of \autoref{definition:lch}. Since $K$ is compact, there exists $\seqf{x_j} \subset K$ such that
     \[
     K \subset \bigcup_{j = 1}^n V_{x_j} \subset U
     \]
-    By \ref{proposition:closure-finite-union},
+
+    By \autoref{proposition:closure-finite-union},
     \[
     \ol{\bigcup_{j = 1}^n V_{x_j}} = \bigcup_{j = 1}^n \overline{V_{x_j}} \subset U
     \]
+
     so $V = \bigcup_{j = 1}^n V_{x_j} \in \cn^o(K)$ is precompact.
 \end{proof}
 
@@ -38,20 +40,22 @@
     Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exists $f \in C_c(X; [0, 1])$ such that $\supp{f} \subset U$.
 \end{lemma}
 \begin{proof}
-    By \ref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that
+    By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that
     \[
     K \subset V \subset \ol{V} \subset W \subset \ol{W} \subset U
     \]
-    As $\ol{W}$ is compact, it is normal by \ref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \ref{proposition:compact-closed}.
 
-    By Urysohn's lemma (\ref{lemma:urysohn}), there exists $f \in C(\ol{V}; [0, 1])$ such that $f|_K = 1$ and $f|_{\ol{W} \setminus V} = 0$. Let
+    As $\ol{W}$ is compact, it is normal by \autoref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \autoref{proposition:compact-closed}.
+
+    By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C(\ol{V}; [0, 1])$ such that $f|_K = 1$ and $f|_{\ol{W} \setminus V} = 0$. Let
     \[
     F: X \to [0, 1] \quad x \mapsto \begin{cases}
     f(x) &x \in W \\
     0 &x \in X \setminus \ol{V}
     \end{cases}
     \]
-    then by the gluing lemma for continuous functions (\ref{lemma:gluing-continuous}), $F \in C_c(X; [0, 1])$ with $F|_{K} = 1$ and $\supp{f} \subset U$.
+
+    then by the \hyperref[gluing lemma for continuous functions]{lemma:gluing-continuous}, $F \in C_c(X; [0, 1])$ with $F|_{K} = 1$ and $\supp{f} \subset U$.
 \end{proof}
 
 \begin{theorem}[Tietze Extension Theorem (LCH)]
@@ -59,16 +63,17 @@
     Let $X$ be a LCH space, $K \subset X$ be compact, $U \in \cn^o(K)$, and $f \in C(K; \real)$, then there exists $F \in C_c(U; \real)$ such that $F|_K = f$.
 \end{theorem}
 \begin{proof}
-    By \ref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$. As $\ol{W}$ is compact, it is normal by \ref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \ref{proposition:compact-closed}.
+    By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$. As $\ol{W}$ is compact, it is normal by \autoref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \autoref{proposition:compact-closed}.
 
-    By the Tietze extension theorem (\ref{theorem:tietze}), there exists $F \in C(\ol{W}; \real)$ such that $F|_K = f$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $\eta \in C_c(X; [0, 1])$ such that $\eta|_K = 1$ and $\supp{\eta} \subset V$. In which case, define
+    By the \hyperref[Tietze extension theorem]{theorem:tietze}, there exists $F \in C(\ol{W}; \real)$ such that $F|_K = f$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\eta \in C_c(X; [0, 1])$ such that $\eta|_K = 1$ and $\supp{\eta} \subset V$. In which case, define
     \[
     \ol F: X \to \real \quad x \mapsto \begin{cases}
         \eta(x) \cdot f(x) &x \in V \\
         0 &x \in X \setminus \supp{\eta}
     \end{cases}
     \]
-    then by the gluing lemma for continuous functions (\ref{lemma:gluing-continuous}), $\ol F \in C_c(X; \real)$ with $\ol F|_K = F|_K = f$ and $\supp{F} \subset \supp{\eta} \subset V \subset U$.
+
+    then by the \hyperref[gluing lemma for continuous functions]{lemma:gluing-continuous}, $\ol F \in C_c(X; \real)$ with $\ol F|_K = F|_K = f$ and $\supp{F} \subset \supp{\eta} \subset V \subset U$.
 \end{proof}
 
 \begin{proposition}[{{\cite[Proposition 4.39]{Folland}}}]
@@ -88,10 +93,11 @@
         \item[(b)] For each $0 \le k < n$, $\overline{U_k} \subset U_{k+1}$.
         \item[(c)] For each $1 \le k \le n$, $U_k \supset \bigcup_{j = 1}^k K_j$.
     \end{enumerate}
-    By \ref{lemma:lch-compact-neighbour}, there exists $U_{n+1} \in \cn^o(\overline{U_n} \cup K_{n+1})$ precompact. In which case, by (c),
+    By \autoref{lemma:lch-compact-neighbour}, there exists $U_{n+1} \in \cn^o(\overline{U_n} \cup K_{n+1})$ precompact. In which case, by (c),
     \[
     U_{n+1} \supset \ol{U_n} \cup K_{n+1} \supset \bigcup_{j = 1}^n K_j \cup K_{n+1} = \bigcup_{j = 1}^{n+1}K_j
     \]
+
     Thus $\bracs{U_j}_0^{n+1}$ satisfies (a), (b), and (c), and $\seq{U_n}$ is an exhaustion of $X$ by compact sets.
 \end{proof}
 
@@ -108,14 +114,16 @@
     \[
     F_j = \bigcup_{\substack{1 \le k \le m \\ N_{x_k} \subset U_j}}N_{x_k}
     \]
+
     then $F_j \subset U_j$ is compact, and $\bigcup_{j = 1}^n F_j \supset K$.
 
-    By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $\seqf{f_j} \subset C_c(X; [0, 1])$ such that for each $1 \le j \le n$, $f_j|_{F_j} = 1$, and $\supp{f_j} \subset U_j$.
+    By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqf{f_j} \subset C_c(X; [0, 1])$ such that for each $1 \le j \le n$, $f_j|_{F_j} = 1$, and $\supp{f_j} \subset U_j$.
 
     By Urysohn's lemma again, there exists $f_{j + 1} \in C(X; [0, 1])$ such that $f_{j+1}|_{K} = 0$ and $\bracs{f_{j+1} = 0} \subset \bigcup_{j = 1}^n \supp{f_j}$. Let $F = \sum_{j = 1}^{n+1}f_j$, then $F(x) > 0$ for all $x \in X$. For each $1 \le j \le n$, let $g_j = f_j/F$, then $g_j \in C_c(X; [0, 1])$ with $\supp{g_j} \subset U_j$. In addition, since $f_{j+1}|_K = 0$,
     \[
     \sum_{j = 1}^n g_j|_K = \frac{\sum_{j = 1}^n f_j}{F} = \frac{\sum_{j = 1}^n f_j}{\sum_{j = 1}^n f_j} = 1
     \]
+
     Therefore $\seqf{g_j}$ is the desired partition of unity.
 \end{proof}
 
@@ -124,17 +132,19 @@
     Let $X$ be a LCH space and $\ce \subset 2^X$ be a locally finite precompact open cover of $X$, then there exists locally finite precompact open covers $\bracs{F_E}_{E \in \ce}, \bracs{G_E}_{E \in \ce} \subset 2^X$ such that for each $E \in \ce$, $F_E \subset \ol{F_E} \subset E \subset \ol{E} \subset G_E$.
 \end{lemma}
 \begin{proof}
-    $(\bracs{F_E}_{E \in \ce})$: For each $E \in \ce$, $\bracs{F \in \ce|F \cap \ol E \ne \emptyset}$ is finite by \ref{lemma:locally-finite-compact}. Let
+    $(\bracs{F_E}_{E \in \ce})$: For each $E \in \ce$, $\bracs{F \in \ce|F \cap \ol E \ne \emptyset}$ is finite by \autoref{lemma:locally-finite-compact}. Let
     \[
     F_E = \bigcup_{\substack{F \in \ce} \\ F \cap \ol E \ne \emptyset}F
     \]
+
     then $F_E \in \cn(\ol{E})$ is precompact.
 
     Let $N \subset X$ and $E \in \ce$. If $N \cap F_E \ne \emptyset$, then there exists $F \in \ce$ such that $N \cap F \ne \emptyset$ and $F \cap \ol{E} \ne \emptyset$. Thus
     \[
     \bracs{E \in \ce|N \cap F_E \ne \emptyset} \subset \bigcup_{\substack{F \in \ce \\ F \cap N \ne \emptyset}}\bracs{E \in \ce|F \cap \ol{E} \ne \emptyset} \subset \bigcup_{\substack{F \in \ce \\ F \cap N \ne \emptyset}}\bracs{E \in \ce|\ol{F} \cap \ol{E} \ne \emptyset}
     \]
-    By \ref{lemma:locally-finite-closure}, $\bracsn{\ol E|E \in \ce}$ is also locally finite. Hence for every $F \in \ce$, $\bracsn{E \in \ce|F \cap \ol{E} \ne \emptyset}$ is finite.
+
+    By \autoref{lemma:locally-finite-closure}, $\bracsn{\ol E|E \in \ce}$ is also locally finite. Hence for every $F \in \ce$, $\bracsn{E \in \ce|F \cap \ol{E} \ne \emptyset}$ is finite.
 
     Let $x \in X$, then there exists $N \in \cn(x)$ such that $\bracs{F \in \ce|N \cap F \ne \emptyset}$ is finite. In which case, $\bracs{E \in \ce|N \cap F_E \ne \emptyset}$ is finite as well. Therefore $\bracs{F_E}_{E \in \ce}$ is locally finite.
 
@@ -149,17 +159,20 @@
     \[
     G_E = \bigcup_{\substack{x \in X_\ce \\ \ol{N_x} \subset E}}N_x
     \]
+
     then $\bracs{G_E}_{E \in \ce}$ is an open cover of $X$. Since $G_E \subset E$ for all $E \in \ce$, $\bracs{G_E}_{E \in \ce}$ is locally finite.
 
     It remains to show that $\ol{G_E} \subset E$. Let $x \in X_F$ such that $N_x \subset E$, then $N_x \cap F \ne \emptyset$. Since $N_x \subset E$, $E \cap F \ne \emptyset$. Thus
     \[
     \bracsn{x \in X_\ce|\ol{N_x} \subset E} \subset \bigcup_{\substack{F \in \ce \\ E \cap F \ne \emptyset}}X_F \subset \bigcup_{\substack{F \in \ce \\ \ol E \cap F \ne \emptyset}}X_F
     \]
-    is finite by \ref{lemma:locally-finite-compact}, so
+
+    is finite by \autoref{lemma:locally-finite-compact}, so
     \[
     \ol{G_E} = \bigcup_{\substack{x \in X_\ce \\ \ol{N_x} \subset E}}\ol N_x \subset E
     \]
-    by \ref{proposition:closure-finite-union}.
+
+    by \autoref{proposition:closure-finite-union}.
 \end{proof}
 
 \begin{proposition}
@@ -177,31 +190,34 @@
 \begin{proof}
     (1) $\Rightarrow$ (2): For each $x \in X$, there exists a precompact open neighbourhood $U_x \in \cn^o(x)$. Since $\bracs{U_x| x \in X}$ is an open cover of $X$, there exists a locally finite refinement $\mathcal{V}$. For each $V \in \mathcal{V}$, there exists $x \in X$ such that $V \subset U_x$. In which case, $\ol{V} \subset \ol{U_x}$ is compact.
 
-    (2) $\Rightarrow$ (3): Let $\cf \subset 2^X$ be a locally finite open cover of $X$ consisting of precompact open sets. By \ref{lemma:lch-locally-finite-precompact-refine}, there exists a locally finite open cover $\bracs{G_F}_{F \in \cf}$ of $X$ consisting of precompact open sets such that $\ol{F} \subset G_F$ for all $F \in \cf$.
+    (2) $\Rightarrow$ (3): Let $\cf \subset 2^X$ be a locally finite open cover of $X$ consisting of precompact open sets. By \autoref{lemma:lch-locally-finite-precompact-refine}, there exists a locally finite open cover $\bracs{G_F}_{F \in \cf}$ of $X$ consisting of precompact open sets such that $\ol{F} \subset G_F$ for all $F \in \cf$.
 
     For each $F \in \cf$, let
     \[
     \mathcal{U}_F = \bracs{U \cap G_F|U \in \mathcal{U}}
     \]
+
     then $\mathcal{U}_F$ is a precompact open cover of $\ol{F}$. By compactness of $\ol{F}$, there exists $\mathcal{V}_F \subset \mathcal{U}_F$ finite such that $\ol{F} \subset \bigcup_{V \in \mathcal{V}_F}V$.
 
     Let $\mathcal{V} = \bigcup_{F \in \cf}\mathcal{V}_F$, then $\mathcal{V}$ is a precompact open cover of $X$. For any $x \in X$, there exists $N \in \cn(x)$ such that $\bracs{F \in \cf|N \cap G_F}$ is finite. Thus
     \[
     \bracs{V \in \mathcal{V}| N \cap V} \subset \bigcup_{\substack{F \in \cf \\ N \cap G_F \ne \emptyset}}\mathcal{V}_F
     \]
+
     is finite, and $\mathcal{V}$ is locally finite.
 
-    (3) $\Rightarrow$ (4): By \ref{lemma:lch-locally-finite-precompact-refine}.
+    (3) $\Rightarrow$ (4): By \autoref{lemma:lch-locally-finite-precompact-refine}.
 
     (4) $\Rightarrow$ (5): Let $\seqi{V}, \seqi{W} \subset 2^X$ be locally finite refinements of $\mathcal{U}$ consisting of precompact open sets such that for each $i \in I$, $\ol{W_i} \subset V_i$.
 
-    By Urysohn's Lemma (\ref{lemma:lch-urysohn}), there exists $\seqi{f} \in C_c(X; [0, 1])$ such that for each $i \in I$, $f_i|_{\ol{W_i}} = 1$ and $\supp{f_i} \subset V_i$.
+    By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqi{f} \in C_c(X; [0, 1])$ such that for each $i \in I$, $f_i|_{\ol{W_i}} = 1$ and $\supp{f_i} \subset V_i$.
 
     Let $F = \sum_{i \in I}f_i$. For each $x \in X$, there exists $N_x \in \cn^o(x)$ such that $\bracs{i \in I|N_x \cap V_i \ne \emptyset}$ is finite. In which case,
     \[
     F|_{N_x} = \sum_{\substack{i \in I \\ N_x \cap V_i \ne \emptyset}}f_i|_{N_x}
     \]
-    thus $F|_{N_x} \in C(N_x; \real)$. By \ref{lemma:gluing-continuous}, $F \in C(X; \real)$.
+
+    thus $F|_{N_x} \in C(N_x; \real)$. By \autoref{lemma:gluing-continuous}, $F \in C(X; \real)$.
 
     Since $\seqi{W}$ is an open cover of $X$, $F(x) > 0$ for all $x \in X$. For each $i \in I$, let $g_i = f_i/F$, then $g_i \in C_c(X; [0, 1])$ with $\supp{g_i} = \supp{f_i} \subset W_i$. For any $x \in X$, there exists $N_x \in \cn^o(x)$ such that $\bracs{i \in I|N_x \cap W_i \ne \emptyset}$ is finite. In which case, $\bracs{i \in I|0 < g_i|_{N_x}}$ is also finite. Thus $\seqi{g}$ is a $C_c$ partition of unity subordinate to $\mathcal{U}$.
 
@@ -217,7 +233,7 @@
     Let $X$ be a $\sigma$-compact LCH space, then $X$ is paracompact.
 \end{proposition}
 \begin{proof}
-    By \ref{proposition:lch-sigma-compact}, there exists an exhaustion $\seq{U_n} \subset 2^X$ of $X$ by precompact open sets. Denote $U_0 = \emptyset$. For each $n \in \natp$, let $V_n = U_{n+1} \setminus \ol{U_{n-1}}$.
+    By \autoref{proposition:lch-sigma-compact}, there exists an exhaustion $\seq{U_n} \subset 2^X$ of $X$ by precompact open sets. Denote $U_0 = \emptyset$. For each $n \in \natp$, let $V_n = U_{n+1} \setminus \ol{U_{n-1}}$.
 
-    Let $x \in X$, then there exists $n \in \natp$ such that $x \in U_n \setminus U_{n-1}$. In which case, if $n > 1$, then $x \in U_{n} \setminus \ol{U_{n - 2}} = V_{n-1}$. If $n = 1$, then $x \in U_{2} = V_1$. Thus $\seq{V_n}$ is an open cover of $X$. In addition, for any $m, n \in \natp$ with $m \le n$, $V_m \cap V_n \ne \emptyset$ implies that $n - m < 2$, so $\seq{V_n}$ is locally finite. By (2) of \ref{proposition:lch-paracompact}, $X$ is paracompact.
+    Let $x \in X$, then there exists $n \in \natp$ such that $x \in U_n \setminus U_{n-1}$. In which case, if $n > 1$, then $x \in U_{n} \setminus \ol{U_{n - 2}} = V_{n-1}$. If $n = 1$, then $x \in U_{2} = V_1$. Thus $\seq{V_n}$ is an open cover of $X$. In addition, for any $m, n \in \natp$ with $m \le n$, $V_m \cap V_n \ne \emptyset$ implies that $n - m < 2$, so $\seq{V_n}$ is locally finite. By (2) of \autoref{proposition:lch-paracompact}, $X$ is paracompact.
 \end{proof}

src/topology/main/local-path-connected.tex

@@ -17,7 +17,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $A \subset X$ be a path component and $x \in A$, then there exists $U \in \cn(x)$ connected. By \ref{proposition:path-connected-union}, $A \cup U \in \cn(x)$ is also connected. Since $A$ is a path-component, $A \cup U \subset A \in \cn(x)$. Thus $A$ is open by \ref{lemma:openneighbourhood}.
+    (1): Let $A \subset X$ be a path component and $x \in A$, then there exists $U \in \cn(x)$ connected. By \autoref{proposition:path-connected-union}, $A \cup U \in \cn(x)$ is also connected. Since $A$ is a path-component, $A \cup U \subset A \in \cn(x)$. Thus $A$ is open by \autoref{lemma:openneighbourhood}.
 
     (2): Let $P$ be a path component in $X$ and $C \supset P$ be its connected components. If $C$ is not path-connected, then $P \subsetneq C$ there exists path-components $\seqi{P} \subset 2^C$ such that $C = \bigsqcup_{i \in I}P_i$. In which case, $C \setminus P$ is open by (1), and $C$ is not connected, which is impossible.
 

src/topology/main/neighbourhoods.tex

@@ -50,6 +50,7 @@
     \[
     \topo = \bracs{U \subset X| U \in \cn(x) \forall x \in U}
     \]
+
     Firstly, $\emptyset$ satisfies the condition vacuously, so $\emptyset \in \topo$.
     For any $x \in X$, $\cn(x)$ is non-empty and there exists $V \in \cn(x)$. Since $X \supset V$, $X \in \topo$ by (F1).
 
@@ -63,11 +64,13 @@
     \[
     U = \bracs{y \in V: V \in \cn(y)}
     \]
+
     then $U \supset U_0$ and $U \in \cn(x)$ by (V1). Let $y \in U$, then (V4) implies that there exists $W \in \cn(y)$ such that $V \in \cn(z)$ for all $z \in W$. Thus $W \subset U$ and $U \in \cn(y)$ by (F1).
 
-    \textit{Uniqueness}: Let $\mathcal{R}$ be a topology such that $\cn_{\mathcal{R}} = \cn$, then by \ref{lemma:openneighbourhood},
+    \textit{Uniqueness}: Let $\mathcal{R}$ be a topology such that $\cn_{\mathcal{R}} = \cn$, then by \autoref{lemma:openneighbourhood},
     \[
     \mathcal{R} = \bracs{U \subset X| U \in \cn_{\mathcal{R}}(x) \forall x \in U} = \bracs{U \subset X| U \in \cn(x) \forall x \in U} = \topo
     \]
+
     so $\topo$ is unique.
 \end{proof}

src/topology/main/normal.tex

@@ -30,16 +30,17 @@
 \begin{proof}
     (1): Let $\seq{q_n}$ be an enumeration of $\rational \cap (0, 1)$. For each $n \in \natp$, let $Q_n = \bracs{0, 1} \cup \bracs{q_k|1 \le k \le n}$.
 
-    Let $U_1 = B^c$. By (2) of \ref{definition:topology-normal}, there exists $U_0 \in \cn(A)$ such that $A \subset U_0 \subset \ol{U_0} \subset B^c$. In which case, for $n = 0$,
+    Let $U_1 = B^c$. By (2) of \autoref{definition:topology-normal}, there exists $U_0 \in \cn(A)$ such that $A \subset U_0 \subset \ol{U_0} \subset B^c$. In which case, for $n = 0$,
     \begin{enumerate}
         \item[(i)] $U_1 = B^c$.
         \item[(ii)] For any $p, q \in Q_k$ with $p < q$, $\overline{U_p} \subset U_q$.
     \end{enumerate}
 
-    Suppose inductively that $\bracs{U_q|q \in Q_n}$ has been constructed, and (ii) holds for $n$. Let $p = \max\bracs{r \in Q_n|r < q_{n+1}}$ and $q = \min\bracs{r \in Q_n|r > r_{n+1}}$. By (2) of \ref{definition:topology-normal}, there exists $U_{q_{n+1}} \in \cn^o(\overline{U_p})$ such that
+    Suppose inductively that $\bracs{U_q|q \in Q_n}$ has been constructed, and (ii) holds for $n$. Let $p = \max\bracs{r \in Q_n|r < q_{n+1}}$ and $q = \min\bracs{r \in Q_n|r > r_{n+1}}$. By (2) of \autoref{definition:topology-normal}, there exists $U_{q_{n+1}} \in \cn^o(\overline{U_p})$ such that
     \[
     U_p \subset \overline{U_p} \subset U_{q_{n+1}} \subset \overline{U_{q_{n+1}}} \subset U_q
     \]
+
     and $\bracs{U_q|q \in Q_{n+1}}$ satisfies (ii) for $n + 1$.
 
     Now suppose that $\bracs{U_q|q \in \rational \cap [0, 1]}$ has been constructed and (ii) holds for all $n \in \nat$. For any $p, q \in \rational \cap [0, 1]$ with $p < q$, there exists $n \in \nat$ such that $p, q \in Q_n$. In which case, $\ol{U_p} \subset U_q$ by (ii). Thus (b) holds.
@@ -48,16 +49,19 @@
     \[
     f: X \to [0, 1] \quad x \mapsto \inf\bracs{q \in [0, 1] \cap \rational| x \in U_q}
     \]
+
     where $f(x) = 1$ if $x \not\in \bigcup_{q \in [0, 1] \cap \rational}U_q$. Since $A \subset \bigcap_{q \in [0, 1] \cap \rational}U_q$ and $U_1 = B^c$, $f|_A = 0$ and $f|_B = 1$.
 
     Let $\alpha \in [0, 1]$, then
     \[
     f^{-1}([0, \alpha)) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q < \alpha}}U_q
     \]
+
     is open. On the other hand, let $x \in X$, then $f(x) > \alpha$ if and only if there exists $q \in (\alpha, f(x)) \cap \rational$ such that $x \not\in U_q$. By (b) of (1), this is equivalent to the existence of $p \in (\alpha, f(x)) \cap \rational$ such that $x \not\in \ol{U_p}$. In which case,
     \[
     f^{-1}((b, 1]) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q > \alpha}}\overline{U_p}^c
     \]
+
     is open.
 \end{proof}
 
@@ -70,18 +74,21 @@
     \[
     R: BC(X; \real) \to BC(A; \real) \quad g \mapsto g|_A
     \]
+
     then $R \in L(BC(X; \real); BC(A; \real))$.
 
     For any $g \in C(A; [0, 1])$, let
     \[
     B = g^{-1}(\norm{g}_u \cdot [0, 1/3]) \quad C = g^{-1}(\norm{g}_u \cdot [2/3, 1])
     \]
+
     then $B, C \subset A$ are closed with $B \cap C = \emptyset$. Since $A$ is closed, $B, C \subset X$ are closed. By Urysohn's lemma, there exists $h \in C(X; [0, 1/3])$ such that $h|_C = 1/3$ and $h|_B = 0$. Thus $g - h|_A \in C(A; [0, 2/3])$.
 
     By linearity, this implies that for any $g \in BC(A; \real)$, there exists $h \in BC(A; \real)$ such that $\norm{h}_u \le \norm{g}_u/3$ and $\norm{g - h|_A}_u \le 2\norm{g}_u/3$.
 
-    Since $\real$ is complete, so is $BC(X; \real)$ by \ref{proposition:set-uniform-complete}. Using successive approximations (\ref{theorem:successive-approximation}), for every $f \in BC(A; \real)$, there exists $F \in BC(X; \real)$ such that $RF = F|_A = f$ and
+    Since $\real$ is complete, so is $BC(X; \real)$ by \autoref{proposition:set-uniform-complete}. Using \hyperref[successive approximations]{theorem:successive-approximation}, for every $f \in BC(A; \real)$, there exists $F \in BC(X; \real)$ such that $RF = F|_A = f$ and
     \[
     \norm{F}_u \le \frac{1}{3} \cdot \frac{1}{1 - 2/3} \cdot \norm{f}_u = \norm{f}_u
     \]
+
 \end{proof}

src/topology/main/para.tex

@@ -15,6 +15,7 @@
     \[
     \bracs{U \in \mathcal{U}|U \cap K \ne \emptyset} \subset \bigcup_{x \in X_K}\bracs{U \in \mathcal{U}|U \cap N_x \ne \emptyset}
     \]
+
 \end{proof}
 
 \begin{lemma}

src/topology/main/path-connected.tex

@@ -18,10 +18,11 @@
 \begin{proof}
     Let $U, V \subset [0, 1]$ be open with $[0, 1] = U \cup V$. If $\sup U = \sup V$, then $\sup U = \sup V = 1$ and $U, V \in \cn^o(1)$, so $U \cap V \ne \emptyset$. If $\sup U < \sup V \le 1$, then $x \not\in U$ and $x \in V$. In which case, $V \in \cn^o(x)$ and $V \cap U \ne \emptyset$. Therefore $[0, 1]$ is connected.
 
-    Fix $x \in X$. For any $y \in X$, let $f_y \in C([0, 1]; X)$ be a path from $x$ to $y$, then $f_y([0, 1])$ is connected with $x, y \in f_y([0, 1])$ by \ref{proposition:connected-image}. By \ref{proposition:connected-union},
+    Fix $x \in X$. For any $y \in X$, let $f_y \in C([0, 1]; X)$ be a path from $x$ to $y$, then $f_y([0, 1])$ is connected with $x, y \in f_y([0, 1])$ by \autoref{proposition:connected-image}. By \autoref{proposition:connected-union},
     \[
     X = \bigcup_{y \in X}f_y([0, 1])
     \]
+
     is connected.
 \end{proof}
 
@@ -37,6 +38,7 @@
         g(2(t - 1/2)) &t \in [1/2, 1]
     \end{cases}
     \]
+
     is a path from $y$ to $z$.
 \end{proof}
 
@@ -45,5 +47,5 @@
     Let $X$ be a topological space and $A \subset X$ be path-connected, then there exists a unique path-connected set $C \supset A$ such that for any $C' \supset A$ path-connected, $C \supset C'$. The set $C$ is the \textbf{path-component} of $A$.
 \end{definition}
 \begin{proof}
-    Let $C$ be the union of all path-connected sets containing $A$, then $C$ is path-connected by \ref{proposition:path-connected-union} and the maximum path-connected set containing $A$ by definition.
+    Let $C$ be the union of all path-connected sets containing $A$, then $C$ is path-connected by \autoref{proposition:path-connected-union} and the maximum path-connected set containing $A$ by definition.
 \end{proof}

src/topology/main/product.tex

@@ -11,6 +11,7 @@
         \cb(\ce) = \bracs{\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_{k}) \bigg | U_{k} \in \topo_{i_k}, \seqf{i_k} \subset I, n \in \nat^+}
         \]
 
+
         is a base for $\topo$.
 
         \item[(U)] For any topological space space $Y$ and $\seqi{f}$ where $f_i \in C(Y; X_i)$ for all $i \in I$, there exists a unique $f \in C(Y; X)$ such that the following diagram commutes
@@ -22,16 +23,18 @@
         }
         \]
 
+
         for all $i \in I$.
     \end{enumerate}
 \end{definition}
 \begin{proof}
-    (1): By (3) of \ref{definition:initial-topology}.
+    (1): By (3) of \autoref{definition:initial-topology}.
 
     (U): Let $f \in \prod_{i \in I}f_i$, then $f$ is the unique function such that the diagrams commute. For each $\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k) \in \topo$,
     \[
     f^{-1}\paren{\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k)} = \bigcap_{k = 1}^n f_{i_k}^{-1}(U_k)
     \]
+
     which is open in $Y$ by (O2).
 \end{proof}
 
@@ -46,6 +49,7 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
     \[
     B \subset \bigcap_{k = 1}^n B_k \subset \bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k) \subset U
     \]
+
     Therefore $\fB$ converges to $x$.
 \end{proof}
 
@@ -62,23 +66,26 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
         }
         \]
 
+
         for all $i \in I$.
         \item If $X$ is T0 and equipped with the initial topology induced by $\seqi{f}$, then $f$ is an embedding.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): By (U) of \ref{definition:product-topology}.
+    (1): By (U) of \autoref{definition:product-topology}.
 
     (2): Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U \in \cn(x)$ such that $y \not\in U$. Let $J \subset I$ finite, $\seqj{U}$, and $\seqj{V}$ such that
     \[
     U = \bigcap_{j \in J}f_j^{-1}(U_j) = f^{-1}\paren{\bigcap_{j \in J}\pi_j^{-1}(U_j)}
     \]
+
     then $f(x) \in \bigcap_{j \in J}\pi_j^{-1}(U_j)$ but $f(y) \not\in \bigcap_{j \in J}\pi_j^{-1}(U_j)$. Thus $f$ is injective.
 
     Let $U \subset X$ be open. Assume without loss of generality that there exists $J \subset I$ finite and $\bracs{U_j}_{j \in J}$ such that $U = \bigcap_{j \in J}f_j^{-1}(U_j)$. In which case, $\bigcap_{j \in J}\pi_j^{-1}(U_j)$ is open in $\prod_{i \in I}Y_i$ and
     \[
     f(U) = f(X) \cap \bigcap_{j \in J}\pi_j^{-1}(U_j)
     \]
+
     is a relatively open set.
 \end{proof}
 
@@ -95,13 +102,14 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
         }
         \]
 
+
         for all $i \in I$.
 
         \item $f$ is an embedding.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): By (U) of \ref{definition:product-topology}/\ref{definition:product-uniform}.
+    (1): By (U) of \autoref{definition:product-topology}/\autoref{definition:product-uniform}.
 
     (2): Consider the following diagram
     \[
@@ -112,11 +120,13 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
     \prod_{i \in I} X_i \ar@{->}[r]_{\pi_i} \ar@{->}[u]^{f} & X_i \ar@{^{(}->}[u]_{f_i}
     }
     \]
+
     Since each $X_i \to Y_i$ is an embedding, the composition
     \[
     \xymatrix{
     \iota_P(\prod_{i \in I}X_i) \ar@{->}[r]^{\pi_i} & Y_i \ar@{->}[r]^{f_i^{-1}} & X_i
     }
     \]
-    is continuous/uniformly continuous. By (U) of \ref{definition:product-topology}/\ref{definition:product-uniform}, $f$ is an embedding.
+
+    is continuous/uniformly continuous. By (U) of \autoref{definition:product-topology}/\autoref{definition:product-uniform}, $f$ is an embedding.
 \end{proof}

src/topology/main/quotient.tex

@@ -35,6 +35,7 @@
         \td X \ar@{->}[r]_{\tilde f} & Y
         }
         \]
+
         \item $\pi$ is a quotient map.
     \end{enumerate}
     The space $(\td X, \pi)$ is the \textbf{quotient} of $X$ by $\sim$.

src/topology/main/regular.tex

@@ -27,11 +27,12 @@
 \begin{proof}
     $(\Rightarrow)$: Let $x \in X$ and $\fF = \bracs{U \cap A| U \in \cn(x)}$, then $f(\fF) = F(\fF)$. By continuity of $F$, $F(\fF)$ converges to $F(x)$, so $\lim_{y \to x, x \in A}f(x)$ exists.
 
-    $(\Leftarrow)$: Let $F(x) = \lim_{y \to x, y \in A}f(y)$, then $F$ is well-defined and $F|_A = f$ by (4) of \ref{definition:hausdorff}.
+    $(\Leftarrow)$: Let $F(x) = \lim_{y \to x, y \in A}f(y)$, then $F$ is well-defined and $F|_A = f$ by (4) of \autoref{definition:hausdorff}.
 
-    Let $x \in X$ and $V \in \cn(F(x))$. Using (2) of \ref{definition:regular}, assume without loss of generality that $V$ is closed. By assumption, there exists $U \in \cn^o(x)$ such that $f(U \cap A) \subset V$. In which case, \ref{lemma:openneighbourhood} implies that $U \in \cn^o(y)$ for all $y \in U$. Since every limit point of a filter is a cluster point,
+    Let $x \in X$ and $V \in \cn(F(x))$. Using (2) of \autoref{definition:regular}, assume without loss of generality that $V$ is closed. By assumption, there exists $U \in \cn^o(x)$ such that $f(U \cap A) \subset V$. In which case, \autoref{lemma:openneighbourhood} implies that $U \in \cn^o(y)$ for all $y \in U$. Since every limit point of a filter is a cluster point,
     \[
     F(y) = \lim_{\substack{z \to y \\ z \in A}}f(z) \in \ol{f(U \cap A)} \subset V
     \]
+
     as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous.
 \end{proof}