Updated content.

.editorconfig

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+root = true
+
+[*]
+end_of_line = lf

refs.bib

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+@book{Taylor,
+  title={Partial Differential Equations: Basic Theory},
+  author={Taylor, M.E.},
+  isbn={9780387946542},
+  lccn={lc95054104},
+  series={Applied mathematical sciences},
+  url={https://books.google.ca/books?id=vISPd8QJfHoC},
+  year={1996},
+  publisher={Springer}
+}
+@book{Evans,
+  title={Partial Differential Equations},
+  author={Evans, L.C.},
+  isbn={9780821807729},
+  lccn={97041033},
+  series={Graduate studies in mathematics},
+  url={https://books.google.ca/books?id=5Pv4LVB_m8AC},
+  year={1998},
+  publisher={American Mathematical Society}
+}
+@book{Chavel,
+  title={Eigenvalues in Riemannian Geometry},
+  author={Chavel, I. and Randol, B. and Dodziuk, J.},
+  isbn={9780080874340},
+  series={Pure and Applied Mathematics},
+  url={https://books.google.ca/books?id=0v1VfTWuKGgC},
+  year={1984},
+  publisher={Academic Press}
+}
+@book{Bourbaki,
+  title={General topology: chapters 1--4},
+  author={Bourbaki, Nicolas},
+  volume={18},
+  year={2013},
+  publisher={Springer Science \& Business Media}
+}
+@book{SchaeferWolff,
+  title={Topological Vector Spaces},
+  author={Schaefer, H.H. and Wolff, M.P.H.},
+  isbn={9780387987262},
+  lccn={98053842},
+  series={Graduate Texts in Mathematics},
+  url={https://books.google.ca/books?id=9kXY742pABoC},
+  year={1999},
+  publisher={Springer New York}
+}
+@book{Lang,
+  title={Real and Functional Analysis},
+  author={Lang, S.},
+  isbn={9780387940014},
+  lccn={92021208},
+  series={Graduate Texts in Mathematics},
+  url={https://books.google.ca/books?id=_3F4opD1X84C},
+  year={1993},
+  publisher={Springer New York}
+}

retag.sh

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+# Rebuilds the tag system. Requires administrative access.
+
+rm ./tags
+rm ./gerby-website/gerby/tools/hello-world.sqlite
+./build.sh

src/cat/cat-func.tex

@@ -0,0 +1,41 @@
+\chapter{Categories and Functors}
+\label{chap:categories}
+
+
+
+\begin{definition}[Category]
+\label{definition:category}
+    A \textbf{category} $\catc$ is a collection of objects $\obj{\catc}$, such that for any $A, B, C \in \obj{\catc}$, there exists sets $\mor{A, B}$, $\mor{B, C}$, and a composition law
+    \[
+    \mor{A, B} \times \mor{B, C} \to \mor{A, C}
+    \]
+    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 $B = B'$.
+        \item[(CAT2)] For any $A \in \obj{\catc}$, there exists $\text{Id}_A \in \mor{A, A}$ such that $f \circ \text{Id}_A = f$ and $\text{Id}_A \circ g = g$ for all $B, C \in \obj{\catc}$, $f \in \mor{A, B}$, and $g \in \mor{C, A}$.
+        \item[(CAT3)] For any $A, B, C, D \in \obj{\catc}$, $f \in \mor{A, B}$, $g \in \mor{B, C}$, and $h \in \mor{C, D}$, $(h \circ g) \circ f = h \circ (g \circ f)$.
+    \end{enumerate}
+    The elements of $\obj{\catc}$ are the \textbf{objects} of $\catc$, and elements of $\mor{A, B}$ are the \textbf{morphisms/arrows} from $A$ to $B$.
+\end{definition}
+
+\begin{definition}[Isomorphism]
+\label{definition:isomorphism}
+    Let $\catc$ be a category, $A, B \in \obj{\catc}$, and $f \in \mor{A, B}$, then $f$ is an \textbf{isomorphism} if there exists $g \in \mor{B, A}$ such that $g \circ f = \text{Id}_A$ and $f \circ g = \text{Id}_B$.
+
+    For any $A, B \in \obj{\catc}$, $A$ and $B$ are \textbf{isomorphic} if there exists an isomorphism $f \in \mor{A, B}$.
+\end{definition}
+
+
+\begin{definition}[Functor]
+\label{definition:functor}
+    Let $\mathfrak{A}$ and $\mathfrak{B}$ be categories. A \textbf{covariant functor} is a rule that assigns each $A \in \obj{\mathfrak{A}}$ to some $\lambda(A) \in \obj{\mathfrak{B}}$, and each $f \in \mor{A, B}$ to some $\lambda(f) \in \mor{\lambda(A), \lambda(B)}$, such that
+    \begin{enumerate}
+        \item[(FN1)] For any $A \in \obj{\mathfrak{A}}$, $\lambda(\text{Id}_A) = \text{Id}_{\lambda(A)}$.
+        \item[(FN2)] For any $A, B, C \in \obj{\mathfrak{A}}$, $f \in \mor{A, B}$, and $g \in \mor{B, C}$, $\lambda(g \circ f) = \lambda(g) \circ \lambda(f)$.
+    \end{enumerate}
+
+    A \textbf{contravariant} functor is a rule that assigns each $A \in \obj{\mathfrak{A}}$ to some $\lambda(A) \in \obj{\mathfrak{B}}$, and each $f \in \mor{A, B}$ to some $\lambda(f) \in \mor{\lambda(B), \lambda(A)}$, that satisfies (FN1) and
+    \begin{enumerate}
+        \item[(FN2')] For any $A, B, C \in \obj{\mathfrak{A}}$, $f \in \mor{A, B}$, and $g \in \mor{B, C}$, $\lambda(g \circ f) = \lambda(f) \circ \lambda(g)$.
+    \end{enumerate}
+\end{definition}

src/cat/index.tex

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

src/cat/universal.tex

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+\chapter{Universal Construction}
+\label{chap:universal}
+
+\begin{definition}[Universal Object]
+\label{definition:universal}
+    Let $\catc$ be a category and $P \in \obj{\catc}$, then $P$ is...
+    \begin{enumerate}
+        \item \textbf{universally attracting} if for every $A \in \obj{\catc}$, there exists a unique $f \in \mor{A, P}$.
+        \item \textbf{universally repelling} if for every $A \in \obj{\catc}$, there exists a unique $f \in \mor{P, A}$.
+    \end{enumerate}
+    If $P$ is universally attracting or repelling, then $P$ is a \textbf{universal object}.
+
+    If $P, Q \in \obj{\catc}$ are both universally attracting/repelling, then they are isomorphic.
+\end{definition}
+\begin{proof}
+    By assumption, there exists morphisms $f \in \mor{P, Q}$ and $g \in \mor{Q, P}$. Since $f \circ g \in \mor{Q, Q}$ and $g \circ f \in \mor{P, P}$ are unique, $f \circ g = \text{Id}_Q$ and $g \circ f = \text{Id}_P$. Thus $f$ is an isomorphism.
+\end{proof}
+
+\begin{definition}[Product]
+\label{definition:product}
+    Let $\catc$ be a category and $\seqi{A} \subset \obj{\catc}$. A \textbf{product} of $\seqi{A}$ is a pair $(P, \seqi{\pi})$ where
+    \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
+
+        \[
+        \xymatrix{
+        C \ar@{->}[rd]^{f_i} \ar@{->}[d]_{f} &  \\
+        P \ar@{->}[r]_{\pi_i} & A_i
+        }
+        \]
+
+        for all $i \in I$.
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Coproduct]
+    \label{definition:coproduct}
+    Let $\catc$ be a category and $\seqi{A} \subset \obj{\catc}$. A \textbf{product} of $\seqi{A}$ is a pair $(P, \seqi{\iota})$ where
+    \begin{enumerate}
+        \item $P \in \obj{\catc}$.
+        \item For each $i \in I$, $\iota_i \in \mor{A_i P}$.
+        \item[\textbf{(U)}] For any pair $(C, \seqi{f})$ satisfying (1) and (2), there exists a unique $f \in \mor{P, C}$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+         & C \\
+        A_i \ar@{->}[r]_{\iota_i} \ar@{->}[ru]^{f_i} & P \ar@{->}[u]_{f}
+        }
+        \]
+
+        for all $i \in I$.
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Directed Set]
+\label{definition:directed-set}
+    Let $I$ be a set and $\lesssim$ be a relation on $I$, then $(I, \lesssim)$ is a \textbf{directed set} if
+    \begin{enumerate}
+        \item For any $i \in I$, $i \lesssim i$.
+        \item For any $i, j, k \in I$ such that $i \lesssim j$ and $j \lesssim k$, $i \lesssim k$.
+    \end{enumerate}
+    and one of the following holds:
+    \begin{enumerate}
+        \item[(3U)] For any $i, j \in I$, there exists $k \in I$ with $i, j \lesssim k$.
+        \item[(3D)] For any $i, j \in I$, there exists $k \in I$ with $k \lesssim i, j$.
+    \end{enumerate}
+    The directed set is \textbf{upward-directed} if it satisfies (3U), and \textbf{downward-directed} if it satisfies (3D).
+\end{definition}
+
+\begin{definition}[Directed System]
+\label{definition:directed-system}
+    Let $\catc$ be a category and $(I, \lesssim)$ be a directed set. A \textbf{directed system} is a pair $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim j})$ such that:
+    \begin{enumerate}
+        \item $\seqi{A} \subset \obj{\catc}$.
+        \item For each $i \in I$, $f^i_i = \text{Id}_{A_i}$.
+        \item For each $i, j \in I$ with $i \lesssim j$, $f^i_j \in \mor{A_i, A_j}$.
+        \item For each $i, j, k \in I$ with $i \lesssim j \lesssim k$, $f^j_k \circ f^i_j = f^i_k$.
+    \end{enumerate}
+    If $I$ is upward/downward-directed, then $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim j})$ is upward/downward-directed.
+\end{definition}
+
+\begin{definition}[Direct Limit]
+\label{definition:direct-limit}
+    Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system, then a \textbf{direct limit} of $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim j})$ is a pair $(A, \bracsn{f^i_A}_{i \in I})$ such that:
+    \begin{enumerate}
+        \item For each $i \in I$, $f^i_A \in \mor{A_i, A}$.
+        \item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
+
+        \[
+        \xymatrix{
+        A_i \ar@{->}[rd]_{f^i_A} \ar@{->}[r]^{f^i_j} & A_j \ar@{->}[d]^{f^j_A} \\
+         & A
+        }
+        \]
+
+        \item[(U)] For any pair $(B, \bracsn{g^i_A}_{i \in I})$ satsifying (1) and (2), there exists a unique $g \in \mor{A, B}$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+        A_j \ar@{->}[d]_{f^j_A} \ar@{->}[rd]^{g^i_B} &  \\
+        A \ar@{->}[r]_{g} & B
+        }
+        \]
+
+        for all $i \in I$.
+    \end{enumerate}
+
+\end{definition}
+
+\begin{definition}[Inverse Limit]
+\label{definition:inverse-limit}
+Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system, then an \textbf{inverse limit} of $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim j})$ is a pair $(A, \bracsn{f^A_i}_{i \in I})$ such that:
+\begin{enumerate}
+    \item For each $i \in I$, $f^A_i \in \mor{A, A_i}$.
+    \item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
+
+    \[
+    \xymatrix{
+    A_i \ar@{->}[r]^{f^i_j} & A_j \\
+    A \ar@{->}[u]^{f^A_i} \ar@{->}[ru]_{f^A_j} &
+    }
+    \]
+
+    \item[(U)] For any pair $(B, \bracsn{g^A_i}_{i \in I})$, there exists a unique $g \in \mor{B, A}$ such that the following diagram commutes
+
+    \[
+    \xymatrix{
+     & A_i \\
+    B \ar@{->}[r]_{g} \ar@{->}[ru]^{g^B_i} & A \ar@{->}[u]_{f^A_i}
+    }
+    \]
+
+    for all $i \in I$.
+\end{enumerate}
+\end{definition}

src/fa/index.tex

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+\part{Functional Analysis}
+\label{part:part-fa}
+
+
+\input{./src/fa/tvs/index.tex}
+\input{./src/fa/lc/index.tex}
+\input{./src/fa/rs/index.tex}

src/fa/lc/continuous.tex

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+\section{Continuous Linear Maps}
+\label{section:tvs-convex-morphism}
+
+
+\begin{proposition}
+\label{proposition:tvs-convex-morphism}
+    Let $E, F$ be locally convex spaces and $T \in \hom(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $T$ is uniformly continuous.
+        \item $T$ is continuous.
+        \item $T$ is continuous at $0$.
+        \item For every continuous seminorm $[\cdot]_F$ on $F$, there exists a continuous seminorm $[\cdot]_E$ on $E$ such that $[Tx]_F \le [x]_E$ for all $x \in E$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    $(1) \Leftrightarrow (2) \Leftrightarrow (3)$: By \ref{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}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tvs-convex-multilinear}
+    Let $\seqf{E_j}$ and $F$ be locally convex spaces, and $T: \prod_{j = 1}^n E_j \to F$ be $n$-linear map, then the following are equivalent:
+    \begin{enumerate}
+        \item $T$ is continuous.
+        \item For every continuous seminorm $[\cdot]_F$ on $F$, there exists continuous seminorms $\seqf{[\cdot]_j}$ on $\seqf{E_j}$, such that for every $x \in \prod_{j = 1}^n E_j$,
+        \[
+        [Tx]_F \le \prod_{j = 1}^n [x_j]_{E_j}
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: By continuity of $T$, there exists continuous seminorms $\seqf{[\cdot]_j}$ on $\seqf{E_j}$ such that for any $x \in \prod_{j = 1}^n E_j$, $\max_{1 \le j \le n}[x_j]_{E_j} < 1$ implies that $[Tx]_F < 1$. In which case, the inequality follows from linearity.
+\end{proof}

src/fa/lc/convex.tex

@@ -0,0 +1,101 @@
+\section{Seminorms}
+\label{section:seminorms}
+
+\begin{definition}[Convex]
+\label{definition:convex}
+    Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is \textbf{convex} if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]} \subset A$.
+\end{definition}
+
+\begin{definition}[Sublinear Functional]
+\label{definition:sublinear-functional}
+    Let $E$ be a vector space over $K \in \RC$, then a \textbf{sublinear functional} is a mapping $\rho: E \to \real$ such that:
+    \begin{enumerate}
+        \item $\rho(0) = 0$.
+        \item For any $x \in E$ and $\lambda \ge 0$, $\rho(\lambda x) = \lambda \rho(x)$.
+        \item For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
+    \end{enumerate}
+\end{definition}
+
+
+\begin{definition}[Seminorm]
+\label{definition:seminorm}
+    Let $E$ be a vector space over $K \in \RC$, then a \textbf{seminorm} on $E$ is a mapping $\rho: E \to [0, \infty)$ such that:
+    \begin{enumerate}
+        \item $\rho(0) = 0$.
+        \item For any $x \in E$ and $\lambda \in K$, $\rho(\lambda x) = \abs{\lambda} \rho(x)$.
+        \item For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
+    \end{enumerate}
+\end{definition}
+
+\begin{lemma}
+\label{lemma:continuous-seminorm}
+    Let $E$ be a TVS over $K \in \RC$ and $[\cdot]: E \times E \to [0, \infty)$ be a seminorm on $E$, then the following are equivalent:
+    \begin{enumerate}
+        \item $[\cdot]$ is uniformly continuous.
+        \item $[\cdot]$ is continuous.
+        \item $[\cdot]$ is continuous at $0$.
+    \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.
+\end{proof}
+
+
+\begin{definition}[Topology Induced by Seminorm]
+\label{definition:seminorm-topology}
+    Let $E$ be a vector space over $K \in \RC$ and $\seqi{[\cdot]}$ be seminorms, then:
+    \begin{enumerate}
+        \item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudo-metric.
+        \item The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
+        \item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous.
+    \end{enumerate}
+    The topology induced by $\seqi{d}$ is the \textbf{vector space topology induced by} $\seqi{[\cdot]}$. In addition,
+    \begin{enumerate}
+        \item[(U)] For any family $\seqj{[\cdot]}$ of seminorms continuous on $E$, the vector space topology induced by $\seqj{[\cdot]}$ is contained in the vector space topology induced by $\seqi{[\cdot]}$.
+    \end{enumerate}
+\end{definition}
+
+
+\begin{definition}[Gauge/Minkowski Functional]
+\label{definition:gauge}
+    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be a radial set, then the mapping
+    \[
+    [\cdot]_A: E \to [0, \infty) \quad x \mapsto \inf\bracsn{\lambda > 0| \lambda^{-1}x \in A}
+    \]
+    is the \textbf{gauge/Minkowski functional} of $A$, and
+    \begin{enumerate}
+        \item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
+        \item If $A$ is convex, then for any $x, y \in E$, $[x + y]_A \le [x]_A + [y]_A$.
+        \item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
+    \end{enumerate}
+    In particular,
+    \begin{enumerate}
+        \item If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
+        \item If $A$ is convex and circled, then $[\cdot]_A$ is a seminorm.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (2): Let $\lambda, \mu > 0$ such that $\lambda^{-1}x, \mu^{-1}y \in A$. By convexity, $t\lambda^{-1} + (1 - t)\mu^{-1}y \in A$ for all $t \in [0, 1]$. Let $t \in [0, 1]$ such that
+    \[
+    (\lambda + \mu)^{-1}(x + y) = t\lambda^{-1}x + (1 - t)\mu^{-1}y
+    \]
+    then $(\lambda + \mu)^{-1} \in A$, and $\lambda + \mu \ge [x + y]_A$. Thus $[x + y]_A \le [x]_A + [y]_A$.
+\end{proof}
+
+\begin{definition}[Locally Convex Space]
+\label{definition:locally-convex}
+    Let $E$ be a TVS over $\RC$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists a fundamental system of neighborhoods at $0$ consisting of convex sets.
+        \item There exists a fundamental system of neighbourhoods at $0$ consisting of convex, circled, and radial sets.
+        \item There exists a family of seminorms $\seqi{[\cdot]}$ that induces the topology on $E$.
+    \end{enumerate}
+    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.
+
+    $(2) \Rightarrow (3)$: 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.
+
+    $(3) \Rightarrow (1)$: For each $i \in I$ and $r > 0$, $\bracs{x \in E| [x]_i < r}$ is convex.
+\end{proof}

src/fa/lc/index.tex

@@ -0,0 +1,6 @@
+\chapter{Locally Convex Spaces}
+\label{chap:lc}
+
+
+\input{./src/fa/lc/convex.tex}
+\input{./src/fa/lc/continuous.tex}

src/fa/rs/bv.tex

@@ -0,0 +1,66 @@
+\section{Functions of Bounded Variation}
+\label{section:bv}
+
+
+\begin{definition}[Total Variation]
+\label{definition:total-variation}
+    Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, $f: [a, b] \to E$, and $P \in \scp([a, b])$ be a partition, then
+    \[
+    V_{\rho, p}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
+    \]
+    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.
+\end{definition}
+
+\begin{definition}[Bounded Variation, {{\cite[Proposition X.1.1]{Lang}}}]
+\label{definition:bounded-variation}
+    Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, and $f: [a, b] \to E$. If $[f]_{\var, \rho} < \infty$, then $f$ is of \textbf{bounded variation} with respect to $\rho$.
+
+    The space $BV([a, b]; E)$ is the set of functions $[a, b] \to E$ of bounded variation with respect to every continuous seminorm on $E$, and
+    \begin{enumerate}
+        \item $BV([a, b]; E)$ is a vector space.
+        \item For each continuous seminorm $\rho$ on $E$, $[\cdot]_{\var, \rho}$ is a seminorm on $BV([a, b]; E)$.
+        \item Let $\fF$ be a filter on $BV([a, b]; E)$ and $f: [a, b] \to E$. If
+        \begin{enumerate}
+            \item $\pi_x(\fF) \to f(x)$ for all $x \in [a, b]$.
+            \item For every continuous seminorm $\rho$ on $E$, there exists $U \in \fF$ such that $\sup_{g \in U}[g]_{\var, \rho} = M_\rho < \infty$.
+        \end{enumerate}
+        then $f \in BV([a, b]; E)$ with $[f]_{\var, \rho} \le M_\rho$.
+        \item For any $f \in BV([a, b]; E)$ and continuous seminorm $\rho$ on $E$, $\sup_{x \in [a, b]}\rho(f(x)) \le \rho(f(a)) + [f]_{\var, \rho}$.
+    \end{enumerate}
+    If $(E, \norm{\cdot}_E)$ is a normed space, then
+    \begin{enumerate}
+        \item[(5)] $f$ has at most countably many discontinuities.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (3): Let $\rho$ be a continuous seminorm on $E$ and $P \in \scp([a, b])$, then by assumption (a),
+    \[
+    V_{\rho, P}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
+    = \lim_{g, \fF}\sum_{j = 1}^n \rho(g(x_j) - g(x_{j - 1}))
+    = \lim_{g \in \fF}V_{\rho, P}(g)
+    \]
+    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
+    \begin{enumerate}
+        \item[(a)] $|E_k| \ge N - k$.
+        \item[(b)] $E_k \subset I_k^o$.
+    \end{enumerate}
+    for $k = 1$.
+
+    Let $k \le N$ and suppose inductively that $E_k, I_k$ have been constructed. Let $x_k \in E_k$, then by (b), there exists $\eps > 0$ such that $[x_k - \eps, x_k + \eps] \subset I_k$ and $|E_k \setminus [x_k - \eps, x_k + \eps]| \ge N - k$. Let $y_k \in [x_k - \eps, x_k + \eps]$ such that $\norm{f(x_k) - f(y_k)} \ge 1/n$, $I_{k + 1} = I_k \setminus [x_k - \eps, x_k + \eps]$, and $E_{k+1} = E_k \setminus [x_k - \eps, x_k + \eps]$, then $I_k$ and $E_k$ satisfies (a) and (b).
+
+    Therefore there exists pairs $\bracs{(x_k, y_k)|1 \le k \le N}$ such that $\norm{f(x_k) - f(y_k)}_E \ge 1/n$ for all $n$, and the smallest interval containing each $(x_k, y_k)$ are pairwise disjoint. Thus $[f]_{\var} \ge N/n$ for all $N \in \nat$, so $[f]_{\var} = \infty$.
+\end{proof}

src/fa/rs/index.tex

@@ -0,0 +1,7 @@
+\chapter{The Riemann-Stieltjes Integral}
+\label{chap:rs-integral}
+
+\input{./src/fa/rs/partition.tex}
+\input{./src/fa/rs/bv.tex}
+\input{./src/fa/rs/rs.tex}
+\input{./src/fa/rs/rs-bv.tex}

src/fa/rs/partition.tex

@@ -0,0 +1,39 @@
+\section{Partitions}
+\label{section:partitions}
+
+\begin{definition}[Partition]
+\label{definition:partition-interval}
+    Let $[a, b] \subset \real$, then a \textbf{partition} of $[a, b]$ is a sequence
+    \[
+    P = \seqfz{x_j} = [a = x_0 \le \cdots \le x_n = b]
+    \]
+    The collection $\scp([a, b])$ is the set of all partitions of $[a, b]$.
+\end{definition}
+
+\begin{definition}[Tagged Partition]
+\label{definition:tagged-partition}
+    Let $[a, b] \subset \real$, then a \textbf{tagged partition} of $[a, b]$ is a pair $(P = \seqfz{x_j}, c = \seqf{c_j})$ such that $c_j \in [x_{j - 1}, x_j]$ for each $1 \le j \le n$.
+
+    The collection $\scp_t([a, b])$ is the set of all tagged partitions of $[a, b]$.
+\end{definition}
+
+\begin{definition}[Mesh]
+\label{definition:mesh}
+    Let $P$ be a partition of $[a, b] \subset \real$, then
+    \[
+    \sigma(P) = \max_{1 \le j \le n}(x_j - x_{j - 1})
+    \]
+    is the \textbf{mesh} of $P$.
+\end{definition}
+
+\begin{definition}[Fine]
+\label{definition:partition-refinement}
+    Let $P = \seqfz[m]{x_j}, Q = \seqfz{y_j} \in \scp([a, b])$, then $Q$ is \textbf{finer} than $P$ if for every $0 \le j \le m$, there exists $0 \le k \le m$ such that $x_j = y_k$. For any $P, Q \in \scp([a, b])$, denote $P \le Q$ if $Q$ is finer than $P$, then
+    \begin{enumerate}
+        \item $\scp([a, b])$/$\scp_t([a, b])$ equipped with $\le$ is a upward-directed set.
+        \item If $P \le Q$, then $\sigma(P) \ge \sigma(Q)$.
+        \item For any $\eps > 0$, there exists $P \in \scp([a, b])$ with $\sigma(P) < \eps$.
+    \end{enumerate}
+
+    If $(P, c), (Q, d) \in \scp_t([a, b])$, then $(Q, d)$ is finer than $(P, c)$ if $Q$ is finer than $P$.
+\end{definition}

src/fa/rs/rs-bv.tex

@@ -0,0 +1,94 @@
+\section{Riemann-Stieltjes Integrals and Functions of Bounded Variationo}
+\label{section:rs-bv}
+
+\begin{proposition}
+\label{proposition:rs-bound}
+    Let $[a, b] \subset \real$, $E_1, E_2, H$ be locally convex spaces, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to E_2$.
+
+    Let $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_1$ on $E_1$ and $[\cdot]_2$ on $E_2$ such that for any $f \in RS([a, b], G)$,
+    \[
+    \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_1 \cdot [g]_{\var, 2}
+    \]
+\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 $(x, y) \in E_1 \times E_2$.
+
+    Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
+    \begin{align*}
+        [S(P, c, f, G)]_H &\le \sum_{j = 1}^n [f(c_j)[G(x_j) - G(x_{j - 1})]]_H \le \sum_{j = 1}^n [f(c_j)]_1[G(x_j) - G(x_{j - 1})]_2 \\
+        &\le \sup_{x \in [a, b]}[f]_1 \cdot V_{2, P}(G) \le \sup_{x \in [a, b]}[f]_1 \cdot [g]_{\var, 2}
+    \end{align*}
+\end{proof}
+
+\begin{proposition}
+\label{proposition:rs-complete}
+    Let $[a, b] \subset \real$, $E_1, E_2, H$ be locally convex spaces, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; E_2)$.
+
+    For each continuous seminorm $\rho$ on $H$ and $f: [a, b] \to E$, define
+    \[
+    [f]_{u, \rho} = \sup_{x \in [a, b]}f(\rho)
+    \]
+    Let $\net{f} \subset RS([a, b], G)$ such that:
+    \begin{enumerate}
+        \item[(a)] $\rho(f_\alpha - f) \to 0$ for all continuous seminorm $\rho$ on $E_1$.
+        \item[(b)] $\lim_{\alpha \in A}\int_a^b f_\alpha dG$ exists.
+    \end{enumerate}
+    then $f \in RS([a, b], G)$ and $\int_a^b f dG = \lim_{\alpha \in A}\int_a^b f_\alpha dG$. In particular,
+    \begin{enumerate}
+        \item If $H$ is complete, then condition (a) may be omitted.
+        \item If $H$ is sequentially complete and $A = \nat$, then condition (b) may be omitted.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
+    \begin{align*}
+        \rho\paren{S(P, c, f, G) - \lim_{\alpha \in A}\int_a^b f_\alpha dG} &\le \rho(S(P, c, f - f_\alpha, G)) \\
+        &+ \rho\paren{\int_a^b f_\alpha dG - \lim_{\alpha \in A}\int_a^b f_\alpha dG} \\
+        &+ \rho\paren{S(P, c, f_\alpha, G) - \int_a^b f_\alpha dG}
+    \end{align*}
+    Let $\rho$ be a continuous seminorm on $H$, and $[\cdot]_1$ and $[\cdot]_2$ be continuous seminorms on $E_1$ and $E_2$ such that $\rho(xy) \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
+
+    Let $\eps > 0$, then by assumption (a) and (b), there exists $\alpha \in A$ such that:
+    \begin{enumerate}
+        \item $[f - f_\alpha]_1 < \eps/(3[G]_{\var, 2})$.
+        \item $\rho\paren{\int_a^b f_\alpha dG - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps/3$.
+    \end{enumerate}
+    Since $f_\alpha \in RS([a, b], G)$, there exists $P_0 \in \scp([a, b])$ such that if $P \ge P_0$,
+    \begin{enumerate}
+        \item[(3)] $\rho\paren{S(P, c, f_\alpha, G) - \int_a^b f_\alpha dG} < \eps/3$.
+    \end{enumerate}
+    Thus for any $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$,
+    \[
+    \rho\paren{S(P, c, f, G) - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps
+    \]
+\end{proof}
+
+\begin{proposition}
+\label{proposition:rs-bv-continuous}
+    Let $[a, b] \subset \real$, $E_1, E_2$ be locally convex spaces, $H$ be a sequentially complete locally convex space, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map.
+
+    Let $f \in C([a, b]; E_1)$, $G \in BV([a, b]; E_2)$, then
+    \begin{enumerate}
+        \item $f \in RS([a, b], G)$.
+        \item For any $\seq{(P_n, t_n)} \subset \scp_t([a, b])$ with $\sigma(P_n) \to 0$,
+        \[
+        \int_a^b fdG = \limv{n}S(P_n, t_n, f, G)
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Let $\rho$ be a continuous seminorm on $H$, and $[\cdot]_1$ and $[\cdot]_2$ be continuous seminorms on $E_1$ and $E_2$ such that $\rho(xy) \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
+
+    Let $(P = \seqfz{x_j}, c = \seqf{c_j}), (Q = \seqfz[m]{y_j}, d = \seqf[m]{d_j}) \in \scp_t([a, b])$ with $Q \ge P$, then
+    \begin{align*}
+        \rho(S(P, c, f, G) - S(Q, d, f, G)) &\le \sum_{j = 1}^n \sum_{y_k \in [x_{j - 1}, x_j]}[f(c_j) - f(d_k)]_1[G(y_k) - G(y_{k - 1})]_2 \\
+        &\le \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \sigma(P) \end{array}}[f(x) - f(y)]_1 \cdot [G]_{\var, 2}
+    \end{align*}
+    Therefore for any two $(P, c), (Q, d) \in \scp_t([a, b])$,
+    \[
+    \rho(S(P, c, f, G) - S(Q, d, f, G)) \le 2 \cdot \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \max(\sigma(P), \sigma(Q)) \end{array}}[f(x) - f(y)]_1 \cdot [G]_{\var, 2}
+    \]
+    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 $\limv{n}S(P_n, t_n, f, G)$.
+\end{proof}

src/fa/rs/rs.tex

@@ -0,0 +1,68 @@
+\section{Riemann-Stieltjes Sums and Integrals}
+\label{section:tvs-rs-integral}
+
+\begin{definition}[Riemann-Stieltjes Sum, {{\cite[Section X.1]{Lang}}}]
+\label{definition:rs-sum}
+    Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to E_2$.
+
+    Let $f: [a, b] \to E_1$ and $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
+    \[
+    S(P, c, f, G) = \sum_{j = 1}^n f(c_j)[G(x_j) - G(x_{j - 1})]
+    \]
+    is the \textbf{Riemann-Stieltjes sum} of $f$ with respect to $G$ and $(P, c)$.
+\end{definition}
+
+\begin{definition}[Riemann-Stieltjes Integral, {{\cite[Section X.1]{Lang}}}]
+\label{definition:rs-integral}
+    Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to E_2$.
+
+    Let $f: [a, b] \to E_2$, then $f$ is \textbf{Riemann-Stieltjes integrable} with respect to $G$ if the limit
+    \[
+    \int_a^b f dG = \int_a^b f(t)dG(t) = \lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)
+    \]
+    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 $G$.
+\end{definition}
+
+\begin{lemma}[Summation by Parts, {{\cite[Proposition 1.4]{Lang}}}]
+\label{lemma:sum-by-parts}
+    Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $f: [a, b] \to E_1$, $G: [a, b] \to E_2$, and $(P, c) \in \scp_t([a, b])$, then
+    \[
+    S(P, c, f, G) + S(P', c', G, f) = f(b)G(b) - f(a)G(a)
+    \]
+    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}
+    Denote $c_0 = a$ and $c_{n+1} = b$, then
+    \begin{align*}
+        S(P, c, f, G) &= \sum_{j = 1}^n f(c_j)[G(x_j) - G(x_{j - 1})]
+        = \sum_{j = 1}^n f(c_j)G(x_j) - \sum_{j = 1}^n f(c_j)G(x_{j - 1}) \\
+        &= f(c_n)G(x_n)- f(c_0)G(x_0) + \sum_{j = 1}^n f(c_{j - 1})G(x_{j-1}) - \sum_{j = 1}^n f(c_j)G(x_{j - 1}) \\
+        &= f(c_n)G(x_n)- f(c_0)G(x_0) - \sum_{j = 1}^n G(x_{j - 1})[f(c_j) - f(c_{j - 1})] \\
+        &= f(c_{n+1})G(x_n) - f(c_0)G(x_0) - \sum_{j = 1}^{n+1}G(x_{j - 1})[f(c_j) - f(c_{j - 1})] \\
+        &= f(b)G(b) - f(a)G(a) - S(P', c', G, f)
+    \end{align*}
+\end{proof}
+
+\begin{theorem}[Integration by Parts]
+\label{theorem:rs-ibp}
+    Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map.
+
+    Let $f: [a, b] \to E_1$ and $G: [a, b] \to E_2$, then $f \in RS([a, b], G)$ if and only if $G \in RS([a, b], f)$. In which case,
+    \[
+    \int_a^b f dG + \int_a^b G df = f(b)G(b) - f(a)G(a)
+    \]
+\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 $\int_a^b fdG - S(Q', d', G, f) \in U$.
+\end{proof}

src/fa/tvs/bounded.tex

@@ -0,0 +1,38 @@
+\section{Bounded Sets}
+\label{section:bounded}
+
+\subsection{Bounded Sets}
+\label{subsection:tvs-bounded}
+
+\begin{definition}[Bounded]
+\label{definition:bounded}
+    Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then $B$ is \textbf{bounded} if for every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$.
+\end{definition}
+
+\begin{proposition}[{{\cite[1.5.1]{SchaeferWolff}}}]
+\label{proposition:bounded-operations}
+    Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$ be bounded, then the following sets are bounded:
+    \begin{enumerate}
+        \item Any $C \subset B$.
+        \item The closure $\ol{B}$.
+        \item $\lambda B$ where $\lambda \in K$.
+        \item $A \cup B$.
+        \item $A + B$.
+    \end{enumerate}
+\end{proposition}
+\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, $\lambda U \supset \ol B$.
+
+    (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 $\lambda' V \supset B$.
+
+    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/completion.tex

@@ -0,0 +1,34 @@
+\section{The Hausdorff Completion}
+\label{section:tvs-complete}
+
+\begin{definition}[Hausdorff Completion of TVS]
+\label{definition:tvs-completion}
+    Let $E$ be a TVS over $K \in \RC$, then there exists $(\wh E, \iota)$ such that:
+    \begin{enumerate}
+        \item $\wh E$ is a complete Hausdorff TVS.
+        \item $\iota \in L(E; \wh E)$.
+        \item[(U)] For any $(F, T)$ satisfying (1) and (2), there exists a unique $\ol{T} \in L(\wh E; F)$ such that the following diagram commutes:
+    \end{enumerate}
+    Moreover,
+    \begin{enumerate}
+        \item[(4)] $\iota(E)$ is dense in $\wh E$.
+    \end{enumerate}
+    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}).
+
+    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
+    \[
+    \xymatrix{
+    \widehat E \times \widehat E \ar@{->}[r] & \widehat E &  & K \times \widehat E \ar@{->}[r] & \widehat E \\
+    E \times E \ar@{->}[u] \ar@{->}[r] & E \ar@{->}[u] &  & K \times E \ar@{->}[u] \ar@{->}[r] & E \ar@{->}[u]
+    }
+    \]
+
+    By continuity and the density of $\iota(E)$ in $E$, $\wh E$ with these operations forms a TVS, and $T$ is linear.
+\end{proof}
+
+\begin{remark}[{{\cite[Section 1.1]{SchaeferWolff}}}]
+	The Hausdorff completion works in general with arbitrary valuated fields. Though the completion yields a TVS over the completion of the field, the field need not to be complete.
+\end{remark}

src/fa/tvs/continuous.tex

@@ -0,0 +1,86 @@
+\section{Continuous Linear Maps}
+\label{section:tvs-linear-maps}
+
+
+\begin{definition}[Continuous Linear Map]
+\label{definition:continuous-linear}
+    Let $E, F$ be TVSs over $K \in \RC$, and $T \in \hom({E, F})$ be a linear map, then the following are equivalent:
+    \begin{enumerate}
+        \item $T$ is uniformly continuous.
+        \item $T$ is continuous.
+        \item $T$ is continuous at $0$.
+    \end{enumerate}
+    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}.
+
+    $(3) \Rightarrow (1)$: 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.
+
+    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}
+
+\begin{definition}[Continuous Multilinear Map]
+\label{definition:continuous-multilinear}
+    Let $\seqf{E}$, $F$ be TVSs over $K \in \RC$, then the set $L^n(E_1, \cdots, E_n; F) = L^n(\seqf{E_j}; F)$ is the space of all continuous $n$-linear maps from $\prod_{j = 1}^n E_j$ to $F$.
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:continuous-bounded}
+    Let $E, F$ be TVSs over $K \in \RC$ and $T \in L(E; F)$, then for any $B \subset E$ bounded, $T(B)$ is also bounded.
+\end{proposition}
+\begin{proof}
+    Let $U \in \cn_F(0)$, then $T^{-1}(U) \in \cn_E(0)$, so there exists $\lambda \in K$ such that $\lambda T^{-1}(U) = T^{-1}(\lambda U) \supset B$ and $\lambda U \supset T(B)$.
+\end{proof}
+
+
+\begin{definition}[Initial Uniformity]
+\label{definition:tvs-initial}
+    Let $E$ be a vector space over $K \in \RC$, $\seqi{F}$ be TVSs, and $\seqi{T}$ where $T_i \in \hom(E; F_i)$ for all $i \in I$, then there exists a uniformity $\fU$ on $E$ such that:
+    \begin{enumerate}
+        \item For each $i \in I$, $T_i \in L(E; F_i)$.
+        \item[(U)] If $\mathfrak{V}$ is a uniformity on $E$ satisfying $(1)$, then $\mathfrak{V} \supset \fU$.
+    \end{enumerate}
+    Moreover,
+    \begin{enumerate}
+        \item[(3)] $\fU$ is translation-invariant.
+        \item[(4)] $E$ equipped with the topology induced by $\fU$ is a topological vector space.
+    \end{enumerate}
+    The uniformity and its induced topology are the \textbf{initial uniformity/topology} induced by $\seqi{T}$.
+\end{definition}
+\begin{proof}
+    (1), (U): By \ref{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 $\abs{\lambda - \lambda'} < \min_{j \in J}\eps$, $(x + y, x' + y'), (\lambda x, \lambda' x') \in V$.
+\end{proof}
+
+\begin{definition}[Product Topology]
+\label{definition:tvs-product}
+    Let $\seqi{E}$ be TVSs over $K \in \RC$ and $E = \prod_{i \in I}E_i$ be their product as a vector space, and $\fU$ be the initial uniformity generated by the projection maps, then
+    \begin{enumerate}
+        \item $E$ equipped with the topology induced by $\fU$ is a topological vector space.
+        \item[(U)] For any TVS $F$ over $K$ and $\seqi{T}$ where $T_i \in L(F; E_i)$ for each $i \in I$, there exists a unique $U \in L(F; E)$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+        F \ar@{->}[rd]^{T_i} \ar@{->}[d]_{T} &  \\
+        \prod_{i \in I}E_i \ar@{->}[r]_{\pi_i} & E_i
+        }
+        \]
+    \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

@@ -0,0 +1,149 @@
+\section{Vector Space Topologies}
+\label{section:tvs-topology}
+
+
+\begin{definition}[Topological Vector Space]
+\label{definition:tvs}
+    Let $E$ be a vector space over $K \in \bracs{\real, \complex}$ and $\topo \subset 2^E$ be a topology. If
+    \begin{enumerate}
+        \item[(TVS1)] $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous.
+        \item[(TVS2)] $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous.
+    \end{enumerate}
+    then the pair $(E, \topo)$ is a \textbf{topological vector space}.
+\end{definition}
+
+\begin{definition}[Translation-Invariant Topology]
+\label{definition:translation-invariant-topology}
+    Let $E$ be a vector space and $\topo$ be a topology on $E$, then $\topo$ is \textbf{translation-invariant} if for any $U \in \topo$ and $y \in E$, $U + y \in \topo$.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:tvs-translation-invariant}
+    Let $E$ be a TVS over $K \in \RC$, then the topology of $E$ is translation-invariant.
+\end{lemma}
+\begin{proof}
+    Let $U \subset E$ open and $y \in E$, then $U + y$ is the preimage of $U$ by the map $x \mapsto x - y$. By (TVS1), $U + y$ is open.
+\end{proof}
+
+\begin{definition}[Translation-Invariant Uniformity]
+\label{definition:translation-invariant-uniformity}
+    Let $E$ be a vector space, $\fU$ be a uniformity on $E$, and $U \in \fU$, then $U$ is \textbf{translation-invariant} if for every $z \in E$,
+    \[
+    U = \bracs{(x + z, y + z)|(x, y) \in U}
+    \]
+    and $\fU$ is \textbf{translation-invariant} if there exists a fundamental system of translation-invariant entourages.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:translation-invariant-symmetric}
+    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.
+\end{proof}
+
+
+\begin{proposition}[{{\cite[1.1.4]{SchaeferWolff}}}]
+\label{proposition:tvs-uniform}
+    Let $E$ be a TVS over $K \in \bracs{\real, \complex}$, then:
+    \begin{enumerate}
+        \item There exists a unique translation-invariant uniformity $\fU$ on $E$ that induces the topology on $E$.
+        \item For each neighbourhood $V \in \cn(0)$, let $U_V = \bracs{(x, y) \in E^2| x - y \in V}$, then for any fundamental system of neighbourhoods $\fB_0$ at $0$, $\fB = \bracs{U_V| V \in \fB_0}$ is a fundamental system of entourages for $\fU$.
+    \end{enumerate}
+    The space $E$ will always be assumed to be equipped with its translation-invariant uniformity.
+\end{proposition}
+\begin{proof}
+    (2): Firstly, for any $V \in \fB_0$, $U_V$ is translation-invariant.
+    \begin{enumerate}
+        \item[(FB1)] For any $V, V' \in \fB_0$, there exists $W \in \fB_0$ such that $W \subset V \cap V'$. In which case, $U_V \cap U_{V'} \supset U_W \in \fB$.
+        \item[(UB1)] For any $V \in \fB_0$, $0 \in V$, so $\Delta \subset U_V$.
+        \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 $E$.
+
+    (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 $V \in \mathfrak{V}$. 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 $\fU \supset \mathfrak{V}$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tvs-closure}
+    Let $E$ be a TVS over $K \in \RC$, $A \subset E$, and $\fB \subset \cn(0)$ be a fundamental system of neighbourhoods, then
+    \[
+    \ol{A} = \bigcap_{U \in \fB}\bracs{A + U| U \in \fB}
+    \]
+\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
+    \[
+    \ol{A} = \bigcap_{V \in \fB}\bracs{U_V(A)| U \in \fB} = \bigcap_{V \in \fB}U + A
+    \]
+\end{proof}
+
+\begin{proposition}[{{\cite[1.1.1]{SchaeferWolff}}}]
+\label{proposition:tvs-set-operations}
+    Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$, then:
+    \begin{enumerate}
+        \item If $A$ is open, then $A + B$ is open.
+        \item If $A$ is closed and $B$ is compact, then $A + B$ is closed.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    $(1)$: For every $x \in B$, $A + x$ is open by \ref{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
+    \[
+    y \in \bigcap_{U \in \fF}\overline{U - B} = \bigcap_{U \in \fF}\overline{(-B) + U}
+    \]
+
+    By \ref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
+    \[
+    y \in \bigcap_{U \in \fF}\overline{(-B) + U} \subset \bigcap_{U \in \fF}[(-B) + U + U]
+    \]
+    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}
+
+\begin{definition}[Balanced/Circled]
+\label{definition:balanced}
+    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$, then $A$ is \textbf{balanced/circled} if $\lambda A \subset A$ for all $\lambda \in K$ with $\abs{\lambda} \le 1$.
+\end{definition}
+
+\begin{definition}[Absorbing/Radial]
+\label{definition:absorbing}
+    Let $E$ be a vector space over $K \in \RC$ and $A, B \subset E$, then $A$ \textbf{absorbs} $B$ if there exists $\lambda \in K$ such that $\lambda A \supset B$, and $A$ is \textbf{absorbing/radial} if it absorbs every point in $E$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:tvs-good-neighbourhood-base}
+    Let $E$ be a topological vector space over $K \in \RC$, then
+    \begin{enumerate}
+        \item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of balanced and absorbing sets.
+        \item The fundamental system of neighbourhoods in $(1)$ can be taken to be open or closed.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Firstly, (TVS2) implies that every neighbourhood of $0$ is balanced.
+
+    By \ref{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.
+\end{proof}

src/fa/tvs/index.tex

@@ -0,0 +1,6 @@
+\chapter{Topological Vector Spaces}
+\label{chap:tvs}
+
+\input{./src/fa/tvs/definition.tex}
+\input{./src/fa/tvs/bounded.tex}
+\input{./src/fa/tvs/continuous.tex}

src/topology/index.tex

@@ -0,0 +1,5 @@
+\part{General Topology}
+\label{part:-part-topology}
+
+\input{./src/topology/main/index.tex}
+\input{./src/topology/uniform/index.tex}

src/topology/main/baire.tex

@@ -0,0 +1,24 @@
+\section{Baire Spaces}
+\label{section:baire}
+
+
+\begin{definition}[Baire Space]
+\label{definition:baire}
+    Let $X$ be a topological space, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $\seq{A_n}$ nowhere dense, $\bigcup_{n \in \nat}A_n \subsetneq X$.
+        \item For any $\seq{A_n}$ closed with empty interior, $\bigcup_{n \in \nat}A_n \subsetneq X$.
+        \item For any $\seq{A_n}$ closed with $\bigcup_{n \in \nat}A_n = X$, there exists $N \in \nat$ such that $\bigcup_{n \le N}A_n$ has non-empty interior.
+        \item For any $\seq{U_n}$ open and dense, $\bigcap_{n \in \nat}U_n$ is dense.
+    \end{enumerate}
+    If the above holds, then $X$ is a \textbf{Baire space}.
+\end{definition}
+
+\begin{theorem}[Baire Category Theorem]
+\label{theorem:baire}
+    Let $X$ be a topological space, then the following are sufficient conditions for $X$ to be Baire:
+    \begin{enumerate}
+        \item $X$ is completely metrisable.
+        \item $X$ is locally compact.
+    \end{enumerate}
+\end{theorem}

src/topology/main/continuity.tex

@@ -0,0 +1,34 @@
+\section{Continuous Maps}
+\label{section:continuity}
+
+
+\begin{definition}[Continuity]
+\label{definition:continuity}
+    Let $X$ and $Y$ be topological spaces, $f: X \to Y$ be a function, and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $V \in \cn(f(x))$, $f^{-1}(V) \in \cn(x)$.
+        \item For each filter base $\fB \subset 2^X$ converging to $x$, $f(\fB)$ converges to $f(x)$.
+    \end{enumerate}
+    If the above holds, then $f$ is \textbf{continuous at} $x \in X$.
+
+    The following are also equivalent:
+    \begin{enumerate}
+        \item For each $U \subset Y$ open, $f^{-1}(U)$ is open in $X$.
+        \item $f$ is continuous at every $x \in X$.
+        \item For each convergent filter base $\fB \subset 2^X$, $f(\fB)$ is convergent.
+    \end{enumerate}
+    If the above holds, then $f$ is \textbf{continuous}.
+
+    The collection $C(X; Y)$ is the space of all continuous functions from $X$ to $Y$.
+\end{definition}
+\begin{proof}
+    Local continuity, $(1) \Rightarrow (2)$: Let $V \in \cn(f(x))$, then $f^{-1}(V) \in \cn(x)$. Since $\fB$ converges to $x$, there exists $B \in \fB$ with $B \subset f^{-1}(V)$. In which case, $f(B) \subset V$ and $V \in \fF(\fB)$ by (F1).
+
+    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, $(2) \Rightarrow (1)$: 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 $f^{-1}(V) \in \cn(x)$.
+
+    Global continuity, $(2) \Leftrightarrow (3)$: Follows from local continuity.
+\end{proof}

src/topology/main/definition.tex

@@ -0,0 +1,112 @@
+\section{Definitions}
+\label{section:top-definitions}
+
+\begin{definition}[Topological Space]
+\label{definition:topspace}
+    Let $X$ be a non-empty set. A \textbf{topology} over $X$ is a family $\topo \subset 2^X$ such that
+    \begin{enumerate}
+        \item[(O1)] $\emptyset \in \topo$ and $X \in \topo$.
+        \item[(O2)] For any $U, V \in \topo$, $U \cap V \in \topo$.
+        \item[(O3)] For any $\seqi{U} \subset \topo$, $\bigcup_{i \in I}U_i \in \topo$.
+    \end{enumerate}
+    The elements of $\topo$ are known as \textbf{open sets}, and the pair $(X, \topo)$ is known as a \textbf{topological space}.
+\end{definition}
+
+\begin{definition}[Closed Set]
+\label{definition:closedset}
+    Let $(X, \topo)$ be a topological space, then $A \subset X$ is \textbf{closed} if $A^c \in \topo$.
+\end{definition}
+
+\begin{definition}[Separation Axioms]
+\label{definition:separation}
+    Let $(X, \topo)$ be a topological space, then $X$ may satisfy the following \textbf{separation axioms}:
+    \begin{enumerate}
+        \item[(T0)] For any $x, y \in X$ with $x \ne y$, there exists $U \in \topo$ with $x \in U$ and $y \not\in U$, or there exists $U \in \topo$ with $x \not\in U$ and $y \in U$.
+        \item[(T1)] For any $x, y \in X$ with $x \ne y$, there exists $U \in \topo$ with $x \in U$ and $y \not\in U$.
+        \item[(T2)] For any $x, y \in X$ with $x \ne y$, there exists $U, V \in \topo$ with $x \in U$, $y \in V$, and $U \cap V = \emptyset$.
+        \item[(T3)] $X$ is (T1), and for any $x \in X$ and $A \subset X$ closed with $x \not\in A$, there exists $U, V \in \topo$ such that $x \in U$, $A \subset V$, and $U \cap V = \emptyset$.
+        \item[(T4)] $X$ is (T1), and for any $A, B \subset X$ closed with $A \cap B = \emptyset$, there exists $U, V \in \topo$ such that $A \subset U$, $B \subset V$, and $U \cap V = \emptyset$.
+    \end{enumerate}
+\end{definition}
+
+\begin{lemma}
+\label{lemma:t1}
+    Let $X$ be a topological space, then $X$ satisfies (T1) if and only if $\bracs{x}$ is closed for all $x \in X$.
+\end{lemma}
+\begin{proof}
+    $(\Rightarrow)$: Let $x \in X$, then for each $y \in X \setminus {x}$, there exists $U_y \subset X$ open such that $x \not\in U_y$. Thus $U^c = \bigcup_{y \in X \setminus \bracs{x}}U_y$ is open.
+
+   $(\Leftarrow)$: Let $x, y \in X$ with $x \ne y$, then $y \in \bracs{x}^c$, $x \not\in \bracs{x}^c$, and $\bracs{x}^c$ is open.
+\end{proof}
+
+
+\begin{definition}[Base]
+\label{definition:base}
+    Let $(X, \topo)$ be a topological space, then a \textbf{base} for $\topo$ is a family $\cb \subset \topo$ such that:
+    \begin{enumerate}
+        \item For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
+        \item For every $x \in X$ and $U \subset X$ open with $x \in U$, there exists $V \in \cb$ such that $x \in V \subset U$.
+    \end{enumerate}
+    In which case,
+    \[
+    \topo = \topo(\cb) = \bracs{\bigcup_{i \in I}U_i \bigg | \seqi{U} \subset \cb, I \text{ index set}}
+    \]
+    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 $x \in U$.
+        \item[(TB2)] For every $x \in X$ and $U, V \in \cb$ such that $x \in U \cap$, there exists $W \in \cb$ such that $x \in W \subset U \cap V$.
+    \end{enumerate}
+    then $\topo(\cb)$ is a topology on $X$, and $\cb$ is a base for $\topo(\cb)$.
+\end{definition}
+\begin{proof}
+    Let $U \in \topo$. If $U = \emptyset$, then $U$ is a union over an empty index set. Otherwise, for each $x \in U$, there exists $V_x \in \cb$ such that $x \in V_x \subset U$. In which case, $U = \bigcup_{x \in U}V_x \in \topo$ and $\topo \subset \topo(\cb)$. On the other hand, since $\cb \subset \topo$, $\topo \supset \topo(\cb)$ by (O3).
+
+    For the converse, $\emptyset \in \topo(\cb)$ as a union over an empty index set and $X = \bigcup_{U \in \cb}U \in \topo(\cb)$ by (TB1). For any $\seqi{U}, \seqj{V} \subset \cb$,
+    \[
+    \paren{\bigcup_{i \in I}U_i} \cap \paren{\bigcup_{j \in J}V_j} = \bigcup_{i \in I}\bigcup_{j \in J}U_i \cap V_j
+    \]
+    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).
+\end{proof}
+
+\begin{definition}[Generated Topology]
+\label{definition:generated-topology}
+    Let $X$ be a set and $\ce \subset 2^X$ such that $\bigcup_{U \in \ce}U = X$, then the smallest topology on $X$ containing $\ce$ is given by
+    \[
+    \topo(\ce) = \bracs{\bigcup_{i \in I}U_i \bigg | U_i \in \cb(\ce)}
+    \]
+    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 $\topo(\ce)$.
+\end{proof}
+
+\begin{definition}[Initial Topology]
+\label{definition:initial-topology}
+    Let $X$ be a set, $\bracsn{(Y_j, \topo_i)}$ be a family of topological spaces, and $\seqi{f}$ be a family of maps such that $f_i: X \to Y_i$ for each $i \in I$, then there exists a topology $\topo$ on $X$ such that:
+    \begin{enumerate}
+        \item For each $i \in I$, $f_i \in C(X; Y_i)$.
+        \item[(U)] If $\mathcal{S}$ is a topology on $X$ satisfying $(1)$, then $\mathcal{S} \supset \topo$.
+        \item The family
+
+        \[
+        \mathcal{B} = \bracs{\bigcap_{j \in J}f_j^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \topo_j}
+        \]
+
+        is a base for $\topo$.
+    \end{enumerate}
+    The topology $\topo$ is known a the \textbf{initial/weak topology} generated by the maps $\seqi{f}$.
+\end{definition}
+\begin{proof}
+    Let $\topo$ be the topology genereated by sets of the form $\ce = \bracs{f_i^{-1}(U_i)| i \in I, U_i \in \topo_i}$. Let $\topo$ be the topology generated by $\ce$, then
+    \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 $\topo$.
+    \end{enumerate}
+\end{proof}

src/topology/main/filters.tex

@@ -0,0 +1,116 @@
+\section{Filters}
+\label{section:filters}
+
+
+\begin{definition}[Filter]
+\label{definition:filter}
+    Let $X$ be a set, a \textbf{filter} $\fF \subset X^2$ is a non-empty family of sets such that:
+    \begin{enumerate}
+        \item[(F1)] For any $E \in \fF$ and $X \supset F \supset E$, $F \in \fF$.
+        \item[(F2)] For any $E, F \in \fF$, $E \cap F \in \fF$.
+        \item[(F3)] $\emptyset \not\in \fF$
+    \end{enumerate}
+\end{definition}
+
+
+\begin{definition}[Filter Base]
+\label{definition:filterbase}
+    Let $X$ be a set, $\fF \subset 2^X$ be a filter, and $\fB \subset \fF$, then $\fB$ is a \textbf{filter base} for $\fF$ if for every $F \in \fF$, there exists $E \in \fB$ such that $E \subset F$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:filterbasecriterion}
+    Let $X$ be a set, $\fF \subset 2^X$ be a filter, and $\fB \subset \fF$ be a filter base, then:
+    \begin{enumerate}
+        \item[(FB1)] For any $E, F \in \fB$, there exists $G \in \fB$ such that $G \subset E \cap F$.
+        \item[(FB2)] $\emptyset \not\in \fB$.
+    \end{enumerate}
+    Conversely, if $\fB \subset 2^X$ is a non-empty collection that satisfies (FB1) and (FB2), then $\fB$ is a base for the filter
+    \[
+    \fF = \bracs{F \subset X| \exists E \in \fB: E \subset F}
+    \]
+\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 $G \subset E \cap F$.
+
+    Filter Base $\Rightarrow$ (FB2): By (F3), $\emptyset \not\in \fB \subset \fU$.
+
+    (FB1) + (FB2) $\Rightarrow$ (F1): Let $E \in \fF$, then there exists $E_0 \in \fB$ such that $E_0 \subset E$. Thus for any $X \supset F \supset E$, $E_0 \subset F \in \fF$.
+
+    (FB1) + (FB2) $\Rightarrow$ (F2): Let $E, F \in \fF$, then there exists $E_0, F_0 \in \fB$ such that $E_0 \subset E$ and $F_0 \subset F$. By (FB1), there exists $G \in \fB$ such that $G \subset E_0 \cap F_0$, then $G \subset E \cap F \in \fF$.
+
+    (FB1) + (FB2) $\Rightarrow$ (F3): Let $E \in \fF$, then there exists $E_0 \in \fB$ such that $E_0 \subset E$. By (FB2), $E_0 \ne \emptyset$, so $E \ne \emptyset$ as well.
+
+    Finally, $\fB$ is a filter base for $\fF$ by definition.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:imagefilterbase}
+    Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^X$ be a filter base, then $f(\fB) = \bracs{f(E)| E \in \fB}$ is also a filter base.
+\end{proposition}
+\begin{proof}
+    (FB1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f(G) \subset f(E \cap F) \subset f(E) \cap f(F)$.
+
+    (FB2): For any $f(E) \in f(\fB)$, $E \ne \emptyset$ implies that $f(E) \ne \emptyset$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:preimage-filterbase}
+    Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^X$ be a filter base. If $\emptyset \not\in f^{-1}(\fB) = \bracs{f^{-1}(E)| E \in \fB}$, then $f^{-1}(\fB)$ is also a filter base.
+\end{proposition}
+\begin{proof}
+    (FB1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f^{-1}(G) \subset f^{-1}(E \cap F) = f^{-1}(E) \cap f^{-1}(F)$.
+
+    (FB2): By assumption, $\emptyset \not \in f^{-1}(\fB)$.
+\end{proof}
+
+
+\begin{definition}[Filter Subbase]
+\label{definition:filtersubbase}
+    Let $X$ be a set and $\fB_0 \subset 2^X$ be a non-empty collection, then $\fB_0$ is a \textbf{filter subbase} if for any $\seqf{E_j} \subset \fB_0$, $\bigcap_{j = 1}^n E_j \ne \emptyset$.
+\end{definition}
+
+\begin{definition}[Generated Filter]
+\label{definition:generatedfilter}
+Let $X$ be a set and $\fB_0 \subset 2^X$ be a filter subbase, then there exists a filter containing $\fB_0$.
+
+The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \textbf{generated by} $\fB_0$, which is given by $\fF(\fB_0) = \bracs{E \subset X| \exists F \in \fB: F \subset E}$, where
+\[
+\fB = \bracs{\bigcap_{j = 1}^n E_j \bigg | \seqf{E_j} \subset \fB_0, n \in \nat^+}
+\]
+\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 $\fF = \bracs{E \subset X| \exists F \in \fB: F \subset E}$.
+
+    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}
+
+\begin{definition}[Convergence]
+\label{definition:filterconvergence}
+    Let $X$ be a topological space, $\fB \subset 2^X$ be a filter base, and $x \in X$, then \textbf{$\fB$ converges} to $x$ if $\cn(x) \subset \fF(\fB)$.
+
+    If $A \subset X$ and $\fB \subset 2^A$, then $\fB$ converges to $x$ if $\fF(\fB) \supset \bracsn{U \cap A| U \in \cn(x)}$.
+\end{definition}
+
+\begin{definition}[Accumulation Point]
+\label{definition:accumulationpoint}
+    Let $X$ be a topological space, $\fB \subset 2^X$ be a filter base, and $x \in X$, then $x$ is an \textbf{accumulation point} of $\fB$ if $x \in \bigcap_{E \in \fB}\ol{E}$.
+\end{definition}
+
+
+\begin{definition}[Limit]
+\label{definition:limit}
+    Let $X, Y$ be topological spaces, $A \subset X$, and $f: A \to Y$ be a function. For any filter base $\fB \subset 2^A$, if $f(\fB)$ converges to $y \in Y$, then $y = \lim_{x, \fB}f(x)$ is a \textbf{limit} of $f$ with respect to $\fB$.
+
+    For any $x_0 \in \overline{A}$, let $\fF = \bracs{U \cap A| U \in \cn(x_0)}$ be the trace of $\cn(x_0)$ on $A$, then $\fF \subset 2^A$ is a filter. If $f(\fF)$ converges to $y \in Y$, then
+    \[
+    y = \lim_{x \to x_0 \\ x \in A}f(x)
+    \]
+    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

@@ -0,0 +1,50 @@
+\section{Hausdorff Spaces}
+\label{section:hausdorff}
+
+
+\begin{definition}[Hausdorff]
+\label{definition:hausdorff}
+    Let $X$ be a topological space, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $x, y \in X$ with $x \ne y$, there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \cap V = \emptyset$.
+        \item For every $x \in X$, $\bracs{x} = \bigcap_{U \in \cn(x)}\overline{U}$.
+        \item Every convergent filter in $X$ has exactly one cluster point.
+        \item Every filter in $X$ converges to at most one point.
+        \item For any index set $I$, the diagonal $\Delta$ is closed in $X^I$.
+        \item The diagonal $\Delta$ is closed in $X \times X$.
+    \end{enumerate}
+    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}, $y \not\in \overline{U} \subset \bigcap_{U \in \cn(x)}\overline{U}$.
+
+    $(2) \Rightarrow (3)$: 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 $i, j \in I$, $x \in \Delta$.
+
+    $(5) \Rightarrow (6)$: Take $I = \bracs{1, 2}$.
+
+    $(6) \Rightarrow (1)$: Let $x, y \in X$ with $x \ne y$, then $\Delta^c \in \cn(x, y)$, and there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $(x, y) \in \pi_1^{-1}(U) \cap \pi_2^{-1}(V) \subset \Delta^c$. Thus $U \cap V = \emptyset$ contain the desired neighbourhoods.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:continuousextensionunique}
+    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}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:product-hausdorff}
+    Let $\seqi{X}$ be Hausdorff spaces, then $\prod_{i \in I}X_i$ is Hausdorff.
+\end{proposition}
+\begin{proof}
+    Let $x, y \in \prod_{i \in I}X_i$ with $x \ne y$, then there exists $i \in I$ such that $\pi_i(x) \ne \pi_i(y)$. In which case, there exists $U \in \cn(\pi_i(x))$ and $V \in \cn(\pi_i(y))$ with $U \cap V = \emptyset$. Thus $\pi_i^{-1}(U) \in \cn(x)$, $\pi_i^{-1}(V) \in \cn(y)$, and $\pi_i^{-1}(U) \cap \pi_i^{-1}(V) = \emptyset$.
+\end{proof}

src/topology/main/index.tex

@@ -0,0 +1,14 @@
+\chapter{Topological Spaces}
+\label{chap:topological-spaces}
+
+\input{./src/topology/main/definition.tex}
+\input{./src/topology/main/filters.tex}
+\input{./src/topology/main/nets.tex}
+\input{./src/topology/main/neighbourhoods.tex}
+\input{./src/topology/main/interiorclosureboundary.tex}
+\input{./src/topology/main/continuity.tex}
+\input{./src/topology/main/product.tex}
+\input{./src/topology/main/hausdorff.tex}
+\input{./src/topology/main/regular.tex}
+\input{./src/topology/main/metric.tex}
+\input{./src/topology/main/baire.tex}

src/topology/main/interiorclosureboundary.tex

@@ -0,0 +1,110 @@
+\section{Interior, Closure, Boundary}
+\label{section:icb}
+
+\begin{definition}[Interior]
+\label{definition:interior}
+    Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
+    \begin{enumerate}
+        \item $A \in \cn(x)$.
+        \item There exists $U \in \fB$ with $U \subset A$.
+        \item There exists $U \in \cn(x)$ with $U \subset A$.
+    \end{enumerate}
+    The set of all points satisfying the above is the \textbf{interior} $A^o$ of $A$, which is the largest open set contained in $A$.
+\end{definition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: Since $\fB$ is a fundamental system of neighbourhoods, there exists $U \in \fB$ with $U \subset A$.
+
+    $(2) \Rightarrow (3)$: $\fB \subset \cn(x)$.
+
+    $(3) \Rightarrow (1)$: 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}.
+\end{proof}
+
+\begin{definition}[Closure]
+\label{definition:closure}
+    Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
+    \begin{enumerate}
+        \item For every $B \supset A$ closed, $x \in B$.
+        \item For every $U \in \cn(x)$, $U \cap A \ne \emptyset$.
+        \item For every $U \in \fB$, $U \cap A \ne \emptyset$.
+        \item There exists a filter $\fF \subset 2^A$ that converges to $\fF$.
+    \end{enumerate}
+
+    The set $\ol{A}$ of all points satisfying the above is the \textbf{closure} of $A$ in $X$. By (1), it is the smallest closed set containing $A$.
+\end{definition}
+\begin{proof}
+    $\neg (2) \Rightarrow \neg (1)$: Let $U \in \cn(x)$ such that $U \cap A = \emptyset$ and $V \in \cn^o(x)$ with $V \subset U$. Since $x \in V$ by (V1), $V^c \supset A$ is closed with $x \not\in V^c$.
+
+    $(2) \Rightarrow (3)$: $\cn(x) \supset \fB$.
+
+    $(3) \Rightarrow (4)$: Let $\fF = \bracs{V \cap A| V \in \cn(x)}$. For any $V \cap A \in \fF$, there exists $W \in \fB$ such that $W \subset V$. Since $W \cap A \ne \emptyset$, $V \cap A \ne \emptyset$ as well. Thus $\emptyset \not\in \fF$, and $\fF$ is a filter in $A$, and $\fF$ converges to $x$ by definition.
+
+    $\neg (1) \Rightarrow \neg (4)$: Let $B \supset A$ such that $x\not\in B$, then $B^c \in \cn^o(x)$ with $B^c \cap A = \emptyset$. Thus there exists no filter on $A$ containing $B^c \cap A$, and no filter on $A$ converging to $x$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:closure-of-image}
+    Let $X, Y$ be topological spaces, $A \subset X$, and $f: X \to Y$ be continuous, then $f(\ol{A}) \subset \ol{f(A)}$.
+\end{proposition}
+\begin{proof}
+    Since $f$ is continuous, $f^{-1}(\ol{f(A)})$ is closed and contains $A$.
+\end{proof}
+
+
+\begin{definition}[Dense]
+\label{definition:dense}
+    Let $X$ be a topologicial space and $A \subset X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\ol{A} = X$.
+        \item For every $\emptyset \ne U \subset X$ open, $A \cap U \ne \emptyset$.
+        \item For every $U \subset X$ open, $\overline{A \cap U} \supset U$.
+    \end{enumerate}
+    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}, $U \cap A \ne \emptyset$.
+
+    $(2) \Rightarrow (3)$: 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}, $x \in \overline{A \cap U}$.
+
+    $(3) \Rightarrow (1)$: $X$ is open.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:dense-product}
+    Let $\seqi{X}$ be topological spaces and $\seqi{A}$ such that each $A_i \subset X_i$ is dense, then $\prod_{i \in I}A_i$ is dense in $\prod_{i \in I}X_i$.
+\end{proposition}
+\begin{proof}
+    Let $x \in \prod_{i \in I}X_i$ and $U \in \cn(x)$, then there exists $J \subset I$ finite and $\seqf{U_j}$ such that $U_j \in \cn(\pi_j(x))$ for each $j \in J$, and $x \in \bigcap_{j \in J}\pi_j^{-1}(U_j) \subset U$. By density of each $A_j$, there exists $y \in X$ such that $y_j \in U_j$ for each $j \in J$. Thus $y \in U$, and $A$ is dense.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:closurecomplement}
+    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 $x \in \ol{A^c}$.
+\end{proof}
+
+
+\begin{definition}[Boundary]
+\label{definition:boundary}
+    Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
+    \begin{enumerate}
+        \item For every $V \in \cn(x)$, $V \cap A \ne \emptyset$ and $V \cap A^c \ne \emptyset$.
+        \item For every $V \in \fB$, $V \cap A \ne \emptyset$ and $V \cap A^c \ne \emptyset$.
+        \item $x \in \overline{A} \setminus A^o$.
+        \item $x \in \overline{A} \cap \overline{A^c}$.
+    \end{enumerate}
+
+    The set $\partial A$ of all points satisfying the above is the \textbf{boundary} of $A$.
+\end{definition}
+\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}, $x \not\in A^o$.
+
+    $(3) \Rightarrow (4)$: By \ref{lemma:closurecomplement}.
+
+    $(4) \Rightarrow (1)$: By (2) of \ref{definition:closure}.
+\end{proof}

src/topology/main/metric.tex

@@ -0,0 +1,50 @@
+\section{Metric Spaces}
+\label{section:metric}
+
+\begin{definition}[Metric]
+\label{definition:metric}
+    Let $X$ be a set, a \textbf{pseudo-metric} is a mapping $d: X \times X \to [0, \infty)$ such that:
+    \begin{enumerate}
+        \item For any $x \in X$, $d(x, x) = 0$.
+        \item For any $x, y \in X$, $d(x, y) = d(y, x)$.
+        \item For any $x, y, z \in X$, $d(x, z) \le d(x, y) + d(y, z)$.
+    \end{enumerate}
+    and the pair $(X, d)$ is a \textbf{pseudo-metric space}. If in addition,
+    \begin{enumerate}
+        \item[(4)] For any $x, y \in X$, $d(x, y) = 0$ if and only if $x = y$.
+    \end{enumerate}
+    then $d$ is a \textbf{metric}.
+\end{definition}
+
+\begin{definition}[Induced Uniformity]
+\label{definition:metric-uniformity}
+    Let $(X, d)$ be a metric space. For each $r > 0$, let
+    \[
+    U_r = \bracs{(x, y) \in X \times Y| d(x, y) < r}
+    \]
+    then there exists a uniformity $\fU$ on $X$ such that the family $\fB = \bracs{U_r| r > 0}$ is a fundamental system of entourages for $\fU$, known as the \textbf{uniformity induced by} $d$.
+
+    Let $\seqi{d}$ be pseudo-metrics on $X$. For each $i \in I$, let $X_i = X$ be equipped with the uniformity induced by $d_i$, then the \textbf{uniformity induced by} $\seqi{d}$ is the initial uniformity generated by the identity $\text{Id}_X: X \to X_i$.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:uniform-first-countable}
+    Let $X$ be a uniform space with a countable fundamental system of entourages, then there exists a countable fundamental system of symmetric entourages $\fB = \bracs{U_{1/2^n}|n \in \nat^+}$ such that $U_{1/2^{n+1}} \circ U_{1/2^{n+1}} \subset U_{1/2^n}$ for all $n \in \nat$.
+\end{lemma}
+\begin{proof}
+    Let $\seq{V}$ be a countable fundamental system of entourages. By \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $U_{1/2} \subset V_1$. Suppose inductively that $U_{1/2^n}$ has been constructed, then by (U2) and \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $U_{1/2^{n+1}} \subset U_{1/2^n} \cap V$. Thus $\fB = \bracs{U_{1/2^n}|n \in \nat^+}$ is the desired family.
+\end{proof}
+
+% Work in progress: pseudo-metrisability of uniform spaces.
+
+\begin{lemma}
+\label{lemma:uniform-urysohn}
+    Let $X$ be a uniform space with a countable fundamental system of entourages, $\mathbb{D}$ be the dyadic rational numbers in $[0, 1]$, then there exists a fundamental system of entourages $\fB = \bracs{}$
+\end{lemma}
+
+
+
+\begin{proposition}
+\label{proposition:uniform-pseudometric}
+    Let
+\end{proposition}

src/topology/main/neighbourhoods.tex

@@ -0,0 +1,74 @@
+\section{Neighbourhoods}
+\label{section:neighbourhoods}
+
+
+\begin{definition}[Neighbourhood]
+\label{definition:neighbourhood}
+    Let $(X, \topo)$ be a topological space and $A \subset X$.
+    A \textbf{neighbourhood} of $A$ is a set $V \supset X$ where there exists $U \in \topo$ such that $A \subset U \subset V \subset X$.
+
+    The set $\cn_{X, \topo}(A) = \cn_X(A) = \cn_\topo(A) = \cn(A)$ denotes the collection of all neighbourhoods of $A$, and $\cn^o(A)$ denotes the set of \textit{open} neighbourhoods of $A$.
+
+    \textit{If $A = \bracs{x}$ is a single point, the above definition and notation applies to $x$ directly}.
+\end{definition}
+
+\begin{definition}[Fundamental System of Neighbourhoods]
+\label{definition:fundamentalneighbourhoods}
+    Let $X$ be a topological space and $A \subset X$. A family $\fB \subset \cn(A)$ is a \textbf{fundamental system of neighbourhoods/neighbourhood base} at $A$ if for every $U \in \cn(A)$, there exists $V \in \fB$ such that $V \subset U$.
+\end{definition}
+
+
+\begin{lemma}[{{\cite[Proposition 1.2.1]{Bourbaki}}}]
+\label{lemma:openneighbourhood}
+    Let $(X, \topo)$ be a topological space, then $U \subset X$ is open if and only if $U \in \cn_\topo(x)$ for all $x \in U$.
+\end{lemma}
+
+\begin{proof}
+    Suppose that $U \in \cn_\topo(x)$ for all $x \in U$. For each $x \in U$, there exists $V_x \in \topo$ with $x \in V_x \subset U$.
+    Thus $U = \bigcup_{x \in U}V_x \in \topo$.
+\end{proof}
+
+
+\begin{proposition}[{{\cite[Proposition 1.2.2]{Bourbaki}}}]
+\label{proposition:neighbourhoodcharacteristic}
+    Let $(X, \topo)$ be a topological space, then for each $x \in X$,
+    \begin{enumerate}
+        \item[(F1)] For every $V \in \cn_\topo(x)$, $W \in \cn_\topo(x)$ for all $X \supset W \supset V$.
+        \item[(F2)] For any $A, B \in \cn_\topo(x)$, $A \cap B \in \cn_\topo(x)$.
+        \item[(V1)] For every $A \in \cn_\topo(x)$, $x \in A$.
+        \item[(V2)] For every $V \in \cn_\topo(x)$, there exists $W \in \cn_\topo(x)$ such that $V \in \cn_\topo(y)$ for all $y \in W$.
+    \end{enumerate}
+    Conversely, if $\cn: X \to 2^X$ is a mapping such that
+    \begin{enumerate}
+        \item $\cn(x) \ne \emptyset$ for all $x \in X$.
+        \item $\cn(x)$ satisfies (F1), (F2), (V1), and (V2).
+    \end{enumerate}
+    then there exists a unique topology $\topo \subset 2^X$ such that $\cn = \cn_\topo$.
+\end{proposition}
+
+\begin{proof}
+    \textit{Existence}: Suppose that $\cn: X \to 2^X$ is a mapping such that (F1), (F2), (V1), and (V2) holds for all $x \in X$. Let
+    \[
+    \topo = \bracs{U \subset X| U \in \cn(x) \forall x \in U}
+    \]
+    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).
+
+    Let $U, V \in \topo$. If $U \cap V \ne \emptyset$, then there exists $x \in U \cap V$, and $U, V \in \cn(x)$. For any $y \in U \cap V$, $U, V \in \cn(y)$ by (F2). Hence $U \cap V \in \cn(y)$ as well. Thus $U \cap V \in \topo$.
+
+    Now, let $\seqi{U} \subset \topo$ and $U = \bigcup_{i \in I}U_i$. For any $x \in U$, there exists $i \in I$ such that $x \in U_i$. In which case, $U \in \cn(x)$ by (F1), and $U \in \cn(x)$. Therefore $U \in \topo$, and $\topo$ forms a topology on $X$.
+
+    \textit{Agreement with $\cn_\topo$: } Fix $x \in X$. Let $V \in \cn_\topo(x)$, then there exists $U \in \topo$ such that $x \in U \subset V$. Since $U \in \topo$ and $x \in U$, $U \in \cn(x)$. By (F1), $V \in \cn(x)$ as well, so $\cn_\topo(x) \subset \cn(x)$.
+
+    Conversely, let $V \in \cn(x)$. By (V4), there exists $U_0 \in \cn(x)$ such that $V \in \cn(y)$ for all $y \in U_0$. Define
+    \[
+    U = \bracs{y \in V: V \in \cn(y)}
+    \]
+    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},
+    \[
+    \mathcal{R} = \bracs{U \subset X| U \in \cn_{\mathcal{R}}(x) \forall x \in U} = \bracs{U \subset X| U \in \cn(x) \forall x \in U} = \topo
+    \]
+    so $\topo$ is unique.
+\end{proof}

src/topology/main/nets.tex

@@ -0,0 +1,22 @@
+\section{Nets}
+\label{section:nets}
+
+\begin{definition}[Net]
+\label{definition:net}
+    Let $X$ be a set and $A$ be an upward-directed set, then a \textbf{net} in $X$ is a mapping $A \to X$, denoted $\net{x} \subset X$.
+\end{definition}
+
+\begin{definition}[Eventually]
+\label{definition:eventually}
+    Let $X$ be a set, $\net{x} \subset X$, and $E \subset X$, then $\net{x}$ is \textbf{eventually in} $E$ if there exists $\alpha_0 \in A$ such that $x_\alpha \in E$ for all $\alpha \gtrsim \alpha_0$.
+\end{definition}
+
+\begin{definition}[Frequently]
+\label{definition:frequently}
+    Let $X$ be a set, $\net{x} \subset X$, and $E \subset X$, then $\net{x}$ is \textbf{frequently in} $E$ if for all $\alpha_0 \in A$, there exists $\alpha \gtrsim \alpha_0$ such that $x_\alpha \in E$.
+\end{definition}
+
+\begin{definition}[Convergence]
+\label{definition:net-convergence}
+    Let $X$ be a topological space and $\net{x} \subset X$, then $\net{x}$ \textbf{converges} to $x$ if for every $U \in \cn(x)$, $\net{x}$ is eventually in $U$, denoted $x_\alpha \to x$.
+\end{definition}

src/topology/main/product.tex

@@ -0,0 +1,122 @@
+\section{Product Spaces}
+\label{section:producttopology}
+
+\begin{definition}[Product Topology]
+\label{definition:product-topology}
+    Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces, then the \textbf{product} topology $\topo$ is the initial topology generated by the projections $\seqi{\pi}$, and:
+    \begin{enumerate}
+        \item The family
+
+        \[
+        \cb(\ce) = \bracs{\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_{k}) \bigg | U_{k} \in \topo_{i_k}, \seqf{i_k} \subset I, n \in \nat^+}
+        \]
+
+        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
+
+        \[
+        \xymatrix{
+        Y \ar@{->}[d]_{f} \ar@{->}[rd]^{f_i} &  \\
+        X \ar@{->}[r]_{\pi_i} & X_i
+        }
+        \]
+
+        for all $i \in I$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): By (3) of \ref{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}
+
+\begin{proposition}
+\label{proposition:productfilterconvergence}
+Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\B$ be a filter base on $\prod_{i \in I}X_i$, then $\fB$ converges to $x \in \prod_{i \in I}X_i$ if and only if $\pi_i(\fB)$ converges to $\pi_i(x)$ for all $i \in I$.
+\end{proposition}
+\begin{proof}
+    $(\Rightarrow)$: Let $i \in I$ and $U \in \cn(\pi_i(x))$, then $\pi_i^{-1}(U) \in \cn(x)$. Since $\fB$ converges to $x$, there exists $B \in \fB$ with $B \subset \pi_i^{-1}(U)$. In which case, $\pi_i(B) \subset U$ and $\pi_i(\fB)$ converges to $\pi_i(x)$.
+
+    $(\Leftarrow)$: Let $U \in \cn(x)$, then there exists $\seqf{i_k}$ and open sets $\seqf{U_k}$ such that $x \in \bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k) \subset U$. For each $1 \le k \le n$, since $\pi_{i_k}(\fB)$ converges to $\pi_{i_k}(x)$ for each $1 \le k \le n$, there exists $B_k \in \fB$ such that $B_k \subset \pi_{i_k}^{-1}(U_k)$. As $\fB$ is a filter base, there exists $B \in \fB$ with
+    \[
+    B \subset \bigcap_{k = 1}^n B_k \subset \bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k) \subset U
+    \]
+    Therefore $\fB$ converges to $x$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:initial-product-topology}
+    Let $X$ be a topological space, $\seqi{Y}$ be topological spaces, $\seqi{f}$ where $f_i \in C(X; Y_i)$ for each $i \in I$, then
+    \begin{enumerate}
+        \item There exists a unique $f \in C(X; \prod_{i \in I}Y_i)$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+         & \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} &  \\
+        X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} &  & Y_i
+        }
+        \]
+
+        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}.
+
+    (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}
+
+\begin{proposition}
+\label{proposition:product-topology-embedding}
+    Let $\seqi{X}$, $\seqi{Y}$ be topological/uniform spaces and $\seqi{f}$ such that $f_i \in C(X_i; Y_i)$ is an embedding for all $i \in I$, then
+    \begin{enumerate}
+        \item There exists a unique $f \in C(\prod_{i \in I}X_i; \prod_{i \in I}Y_i)$/$f \in UC(\prod_{i \in I}X_i; \prod_{i \in I}Y_i)$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+        \prod_{i \in I}Y_i \ar@{->}[r]^{\pi_i} & Y_i \\
+        \prod_{i \in I} X_i \ar@{->}[r]_{\pi_i} \ar@{->}[u]^{f} & X_i \ar@{^{(}->}[u]_{f_i}
+        }
+        \]
+
+        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}.
+
+    (2): Consider the following diagram
+    \[
+    % https://darknmt.github.io/res/xypic-editor/#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
+    \xymatrix{
+    \prod_{i \in I}Y_i \ar@{->}[rd]^{\pi_i} &  \\
+    \iota_P(\prod_{i \in I}X_i) \ar@{->}[u] & Y_i \\
+    \prod_{i \in I} X_i \ar@{->}[r]_{\pi_i} \ar@{->}[u]^{f} & X_i \ar@{^{(}->}[u]_{f_i}
+    }
+    \]
+    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.
+\end{proof}

src/topology/main/regular.tex

@@ -0,0 +1,37 @@
+\section{Regular Spaces}
+\label{section:regularspaces}
+
+\begin{definition}[Regular Space, {{\cite[Proposition 1.4.11]{Bourbaki}}}]
+\label{definition:regular}
+    Let $X$ be a topological space, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $x \in X$ and $A \subset X$ closed with $x \not\in A$, there exists $U \in \cn(x)$ and $V \in \cn(A)$ such that $U \cap V = \emptyset$.
+        \item For each $x \in X$, the closed neighbourhoods of $x$ forms a fundamantal system of neighbourhoods at $x$.
+    \end{enumerate}
+    If $X$ is a T1 space such that the above holds, then $X$ is \textbf{regular}.
+\end{definition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: Let $U \in \cn^o(x)$, then $U^c$ is closed with $x \not\in U^c$ by (V1). Thus there exists $V \in \cn^o(U^c)$ such that $x \not\in V$, so $V^c \in \cn(x)$ is closed with $V^c \subset U$.
+
+    $(2) \Rightarrow (1)$: Since $X$ is Hausdorff, $X$ is T1 as well. Let $x \in X$ and $A \subset X$ closed such that $x \not\in A$, then $A^c \in \cn(x)$. So there exists $K \in \cn(x)$ closed such that $x \in K \subset A^c$. Thus $K \in \cn(x)$ and $K^c \in \cn(A)$ are the desired neighbourhoods.
+\end{proof}
+
+\begin{theorem}[{{\cite[Theorem 1.8.1]{Bourbaki}}}]
+\label{theorem:regularextension}
+    Let $X$ be a topological space, $Y$ be a regular space, $A \subset X$, and $f \in C(A; Y)$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists $F \in C(X; Y)$ such that $F|_A = f$.
+        \item For each $x \in X$, $\lim_{y \to x, y \in A}f(y)$ exists.
+    \end{enumerate}
+\end{theorem}
+\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}.
+
+    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,
+    \[
+    F(y) = \lim_{\substack{z \to y \\ z \in A}}f(z) \in \ol{f(U \cap A)} \subset V
+    \]
+    as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous.
+\end{proof}

src/topology/uniform/complete.tex

@@ -0,0 +1,158 @@
+\section{Completeness}
+\label{section:completeness}
+
+
+\begin{definition}[Small]
+\label{definition:small}
+    Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A \subset X$, then $A$ is \textbf{$V$-small} if $A \times A \subset V$.
+\end{definition}
+
+\begin{lemma}[{{\cite[Proposition 2.3.1]{Bourbaki}}}]
+\label{lemma:small-intersect}
+    Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A, B \subset X$ such that:
+    \begin{enumerate}
+        \item $A, B$ are $V$-small.
+        \item $A \cap B \ne \emptyset$.
+    \end{enumerate}
+    then $A \cup B$ is $V \circ V$-small.
+\end{lemma}
+\begin{proof}
+    Since $V \circ V \supset V$, it is sufficient to consider the case where $x \in A$ and $y \in B$. By assumption (2), there exists $z \in A \cap B$. By assumption (1), $(x, z), (z, y) \in V$, so $(x, y) \in V \circ V$.
+\end{proof}
+
+\begin{definition}[Cauchy Filter]
+\label{definition:cauchyfilter}
+    Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a filter on $X$, then $\fF$ is \textbf{Cauchy} if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small.
+\end{definition}
+
+\begin{proposition}[Cauchy Criterion]
+\label{proposition:cauchycriterion}
+    Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a convergent filter, then $\fF$ is Cauchy.
+\end{proposition}
+\begin{proof}
+    Let $x \in X$ such that $\cn(x) \subset \fF$ and $V \in \fU$. By \ref{lemma:symmetricfundamentalentourage}, assume without loss of generality that $V \in \fU$. Since $\cn(x) \subset \fF$, then $V(x) \in \fU$ and $V(x) \times V(x) \subset V$.
+\end{proof}
+
+\begin{proposition}[{{\cite[Proposition 2.3.3]{Bourbaki}}}]
+\label{proposition:imagecauchy}
+    Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces, $f \in UC(X; Y)$, and $\fB \subset 2^X$ be a Cauchy filter base, then $f(\fB) \subset 2^Y$ is a Cauchy filter base.
+\end{proposition}
+\begin{proof}
+    Let $V \in \mathfrak{V}$, then there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$. Given that $\fB$ is a Cauchy filter base, there exists $E \in \fB$ such that $E \times E \subset V'$, so $f(E) \times f(E) \subset V$.
+\end{proof}
+
+\begin{definition}[Minimal Cauchy Filter]
+\label{definition:minimalcauchy}
+    Let $X$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then $\fF$ is \textbf{minimal} if it is minimal with respect to inclusion.
+\end{definition}
+
+\begin{proposition}[{{\cite[Proposition 2.3.5]{Bourbaki}}}]
+\label{proposition:minimalcauchyexistence}
+    Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then:
+    \begin{enumerate}
+        \item There exists a unique minimal Cauchy filter $\fF_0$ such that $\fF_0 \subset \fF$.
+        \item If $\fB \subset \fF$ is a base of $\fF$, and $\fV \subset \fU$ is a fundamental system of symmetric entourages, then $\mathfrak{M} = \bracs{V(M): V \in \fV, M \in \fB}$ is a base for $\fF_0$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_1, V_2 \in \fV$ and $M_1, M_2 \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_0 \in \fV$ and $M_0 \in \fB$ such that $V_0 \subset V_1 \cap V_2$ and $M_0 \subset M_1 \cap M_2$. So $V_0(M_0) \subset V_1(M_1) \cap V_2(M_2)$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By \ref{proposition:filterbasecriterion}, $\mathfrak{M}$ is a filter base for
+    \[
+    \fF_0 = \bracs{E \subset X| \exists V \in \fV, M \in \fB: V(M) \subset E} \subset \fF
+    \]
+
+    To see that $\mathfrak{M}$ is a Cauchy filter base, let $V \in \fV$. By (U2), there exists $W \in \fV$ such that $W \circ W \circ W \subset V$. Since $\fB$ is a Cauchy filter base, there exists $M \in \fB$ such that $M \times M \subset W$. Let $(x, y) \in W(M) \times W(M)$, then there exists $(p, q) \in M \times M$ such that $(p, x), (q, y) \in W$. As $M \times M$ \subset $W$, $(x, y) \in W \circ W \circ W$ by symmetry of $W$ and \ref{lemma:compositiongymnastics}. Therefore $W(M) \times W(M) \subset V$ and $\mathfrak{M}$ is a Cauchy filter base.
+
+    Finally, let $\fF' \subset \fF$ be a Cauchy filter. Let $M \in \fB$ and $V \in \fV$, then there exists $N \in \fF'$ such that $N \times N \subset V$. Since $N \in \fF$, $M \cap N \ne \emptyset$, and there exists $y \in M \cap N$. For any $x \in N$, $N \times N \subset V$ implies that $x \in V(y)$. Hence $N \subset V(M)$ and $\fF' \supset \fF_0$.
+\end{proof}
+
+\begin{proposition}[{{\cite[Corollary 2.3.4]{Bourbaki}}}]
+\label{proposition:cauchyinterior}
+    Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a minimal Cauchy filter, then $\fF$ admits a base consisting of open sets.
+\end{proposition}
+\begin{proof}
+    Let $\fV \subset \fU$ be the set of all symmetric, open entourages. By \ref{proposition:goodentourages}, $\fV$ is a fundamental system of entourages. By \ref{proposition:minimalcauchyexistence}, $\mathfrak{M} = \bracs{U(M)| U \in \fV, M \in \fF}$ is a base for $\fF$.
+
+    Let $V \in \fU$, then there exists $U \in \fV$ such that $U \subset V$. For any $M \in \fF$, $U(M)$ is open by \ref{lemma:openentourageneighbourhoods}. Thus $\mathfrak{M}$ consists of open sets.
+\end{proof}
+
+\begin{proposition}[{{\cite[Corollary 2.3.1, 2.3.2]{Bourbaki}}}]
+\label{proposition:cauchyfilterlimit}
+    Let $(X, \fU)$ be a uniform space, then:
+    \begin{enumerate}
+        \item For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter.
+        \item For each Cauchy filter $\fF \subset 2^X$ and $x \in X$, $x$ is an accumulation point of $\fF$ if and only if $\fF$ converges to $x$.
+        \item For any $x \in X$, filter $\fF \subset 2^X$ converging to $x$, and Cauchy filter $\mathfrak{G} \subset \fF$, $\mathfrak{G}$ converges to $x$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\fF = \bracs{E \subset X: x \in E}$, then $\fF$ is a filter generated by the filter base $\fB = \bracs{\bracs{x}}$. By \ref{proposition:minimalcauchyexistence}, $\bracs{U(x): U \in \fU} = \cn(x)$ is the unique minimal filter contained in $\fF$.
+
+    (2): Let $x \in X$ be an accumulation point of $\fF$, then $U(x) \cap E \ne \emptyset$ for all $U \in \fU$ and $E \in \fF$. By \ref{definition:generatedfilter}, there exists a filter $\fF' \supset \fF \cup \cn(x)$. Since $\fF$ is a Cauchy filter, so is $\fF'$. By minimality of $\cn(x)$, $\fF \supset \cn(x)$.
+
+    (3): Since $\fF$ converges to $x$, $x$ is a cluster point of $\fF$ and $\mathfrak{G}$. By (2), $x$ is also a limit point of $\mathfrak{G}$.
+\end{proof}
+
+\begin{definition}[Complete Space]
+\label{definition:completespace}
+    Let $(X, \fU)$ be a uniform space, then $X$ is \textbf{complete} if every Cauchy filter converges to at least one point.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:product-complete}
+    Let $\seqi{X}$ be complete uniform spaces, then $\prod_{i \in I}X_i$ is complete.
+\end{proposition}
+\begin{proof}
+    For any Cauchy filter $\fF$ in $\prod_{i \in I}\wh X_i$, $\pi_i(\fF)$ is a Cauchy filter base by (1) of \ref{definition:product-uniform} and \ref{proposition:imagecauchy}. For each $i \in I$, $\wh X_i$ is complete, so there exists $x_i \in \wh X_i$ such that $\pi_i(\fF)$ converges to $x_i$. Let $x \in X$ such that $\pi_i(x) = x_i$ for all $i \in I$, then $\fF$ converges to $x$ by \ref{proposition:productfilterconvergence}. Therefore $\prod_{i \in I}X_i$ is complete.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:complete-closed}
+    Let $X$ be a uniform space and $A \subset X$ be closed, then $A$ is complete.
+\end{proposition}
+\begin{proof}
+    Let $\fF$ be a Cauchy filter on $A$, then there exists $x \in X$ such that $\fF$ converges to $x$. Since $A$ is closed, $x \in X$ by (4) of \ref{definition:closure}.
+\end{proof}
+
+
+
+\begin{lemma}
+\label{lemma:complete-dense}
+    Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $X$ is \textbf{complete} if and only if every Cauchy filter on $A$ converges to at least one point in $X$.
+\end{lemma}
+\begin{proof}
+    Let $\fF \subset X$ be a Cauchy filter. Using \ref{proposition:minimalcauchyexistence}, assume without loss of generality that $\fF$ is minimal. By \ref{proposition:cauchyinterior}, $\fF$ admits a base consisting of open sets. Hence $E \cap A \ne \emptyset$ for all $E \in \fF$. Let $\fF_A = \bracs{E \cap A| E \in \fF}$, then $\fF_A$ is a filter base on $A$.
+
+    By assumption, there exists $x \in X$ such that $\fF_A$ converges to $x$. Since $\fF$ is a Cauchy filter contained in the filter generated by $\fF_A$ in $X$, $x$ is also a limit point of $\fF$ by \ref{proposition:cauchyfilterlimit}.
+\end{proof}
+
+
+\begin{proposition}[{{\cite[Proposition 2.3.11]{Bourbaki}}}]
+\label{proposition:uniformextension}
+    Let $X$ be a topological space, $Y$ be a complete Hausdorff uniform space, $A \subset X$ be a dense subset, and $f \in C(A; Y)$ be a continuous function, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists a continuous function $F \in C(X; Y)$ such that $F|_A = f$.
+        \item For each $x \in X$, $f(\bracs{U \cap A| U \in \cn(x)})$ is a Cauchy filter base.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    By \ref{proposition:uniform-neighbourhoods}, $Y$ is regular. Since $Y$ is complete, (2) is equivalent to (2) of \ref{theorem:regularextension}. Therefore the proposition follows from \ref{theorem:regularextension}.
+\end{proof}
+
+
+\begin{theorem}[Extension of Uniformly Continuous Functions, {{\cite[Theorem 1.3.2]{Bourbaki}}}]
+\label{theorem:uniform-continuous-extension}
+    Let $(X, \fU)$, $(Y, \fV)$ be uniform spaces, $A \subset Y$ be a dense subset, and $f \in UC(A; Y)$, then:
+    \begin{enumerate}
+        \item There exists a unique $F \in UC(X; Y)$ such that $F|_A = f$.
+        \item $F \in UC(X; Y)$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): By \ref{proposition:imagecauchy}, $f(\bracs{U \cap A| U \in \cn(x)})$ is a Cauchy filter base for all $x \in X$. Thus the existence and uniqueness of the extension follows from \ref{proposition:uniformextension}.
+
+    (2): Let $V \in \fV$. Using \ref{proposition:goodentourages}, assume without loss of generality that $V$ is symmetric and closed. Since $f \in UC(A; Y)$, there exists $U \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in U$. Using \ref{proposition:subspace-entourage}, assume without loss of generality that $U = \overline{U} = \ol{U \cap (A \times A)}$. In which case, since $F$ is continuous,
+    \[
+    F(U) = F(\overline{U}) = F(\ol{U \cap (A \times A)}) \subset \overline{F(U \cap (A \times A))} = \overline{f(U \cap (A \times A))} \subset V
+    \]
+    by \ref{proposition:closure-of-image}.
+\end{proof}

src/topology/uniform/completion.tex

@@ -0,0 +1,226 @@
+\section{The Hausdorff Completion}
+\label{section:completion}
+
+
+\begin{definition}[Hausdorff Completion, {{\cite[Theorem 2.3.3]{Bourbaki}}}]
+\label{definition:hausdorff-completion}
+    Let $(X, \fU)$ be a uniform space, then there exists a $(\wh X, \iota)$ such that:
+    \begin{enumerate}
+        \item $(\wh X, \wh \fU)$ is a complete Hausdorff uniform space.
+        \item $\iota \in UC(X; \wh X)$.
+        \item[(U)] For any pair $(Y, f)$ satisfying (1) and (2), there exists unique $F \in C(\wh X; Y)$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+         & \wh X \ar@{->}[rd]^{F} &  \\
+        X \ar@{->}[rr]_{f} \ar@{->}[ru]^{\iota} &  & Y
+        }
+        \]
+
+        and $F \in UC(\wh X; Y)$.
+    \end{enumerate}
+    Moreover,
+    \begin{enumerate}
+        \item[(4)] For any symmetric entourage $V \in \fU$, let
+        \[
+        \wh V = \bracsn{(\fF, \mathfrak{G}) \in \wh X| \exists U \in \fF \cap \mathfrak{G}: U \times U \subset V}
+        \]
+        then[(5)] The family $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$ forms a fundamental system of entourages for $\wh X$. In particular, $\bracs{\wh V \cap \iota(X)| V \in \fU, V \text{ symmetric}}$ is a fundamental system of entuorages for $\iota(X)$.
+
+        \item $\iota(X)$ is dense in $\wh X$.
+    \end{enumerate}
+
+
+
+    The pair $(\wh X, \iota)$ is the \textbf{Hausdorff completion} of $X$.
+\end{definition}
+\begin{proof}
+    (1, Uniform), (4): Let $\wh X$ be the set of all minimal Cauchy filters on $X$. For each $V \in \fU$, let
+    \[
+    \wh V = \bracsn{(\fF, \mathfrak{G}) \in \wh X| \exists U \in \fF \cap \mathfrak{G}: U \times U \subset V}
+    \]
+    and $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$, then
+    \begin{enumerate}
+        \item[(FB1)] Let $\wh U, \wh V \in \wh \fB$. By \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $W \in \fU$ such that $W \subset U \cap V$. In which case, for any $(\fF, \mathfrak{G}) \in \wh W$, there exists $E \in \fF \cap \mathfrak{G}$ with $E \times E \subset W \subset U \cap V$. Thus $\wh W \subset \wh U \cap \wh V$.
+        \item[(UB1)] Let $\wh U \in \wh \fB$ and $\fF \in \wh X$, then since $\fF$ is Cauchy, there exists $E \in \fF$ such that $E \times E \subset U$, so $(\fF, \fF) \in \wh U$.
+        \item[(UB2)] Let $\wh U \in \wh \fB$. By \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $W \in \fU$ such that $W \circ W \subset U$. For any $\fF, \mathfrak{G}, \mathfrak{H} \in \wh X$ such that $(\fF, \mathfrak{G}), (\mathfrak{G}, \mathfrak{H}) \in \wh W$, there exists $W$-small sets $E \in \fF \cap \mathfrak{G}$ and $F \in \fF \cap \mathfrak{H}$. Since $\mathfrak{G}$ is a filter, $E \cap F \ne \emptyset$ by (F2) and (F3). By \ref{lemma:small-intersect}, $E \cup H$ is $W \circ W$-small and thus $U$-small. Using (F1), $E \cup H \in \fF \cap \mathfrak{H}$, so $(\fF, \mathfrak{H}) \in \wh U$. Therefore $\wh W \circ \wh W \subset U$.
+    \end{enumerate}
+    By \ref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\wh \fU \supset \wh \fB$. Moreover, $\wh \fB$ consists of symmetric entourages by construction.
+
+    (1, Hausdorff): It is sufficient to show that $\Delta$ is closed and use (6) of \ref{definition:hausdorff}. Let $(\fF, \mathfrak{G}) \in \ol{\Delta}$, then $(\fF, \mathfrak{G}) \in U$ for all $U \in \fU$ closed. Let $\fB = \bracs{F \cup G| F \in \fF, G \in \mathfrak{G}}$, then
+    \begin{enumerate}
+        \item[(FB1)] For any $F \cup G, F' \cup G' \in \fB$, $(F \cup G) \cap  (F' \cup G') \supset (F \cap F') \cup (G \cap G') \in \fB$.
+        \item[(FB2)] By (F3), $\emptyset \not\in \fF \cup \mathfrak{G}$, so $\emptyset \not\in \fB$.
+    \end{enumerate}
+    Thus $\fB$ is a filter base by \ref{proposition:filterbasecriterion}, and the filter $\mathfrak{H}$ generated by $\fB$ is contained in $\fF$ and $\mathfrak{G}$. By \ref{proposition:goodentourages}, for every $U \in \fU$, there exists a $U$-small set $E \in \fF \cap \mathfrak{G} \subset \fB \subset \mathfrak{H}$. So $\mathfrak{H} \subset \fF, \mathfrak{G}$ is a Cauchy filter. By minimality of $\fF$ and $\mathfrak{G}$, $\fF = \mathfrak{G} = \mathfrak{H}$.
+
+    (2): For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter by (1) of \ref{proposition:cauchyfilterlimit}. Define $\iota: X \to \wh X$ by $x \mapsto \cn(x)$. Let $\wh U \in \wh \fU$ and $(\cn(x), \cn(y)) \in \wh U$, then there exists a $U$-small set $E \in \cn(x) \cap \cn(y)$. By (V1), $(x, y) \in E \times E \in U$.
+
+    Conversely, if $(x, y) \in U$, then $E = U(x) \cap U(y) \in \cn(x) \cap \cn(y)$ by (F2), and $E$ is $U$-small by symmetry of $U$.
+
+    Thus $(\iota \times \iota)^{-1}(\wh U) = U$, $(\iota \times \iota)^{-1}: \wh \fU \to \fU$ is a bijection, and $\iota \in UC(X; \wh X)$.
+
+    (4): Let $\fF \in \wh X$ and $\wh U \in \wh \fU$. Since $\fF$ is a Cauchy filter, there exists a $U$-small set $E \in \fF$. Using \ref{proposition:cauchyinterior}, assume without loss of generality that $E$ is open. Let $x \in E$, then $E \in \cn(x)$, so $(\fF, \iota(x)) \in \wh \fU$, and $X$ is dense in $\wh X$ by (3) of \ref{definition:closure}.
+
+    (1, Complete): Using \ref{lemma:complete-dense}, it is sufficient to show that every Cauchy filter in $\iota(X)$ converges in $\wh X$.
+
+    Let $\wh \fF \in 2^{\iota(X)}$ be a Cauchy filter, then since $\iota: X \to \iota(X)$ is surjective, $\iota^{-1}(\wh \fF)$ is a filter base by \ref{proposition:preimage-filterbase}.
+
+    For any $U \in \fU$, there exists a $\wh U$-small set $\wh E \in \wh \fF$. Since $U = (\iota \times \iota)^{-1}(\wh U)$, $\iota^{-1}(\wh E) \subset U$, so $\iota^{-1}(\wh \fF)$ is a Cauchy filter base.
+
+    Let $\mathfrak{G} \subset \iota^{-1}(\wh \fF)$ be a minimal Cauchy filter, then $\iota(\mathfrak{G})$ is a Cauchy filter base by \ref{proposition:imagecauchy}.
+
+    Let $U \in \fU$. By \ref{proposition:cauchyinterior}, there exists a $U$-small open set $E \in \mathfrak{G}$. For any $x \in E$, $E \in \cn(x)$ by \ref{lemma:openneighbourhood}, so $(\cn(x), \mathfrak{G}) = (\iota(x), \mathfrak{G}) \in \wh U$, and $\iota(E) \subset \wh U(\mathfrak{G})$. Since $E \in \mathfrak{G}$, $\wh U(\mathfrak{G}) \supset \iota(E) \in \iota(\mathfrak{G})$, so $\iota(\mathfrak{G})$ converges to $\mathfrak{G}$.
+
+    Now, given that $\mathfrak{G} \subset \iota^{-1}(\wh \fF)$, $\iota(\mathfrak{G}) \subset \iota(\iota^{-1}(\fF)$, so $\iota(\iota^{-1}(\fF))$ converges to $\mathfrak{G}$ as well. Since $\fF$ is a filter on $\iota(X)$, $\iota(\iota^{-1}(\fF)) = \fF$, thus $\fF$ is convergent.
+
+    (U): Let $(Y, \mathfrak{V})$ be a complete Hausdorff uniform space and $f \in UC(X; Y)$. For each $\fF \in \wh X$, let $F(\fF) = \lim_{x, \fF}f(x)$.  For any $x \in X$, $F(\iota(x)) = \lim_{y \to x}f(y) = f(x)$, so $f = F \circ \iota$. Since $\fF$ is a Cauchy filer, $f \in UC(X; Y)$, and $Y$ is a complete Hausdorff uniform space, the limit exists and is unique. Thus $F: \iota(X) \to Y$ is well-defined.
+
+    Let $U \in \mathfrak{V}$, then there exists a symmetric entourage $V \in \fU$ such that $(f(x), f(y)) \in U$ for all $(x, y) \in V$, then for any $(\iota(x), \iota(y)) \in \wh V \cap (\iota(X) \times \iota(X))$,  $(F(\iota(x)), F(\iota(y))) \in U$, so $F$ is uniformly continuous. By \ref{theorem:uniform-continuous-extension}, there exists a unique $\td F \in C(\wh X; Y)$ such that $\td F|_{\iota(X)} = F|_{\iota(X)}$, and any such $\td F$ is uniformly continuous.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:completion-of-hausdorff}
+    Let $(X, \fU)$ be a Hausdorff uniform space, then the canonical map $\iota: X \to \wh X$ is an embedding.
+\end{lemma}
+\begin{proof}
+    Since the mapping $(\iota \times \iota)^{-1}$ is a bijection between two bases of uniformities of $X$ and $\wh X$, it is sufficient to show that $\iota$ is injective. To this end, observe that for any $x, y \in X$, $\cn(x) = \cn(y)$ if and only if $x = y$ by (4) of \ref{definition:hausdorff}.
+\end{proof}
+
+\begin{definition}[Associated Hausdorff Uniform Space, {{\cite[Proposition 2.8.16]{Bourbaki}}}]
+\label{definition:associated-hausdorff}
+    Let $(X, \fU)$ be a uniform space, then there exists $(X', i)$ such that:
+    \begin{enumerate}
+        \item $X'$ is a Hausdorff uniform space.
+        \item $i \in UC(X; X')$.
+        \item For any pair $(Y, f)$ satisfying (1) and (2), there exists a unique $f' \in UC(X'; Y)$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+        X' \ar@{->}[rd]^{F} &  \\
+        X \ar@{->}[r]_{f} \ar@{->}[u]^{i} & Y
+        }
+        \]
+    \end{enumerate}
+    known as the \textbf{Hausdorff uniform space associated with} $(X, \fU)$.
+\end{definition}
+\begin{proof}
+    Let $(\wh X, \iota)$ be the Hausdorff completion of $X$, $X' = \iota(X)$, and $i = \iota$, then $(X', i)$ satisfies (1) and (2).
+
+    (U): Let $(\wh Y, \iota)$ be the Hausdorff completion of $Y$. Using \ref{lemma:completion-of-hausdorff}, identify $Y$ as a subspace of $\wh Y$. By (U) of the Hausdorff completion, there exists a unique $\ol F \in UC(\wh X; \wh Y)$ such that the following diagram commutes:
+
+    \[
+    \xymatrix{
+    \wh X \ar@{->}[r]^{\overline{F}} & \wh Y \\
+    X' \ar@{->}[u]^{\iota} \ar@{->}[r]^{F} & Y' \ar@{->}[u]_{\iota} \\
+    X \ar@{->}[r]_{f} \ar@{->}[u]^{i} & Y \ar@{=}[u]
+    }
+    \]
+
+    Since $\ol F(X') \subset \iota(Y) = Y'$, $F = \ol F|_{X'} \in UC(X'; Y')$ is continuous. By \ref{lemma:completion-of-hausdorff}, $\iota \in UC(Y; Y')$ is a homeomorphism. Upon identifying $Y$ with $Y'$, $F \in UC(X'; Y)$ is the desired map.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:hausdorff-uniform-factor}
+    Let $X, Y$ be uniform spaces, $X', Y'$ be their associated Hausdorff uniform spaces, and $\wh X, \wh Y$ be their Hausdorff completions, then there exists a unique $F \in UC(X'; Y')$ and $\ol F \in UC(\wh X; \wh Y)$ such that the following diagram commutes:
+
+    \[
+    \xymatrix{
+    \wh X \ar@{->}[r]^{\overline{F}} & \wh Y \\
+    X' \ar@{->}[u] \ar@{->}[r]^{F} & Y' \ar@{->}[u] \\
+    X \ar@{->}[r]_{f} \ar@{->}[u] & Y \ar@{->}[u]
+    }
+    \]
+\end{proposition}
+\begin{proof}
+    By (U) of \ref{definition:hausdorff-completion} and \ref{definition:associated-hausdorff}.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:initial-completion}
+    Let $X$ be a set, $\seqi{Y}$ be uniform spaces, $\seqi{f}$ with $f_i: X \to Y_i$ for each $i \in I$. Let $\fU$ be the initial uniformity on $X$ induced by $\seqi{f}$, and $(\wh X, \iota_X)$ be the Hausdorff completion of $X$.
+    \begin{enumerate}
+        \item For each $i \in I$, let $(\wh Y_i, \iota_i)$ be the Hausdorff completion of $Y_i$, then there exists a unique $F_i \in UC(\wh X; \wh Y_i)$ such that the following diagram commutes
+        \[
+        \xymatrix{
+        \wh X \ar@{->}[r]^{F_i} & \wh Y \\
+        X \ar@{->}[r]_{f_i} \ar@{->}[u] & Y \ar@{->}[u]
+        }
+        \]
+        \item The uniformity of $\wh X$ is the initial uniformity induced by $\seqi{F}$.
+        \item There exists a unique $F \in UC(X; \prod_{i \in I}\wh Y_i)$ and $\ol{F} \in UC(\wh X; \prod_{i \in I}\wh Y_i)$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+         & \prod_{i \in I} \wh Y_i \ar@{->}[rd]^{\pi_i} &  \\
+         & \wh X \ar@{->}[r]^{F_i} \ar@{->}[u]_{\overline{F}} & \wh Y_i \\
+        X \ar@{->}[ruu]^{F} \ar@{->}[rr]_{f_i} \ar@{->}[ru]_{\iota_X} &  & Y_i \ar@{->}[u]_{\iota_i}
+        }
+        \]
+
+        Moreover, $\ol{F}(\wh X) = \overline{F(X)}$, and $\ol{F}$ is an embedding.
+    \end{enumerate}
+    In particular, by \ref{proposition:dense-product}, there is a natural isomorphism
+    \[
+    \prod_{i \in I}\wh X_i \iso \wh{\prod_{i \in I}X_i}
+    \]
+    induced by extending the identity on $\pi_{i \in I}X_i$.
+\end{proposition}
+\begin{proof}
+    (1): By (U) of \ref{definition:hausdorff-completion}.
+
+    (2), (3): By (U) of the product (\ref{definition:product-topology}), there exists $f \in UC(X; \prod_{i \in I}Y_i)$ such that the following diagram commutes:
+
+    \[
+    \xymatrix{
+     & \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} &  \\
+    X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} &  & Y_i
+    }
+    \]
+
+    First suppose that $X$ and $\seqi{Y}$ are Hausdorff. For each $i \in I$, $\iota_i \circ \pi_i \in UC(\prod_{i \in I} Y_i; \wh Y_i)$, so by (U) of \ref{proposition:product-complete}, there exists a unique $\iota_P \in UC(\prod_{i \in I} Y_i; \prod_{i \in I} \wh Y_i)$ such that the following diagram commutes
+
+    \[
+    \xymatrix{
+    \prod_{i \in I}\wh Y_i \ar@{->}[r]^{\pi_i} & \wh Y_i \\
+    \prod_{i \in I}Y_i \ar@{->}[r]^{\pi_i} \ar@{^{(}->}[u]^{\iota_P} & Y_i \ar@{^{(}->}[u]_{\iota_i}
+    }
+    \]
+
+    for all $i \in I$. By \ref{lemma:completion-of-hausdorff}, each $\iota_i \in UC(Y_i; \wh Y_i)$ is an embedding, so \ref{proposition:product-topology-embedding} implies that $\iota_P$ is an embedding as well.
+
+    Since $X$ has the initial topology, $f: X \to \prod_{i \in I}Y_i$ is an embedding by \ref{proposition:initial-product-topology}. Thus the composition $\iota_P \circ f$ is an embedding. As $\prod_{i \in I}\wh Y_i$ is complete by \ref{proposition:product-complete} and \ref{proposition:product-hausdorff}, $\overline{\iota_P \circ f(X)} \subset \prod_{i \in I}\wh Y_i$ is a complete Hausdorff uniform space. Let $Z$ be a complete Hausdorff uniform space and $g \in UC(X; Z)$, then \ref{theorem:uniform-continuous-extension} implies that there exists a unique $G \in UC(\overline{\iota_P \circ f(X)}; Z)$ such that the following diagram commutes
+
+    \[
+    \xymatrix{
+    \prod_{i \in I}\wh Y_i &  \\
+    \overline{\iota_P \circ f(X)} \ar@{^{(}->}[u] \ar@{->}[rd]^{G} &  \\
+    X \ar@{^{(}->}[u]^{\iota_P\circ f} \ar@{->}[r]_{g} & Z
+    }
+    \]
+
+    Thus $(\overline{\iota_P \circ f(X)}, \iota \circ f)$ satisfies (1), (2), and (U) of the Hausdorff completion (\ref{definition:hausdorff-completion}). Therefore $\wh X$ may be identified as a subspace of $\prod_{i \in I}\wh Y_i$ as follows:
+
+    \[
+    % https://darknmt.github.io/res/xypic-editor/#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
+    \xymatrix{
+    \prod_{i \in I}\wh Y_i \ar@{->}[r]^{\pi_i} & \wh Y_i \\
+    \overline{\iota_P \circ f(X)} \ar@{^{(}->}[u] \ar@{=}[r] & \wh X \\
+    X \ar@{^{(}->}[u]^{\iota_P\circ f} \ar@{->}[ru]_{\iota_X} &
+    }
+    \]
+
+    In which case, $\wh X$ must be equipped with the initial topology induced by the projection maps.
+
+    Now assume that $X$ and $Y$ are arbitrary. Let $X'$ and $\seqi{Y'}$ be the Hausdorff spaces associated with $X$ and $\seqi{Y}$, respectively. By \ref{proposition:hausdorff-uniform-factor}, there exists $\seqi{f'}$ such that the following diagram commutes
+
+    \[
+    \xymatrix{
+    X' \ar@{->}[r]^{f_i'} & Y_i' \\
+    X \ar@{->}[u] \ar@{->}[r]_{f_i} & Y_i \ar@{->}[u]
+    }
+    \]
+
+    for all $i \in I$. By (5) of \ref{definition:hausdorff-completion}, there is a correspondence between the uniformities of $X$ and $X'$, and $Y$ and $Y'$. Thus $X'$ is equipped with the initial uniformity genereated by $\seqi{f'}$.
+\end{proof}

src/topology/uniform/definition.tex

@@ -0,0 +1,277 @@
+\section{Uniform Structures}
+\label{section:uniformstructures}
+
+
+\begin{definition}[Inversion]
+\label{definition:inversion}
+    Let $X$ be a set and $U \subset X \times X$, then the \textbf{inversion} of $U$ is the set
+    \[
+    U^{-1} = \bracs{(y, x)| (x, y) \in U}
+    \]
+    A set $U \subset X \times X$ is \textbf{symmetric} if $U = U^{-1}$.
+\end{definition}
+
+\begin{definition}[Composition]
+\label{definition:composition}
+    Let $X$ be a set and $U, V \subset X \times X$, then the \textbf{composition} of $U$ and $V$ is the set
+    \[
+    U \circ V = \bracs{(x, z) \in X \times X| \exists y \in Y: (x, y) \in U, (y, z) \in V}
+    \]
+\end{definition}
+
+\begin{definition}[Slice]
+\label{definition:slice}
+    Let $X, Y$ be sets, $U \subset X \times Y$, and $A \subset X$, then
+    \[
+    U(A) = \bracs{y \in Y: (x, y) \in U, x \in A}
+    \]
+    is the \textbf{slice} of $U$ at $A$.
+\end{definition}
+
+
+\begin{definition}[Uniformity]
+\label{definition:uniformity}
+    Let $X$ be a set, then a non-empty family $\fU \subset X \times X$ is a \textbf{uniformity} on $X$ if:
+    \begin{enumerate}
+        \item[(F1)] For any $U \in \fU$ and $V \supset U$, $V \in \fU$.
+        \item[(F2)] For any $U, V \in \fU$, $U \cap V \in \fU$.
+        \item[(U1)] For every $U \in \fU$, $U \supset \Delta = \bracs{(x, x)| x \in X}$.
+        \item[(U2)] For any $U \in \fU$, $U^{-1} \in \fU$.
+        \item[(U3)] For any $U \in \fU$, there exists $V \in \fU$ such that $V \circ V \subset U$.
+    \end{enumerate}
+    The elements of $\fU$ are called the \textbf{entourages} of $\fU$, and the pair $(X, \fU)$ is a \textbf{uniform space}.
+
+    For any $x, y \in X$ and $U \in \fU$, $x$ and $y$ are \textbf{$U$-close} if $(x, y) \in U$.
+\end{definition}
+
+\begin{definition}[Subspace Uniformity]
+\label{definition:subspaceuniform}
+    Let $(X, \fU)$ be a uniform space and $A \subset X$, then the family
+    \[
+    \fU_A = \bracs{U \cap (A \times A)| U \in \fU}
+    \]
+    forms a uniformity on $A$, known as the \textbf{subspace uniformity} induced on $A$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:subspace-uniform-compatible}
+    Let $(X, \fU)$ be a uniform space and $A \subset X$, then the subspace topology of $A$ coincides with the topology induced by the subspace uniformity on $A$.
+\end{proposition}
+\begin{proof}
+    Let $x \in A$ and $U \in \fU$, then $U(x) \cap A = [U \cap (A \times A)](x)$. Thus $V \subset X$ is a neighbourhood of $x$ with respect to the subspace topology if and only if it is a neighbourhood of $x$ with respect to the topology on $A$ induced by the subspace uniformity.
+\end{proof}
+
+
+
+\begin{definition}[Fundamental System of Entourages]
+\label{definition:fundamentalentourage}
+    Let $(X, \fU)$ be a uniform space, then a family $\fB \subset \fU$ is a \textbf{fundamental system of entourages} for $\fU$ if for every $U \in \fU$, there exists $V \in \fB$ such that $V \subset U$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:fundamental-entourage-criterion}
+    Let $X$ be a set and $\fB \subset 2^{X \times X}$ be a non-empty family of sets. If
+    \begin{enumerate}
+        \item[(FB1)] For each $U, V \in \fB$, there exists $W \in \fB$ such that $W \subset U \cap V$.
+        \item[(UB1)] For each $V \in \fB$, $\Delta \subset V$.
+        \item[(UB2)] For each $V \in \fB$, there exists $W \in \fB$ such that $W \circ W \subset V$.
+    \end{enumerate}
+    then there exists a unique uniformity $\fU \subset 2^{X \times X}$, which is given by
+    \[
+    \fU = \bracs{U \subset X \times X| \exists V \in \fB: V \subset U}
+    \]
+\end{proposition}
+\begin{proof}
+    (F1): By definition of $\fU$.
+
+    (F2): For any $U, V \in \fU$, there exists $U_0, V_0 \in \fB$ such that $U_0 \subset U$ and $V_0 \subset V$. By (FB1), there exists $W \in \fB$ with $W \subset U_0 \cap V_0 \subset U \cap V$. Thus $U \cap V \in \fU$.
+
+    (U1) and (U2): For any $U \in \fU$, there exists $U_0 \in \fB$ with $U_0 \subset U$. By (UB1), $\Delta \subset U_0 \subset U$. By (UB2), there exists $V_0 \in \fB \subset \fU$ with $V_0 \circ V_0 \subset U_0 \subset U$.
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:symmetricfundamentalentourage}
+    Let $(X, \fU)$ be a uniform space, $\fB \subset \fU$ be a fundamental system of entourages, then
+    \[
+    \fB_S = \bracsn{U \cap U^{-1}| U \in \fB}
+    \]
+    is also a fundamental system of entourages.
+\end{lemma}
+\begin{proof}
+    By (F2), $\fB \subset \fU$. For any $U \in \fU$, there exists $V \in \fB$ such that $V \subset U$. In which case,
+    \[
+    U \supset V \supset V \cap V^{-1} \in \fB_S
+    \]
+    so $\fb_S$ is a fundamental system of entourages.
+\end{proof}
+
+\begin{definition}[Topology of a Uniform Space, {{\cite[Proposition 2.1.2]{Bourbaki}}}]
+\label{definition:uniformtopology}
+    Let $(X, \fU)$ be a uniform space and
+    \[
+    \cn: X \to 2^X \quad x \mapsto \bracsn{U(x)| U \in \fU}
+    \]
+    then there exists a unique topology $\topo \subset 2^X$ such that $\cn_\topo = \cn$, known as the \textbf{topology induced by the uniform structure $\fU$}.
+\end{definition}
+\begin{proof}
+Using \ref{proposition:neighbourhoodcharacteristic}, it is sufficient to show that $\cn(x)$ is non-empty for all $x \in X$, and that it satisfies (F1), (F2), (V1), and (V2). Firstly, since $\fU \ne \emptyset$, $\cn(x) \ne \emptyset$ for all $x \in X$.
+
+(F1): Let $U \in \fU$ and $V \supset U(x)$, then
+\[
+W = U \cup (\bracs{x} \times V) \supset U
+\]
+As $\fU$ satisfies (F1), $W \in \fU$. Thus
+\[
+V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
+\]
+
+(F2): Let $U, V \in \fU$, then $U(x) \cap V(x) = (U \cap V)(x)$. As $\fU$ satisfies (F2), $U \cap V \in \fU$ and $(U \cap V)(x) \in \cn(x)$.
+
+(V1): Let $U \in \fU$. By (U1), $\Delta \subset U$, so $x \in U(x)$.
+
+(V2): Let $U \in \fU$ and $x \in X$. By (U2), there exists $W \in \fU$ such that $W \circ W \subset U$. Let $y \in W$, then for any $z \in W(y)$, $(x, z) \in V$. Hence $W(y) \subset V(x)$ for all $y \in W$, so $V(x) \in \cn(y)$ for all $y \in W$.
+\end{proof}
+
+
+
+\begin{lemma}
+\label{lemma:compositiongymnastics}
+    Let $(X, \fU)$ be a uniform space, $V \in \fU$ be a symmetric entourage, and $M \subset X \times X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $(x, y) \in V \circ M \circ V$.
+        \item There exists $(p, q) \in M$ such that $(x, p) \in V$ and $(q, y) \in V$.
+        \item There exists $(p, q) \in M$ such that $x \in V(p)$ and $y \in V(q)$.
+        \item There exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q)$.
+        \item There exists $(p, q) \in M$ such that $p \in V(x)$ and $q \in V(y)$.
+        \item There exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    $(1) \Rightarrow (2)$: By definition of composition, there exists $q \in X$ such that $(x, q) \in M$ and $(q, y) \in V$. Similarly, there exists $p \in X$ such that $(x, p) \in V$, $(p, q) \in M$, and $(q, y) \in V$.
+
+    $(2) \Rightarrow (3)$: By symmetry, $(p, x) \in V$. Thus $p \in V(x)$ and $q \in V(y)$.
+
+    $(3) \Leftrightarrow (4)$, $(5) \Leftrightarrow (6)$: By definition of product.
+
+    $(3) \Rightarrow (5)$: By symmetry, $a \in V(b)$ if and only if $b \in V(a)$.
+
+    $(5) \Rightarrow (1)$: By symmetry, $p \in V(x)$ implies that $(x, p) \in V$, and $(q, y) \in V$.
+\end{proof}
+
+
+\begin{proposition}[{{\cite[Proposition 2.1.2]{Bourbaki}}}]
+\label{proposition:uniformneighbourhood}
+    Let $(X, \fU)$ be a uniform space, $V \in \fU$ be a symmetric entourage, and $M \subset X \times X$, then:
+    \begin{enumerate}
+        \item $V \circ M \circ V \in \cn(M)$.
+        \item Let $\fB$ be the set of all symmetric entourages, then $\ol{M} = \bigcap_{V \in \fB}V \circ M \circ V$.
+    \end{enumerate}
+    with respect to the product topology on $X \times X$.
+\end{proposition}
+\begin{proof}
+    (1): Let $(x, y) \in V \circ M \circ V$. By \ref{lemma:compositiongymnastics}, there exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q) \in \cn(p, q)$. In particular, if $(x, y) \in M$, then this implies that $V \circ M \circ V \in \cn(x, y)$. Thus $V \circ M \circ V \in \cn(M)$.
+
+    (2): Let $(x, y) \in X \times X$. By \ref{lemma:compositiongymnastics}, the following are equivalent:
+    \begin{enumerate}
+        \item[(a)] $(x, y) \in V \circ M \circ V$ for all $V \in \fB$.
+        \item[(b)] For every $V \in \fB$, there exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
+    \end{enumerate}
+
+    As $\bracsn{V(x) \times V(y)| V \in \fB}$ is a fundamental system of neighbourhoods at $(x, y)$, (b) is equivalent to $(x, y) \in \ol{M}$. Therefore
+
+    \[
+    \ol{M} = \bigcap_{V \in \fB}V \circ M \circ V
+    \]
+\end{proof}
+
+\begin{proposition}[{{\cite[Corollary 2.1.1]{Bourbaki}}}]
+\label{proposition:uniformclosure}
+    Let $(X, \fU)$ be a uniform space and $A \subset X$, then:
+    \begin{enumerate}
+        \item For any symmetric entourage $V \in \fU$, $V(A) \in \cn(A)$.
+        \item Let $\fB \subset \fU$ be the family of all symmetric entourages, then $\ol{A} = \bigcap_{U \in \fB}U(A)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $x \in A$, $x \in V(x) \subset V(A)$. By (F1) of $\cn(x)$, $V(A) \in \cn(x)$ for all $x \in A$. Thus $V(A) \in \cn(A)$.
+
+    (2): Let $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item[(a)] For all $V \in \fB$, $x \in V(A)$.
+        \item[(b)] For all $V \in \fB$, there exists $y \in A$ such that $x \in V(y)$.
+        \item[(c)] For all $V \in \fB$, $V(x) \cap A \ne \emptyset$.
+    \end{enumerate}
+
+    Since $\bracs{V(y): V \in \fB}$ is a fundamental system of neighbourhoods at $y$ (\ref{lemma:symmetricfundamentalentourage}), (c) is equivalent to $x \in \overline{A}$. Therefore $\ol{A} = \bigcap_{U \in \fB}U(A)$.
+\end{proof}
+
+\begin{proposition}[{{\cite[Corollary 2.1.2]{Bourbaki}}}]
+\label{proposition:goodentourages}
+    Let $(X, \fU)$ be a uniform space, then the following famillies of sets form fundamental systems of entourages for $\fU$:
+    \begin{enumerate}
+        \item $\mathfrak{O} = \bracs{U^o| U \in \fU}$
+        \item $\mathfrak{K} = \bracsn{\overline{U}| U \in \fU}$.
+    \end{enumerate}
+    By \ref{lemma:symmetricfundamentalentourage}, there exists fundamental systems of entourages for $\fU$ consisting of symmetric and open/closed sets.
+\end{proposition}
+\begin{proof}
+    Let $U \in \fU$, then there exists a symmetric entourage $V \in \fU$ such that $V \circ V \circ V \subset U$ by (U2) and \ref{lemma:symmetricfundamentalentourage}. By (1) of \ref{proposition:uniformneighbourhood}, $V \circ V \circ V \in \cn(V)$. Since
+    \[
+    V \subset (V \circ V \circ V)^o \subset V \circ V \circ V \subset U
+    \]
+    the interior $(V \circ V \circ V)^o \in \fU$, and $U$ contains the interior of an entourage. Thus (1) is a fundamental system of entourages.
+
+    On the other hand, by (2) of \ref{proposition:uniformneighbourhood},
+    \[
+    V \subset \overline{V} \subset V \circ V \circ V \subset U
+    \]
+    So $\overline{V} \in \fU$ and is contained in $U$. Therefore (2) is also a fundamental system of entourages.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:openentourageneighbourhoods}
+    Let $(X, \fU)$ be a uniform space, $U \in \fU$ be a symmetric, open entourage, and $M \subset X$, then $U(M)$ is open.
+\end{lemma}
+\begin{proof}
+    Let $U \in \fV$ and $y \in U(M)$. Since $U \subset X \times X$ is open, there exists $x \in M$, $V \in \cn(x)$, and $V' \in \cn(y)$ such that $(x, y) \subset V \times V' \subset U$. In which case, $U(M) \supset V' \in \cn(y)$. Hence $U(M)$ is open.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:uniform-neighbourhoods}
+    Let $X$ be a uniform space and $x \in X$, then the closed neighbourhoods of $x$ form a fundamental system of neighbourhoods at $x$.
+\end{proposition}
+\begin{proof}
+    By \ref{proposition:goodentourages} and \ref{lemma:openentourageneighbourhoods}, the closed neighbourhoods form a fundamental system of neighbourhoods.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:uniform-regular}
+    Let $(X, \fU)$ be a uniform space, then the following are equivalent:
+    \begin{enumerate}
+        \item $X$ is T0.
+        \item $X$ is T1.
+        \item $X$ is Hausdorff.
+        \item $X$ is regular.
+        \item $\Delta = \bigcap_{U \in \fU}U$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    $(1) \Rightarrow (5)$: Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U(x) \in \cn(x)$ such that $y \not\in U$. In which case, $(x, y) \not\in U$ and $\Delta \supset \bigcap_{U \in \fU}U$.
+
+    $(5) \Rightarrow (2)$: By \ref{proposition:goodentourages}, $\ol \Delta \subset \bigcap_{U \in \fU}\ol U = \Delta$, so $\ol \Delta$ is closed. By (6) of \ref{definition:hausdorff}, $X$ is Hausdorff.
+
+    $(1) \Rightarrow (4)$: $X$ is T1 and satisfies (2) of \ref{definition:regular} by \ref{proposition:uniform-neighbourhoods}, so $X$ is regular.
+
+    $(4) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$:  (T3) $\Rightarrow $ (T2) $\Rightarrow$ (T1) $\Rightarrow$ (T0).
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:subspace-entourage}
+    Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $\bracsn{\overline{U}: U \in \fU_A}$ forms a fundamental system of entourages for $X$.
+\end{proposition}
+\begin{proof}
+    Let $U \in \fU$ be an open entourage, then by (3) of \ref{definition:dense}, $\overline{U \cap (A \times A)} = \overline{U}$ for all $U \in \fU$, so $\overline{U \cap (A \times A)}$ is an entourage. By \ref{proposition:goodentourages}, every closed entourage of $X$ contains an element of $\bracsn{\overline{U}: U \in \fU_A}$.
+\end{proof}

src/topology/uniform/index.tex

@@ -0,0 +1,7 @@
+\chapter{Uniform Spaces}
+\label{chap:uniform-spaces}
+
+\input{./src/topology/uniform/definition.tex}
+\input{./src/topology/uniform/uc.tex}
+\input{./src/topology/uniform/complete.tex}
+\input{./src/topology/uniform/completion.tex}

src/topology/uniform/uc.tex

@@ -0,0 +1,149 @@
+\section{Uniform Continuity}
+\label{section:uc}
+
+\begin{definition}[Uniform Continuity]
+\label{definition:uniformcontinuity}
+    Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f: X \to Y$, then the following are equivalent:
+    \begin{enumerate}
+        \item For every $V \in \mathfrak{V}$, there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$.
+        \item For every $V \in \mathfrak{V}$, $(f \times f)^{-1}(V) \in \fU$.
+    \end{enumerate}
+    If the above holds, then $f$ is a \textbf{uniformly continuous} function.
+
+    The collection $UC(X; Y)$ denotes the set of all uniformly continuous functions from $X$ to $Y$.
+\end{definition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: Since $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$, $V' \subset f^{-1}(V)$. By (F1) of $\fU$, $f^{-1}(V) \in \fU$.
+
+    $(2) \Rightarrow (1)$: Take $V' = (f \times f)^{-1}(V)$, then $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:uniform-continuous}
+    Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f \in C(X; Y)$.
+\end{proposition}
+\begin{proof}
+    Let $x \in X$ and $V(f(x)) \in \cn(f(x))$, then since $f^{-1}(V(f(x))) =[f^{-1}(V)](x)$, by (2) of \ref{definition:uniformcontinuity}, $f^{-1}(V(f(x))) \in \cn(x)$. As this holds for all $x \in X$, $f$ is continuous.
+\end{proof}
+
+
+
+\begin{definition}[Initial Uniformity]
+\label{definition:initial-uniformity}
+    Let $X$ be a set, $\bracsn{(Y_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, and $\seqi{f}$ be a family of maps such that $f_i: X \to Y_i$ for each $i \in I$, then there exists a uniformity $\fU$ on $X$ such that:
+    \begin{enumerate}
+        \item For each $i \in I$, $f_i \in UC(X; Y_i)$.
+        \item[(U)] If $\mathfrak{V}$ is a uniformity on $X$ satisfying $(1)$, then $\mathfrak{V} \supset \fU$.
+    \end{enumerate}
+    Moreover,
+    \begin{enumerate}
+        \item[(3)] The family
+
+        \[
+        \fB = \bracs{\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \fU_j}
+        \]
+
+        is a fundamental system of entourages for $\fU$.
+    \end{enumerate}
+
+
+    known as the \textbf{initial uniformity} on $X$ generated by $\seqi{f}$.
+\end{definition}
+\begin{proof}
+    (3): Since the diagonal is mapped to the diagonal and $\fB$ is closed under intersections, it is sufficient to verify (UB3) for $\fB$. Let $J \subset I$ be finite and $\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \in \fB$, then there exists $\bracs{V_j}_{j \in J}$ such that $V_j \circ V_j \subset U_j$ for each $j \in J$. In which case, for any $(x, y), (y, z) \in f_j^{-1}(V_j)$, $(f(x), f(y)), (f(y), f(z)) \in V_j$ and $(f(x), f(z)) \in U_j$. Thus $(f_j \times f_j)^{-1}(V_j) \circ (f_j \times f_j)^{-1}(V_j) \subset (f_j \times f_j)^{-1}(U_j)$, and
+    \[
+    \paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(V_j)} \circ \paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(V_j)} \subset \bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j)
+    \]
+    By \ref{proposition:fundamental-entourage-criterion}, there exists a uniformity $\fU$ such that $\fB$ is a fundamental system of entourages for $\fU$.
+
+    (1): $\fU \supset (f_i \times f_i)^{-1}(\fU_i)$ for all $i \in I$.
+
+    (U): For any $i \in I$, $\mathfrak{V} \supset (f_i \times f_i)^{-1}(\fU_i)$. By (F2), $\mathfrak{V} \supset \fB$, so $\mathfrak{V} \supset \fU$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:initial-uniform-topology}
+    Let $X$ be a set, $\bracsn{(Y_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, and $\seqi{f}$ where $f_i: X \to Y_i$ for each $i \in I$, then the initial topology $\topo$ on $X$ coincides with the topology $\topo_U$ induced by the initial uniformity.
+\end{proposition}
+\begin{proof}
+    By \ref{proposition:uniform-continuous} and (1) of \ref{definition:initial-uniformity}, $f_i: X \to Y_i$ is continuous with respect to $\topo_U$ for all $i \in I$. By (U) of \ref{definition:initial-topology}, $\topo_U \supset \topo$.
+
+    On the other hand, let $x \in X$ and $U \in \cn_{\topo_U}(x)$, then there exists an entourage $V$ such that $U = V(x)$. By (3) of \ref{definition:initial-uniformity}, there exists $J \subset I$ finite and $\seqj{V}$ such that $\bigcap_{j \in J}(f_j \times f_j)^{-1}(V_j) \subset V$ and each $V_j \in \fU_j$. In which case, $f_j^{-1}(V_j(f_j(x))) \in \cn(x)$ for all $j \in J$. Using (F2),
+    \[
+    W = \bigcap_{j \in J}f_j^{-1}(V_j(f_j(x))) \in \cn(x)
+    \]
+    where for any $y \in W$, $(f_j(x), f_j(y)) \in V_j$ for all $j \in J$. Thus $f(y) \in V(x) = U$, and $\topo_U \subset \topo$.
+\end{proof}
+
+
+\begin{definition}[Product Uniformity]
+\label{definition:product-uniform}
+    Let $\bracs{(X_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, then the \textbf{product uniformity} $\fU$ on $X = \prod_{i \in I}X_i$ is the initial uniformity generated by the projections $\seqi{\pi}$, and:
+    \begin{enumerate}
+        \item The family
+
+        \[
+        \fB = \bracs{\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \fU_j}
+        \]
+
+        forms a fundamental system of entourages for $\fU$.
+        \item[(U)] For any uniform space $(Y, \mathfrak{V})$ and $\seqi{f}$ where $f_i \in UC(Y; X_i)$ for all $i \in I$, there exists a unique $f \in UC(Y; X)$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+        Y \ar@{->}[d]_{f} \ar@{->}[rd]^{f_i} &  \\
+        X \ar@{->}[r]_{\pi_i} & X_i
+        }
+        \]
+
+        for all $i \in I$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): By (3) of \ref{definition:initial-uniformity}.
+
+    (U): Let $f = \prod_{i \in I}f_i$, then $f: Y \to X$ is the unique function such that the diagram commutes for all $i \in I$.
+
+    For each $J \subset I$ finite and $\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j) \in \fU$,
+    \[
+    (f \times f)^{-1}\paren{\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_i)} = \bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \in \mathfrak{V}
+    \]
+    by (F2) of $\mathfrak{V}$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:product-uniform-topology}
+    Let $\bracs{(X_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, then the topology $\topo_U$ induced by the product uniformity coincides the product topology $\topo_P$.
+\end{proposition}
+\begin{proof}
+    By \ref{proposition:initial-uniform-topology}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:initial-product-uniformity}
+    Let $X$, $\seqi{Y}$ be uniform spaces, $\seqi{f}$ where $f_i \in UC(X; Y_i)$ for each $i \in I$, then
+    \begin{enumerate}
+        \item There exists a unique $f \in UC(X; \prod_{i \in I}Y_i)$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+         & \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} &  \\
+        X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} &  & Y_i
+        }
+        \]
+
+        for all $i \in I$.
+        \item If $X$ is T0 and equipped with the initial uniformity induced by $\seqi{f}$, then $f$ is an isomorphism onto its image.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): By (U) of \ref{definition:product-uniform}.
+
+    (2): By (2) of \ref{proposition:initial-product-topology}, $f$ is injective.
+
+    Let $U$ be an entourage in $X$. Assume without loss of generality that there exists $J \subset I$ finite and $\bracs{U_j}_{j \in J}$ such that $U = \bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j)$. In which case, $\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j)$ is open in $\prod_{i \in I}Y_i$ and
+    \[
+    (f \times f)(U) = (f \times f)(X) \cap \bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j)
+    \]
+\end{proof}