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Added the Mean Value Theorem.

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Bokuan Li
2026-02-03 00:54:44 -05:00
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src/dg/derivative/index.tex
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\input{./src/dg/derivative/derivative.tex}
\input{./src/dg/derivative/sets.tex}
\input{./src/dg/derivative/mvt.tex}

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src/dg/derivative/mvt.tex Normal file
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\section{The Mean Value Theorem}
\label{section:mean-value-theorem}
\begin{definition}[Right-Differentiable]
\label{definition:right-differentiable-mvt}
Let $-\infty < a < b < \infty$, $E$ be a separated topological vector space, $f: [a, b] \to E$, and $x \in [a, b)$, then $f$ is \textbf{right-differentiable at $x$} if
\[
D^+f(x) = \lim_{t \downto 0} \frac{f(x + t) - f(x)}{t}
\]
exists.
\end{definition}
\begin{lemma}[{{\cite[Lemma 4.6.2]{Bogachev}}}]
\label{lemma:right-differentiable-inequality}
Let $-\infty < a < b < \infty$, $f, g \in C([a,b]; \real)$, and $N \subset [a, b]$ be at most countable such that:
\begin{enumerate}
\item[(a)] $f, g$ are right-differentiable on $[a, b] \setminus N$.
\item[(b)] For every $x \in [a, b] \setminus N$, $D^+f(x) \le D^+g(x)$.
\end{enumerate}
then for any $x \in [a, b]$, $f(x) - f(a) \le g(x) - g(a)$.
\end{lemma}
\begin{proof}
First assume that for every $x \in [a, b] \setminus N$, $D^+f(x) < D^+g(x)$.
Let $\seq{x_n}$ be an enumeration of $N$. For each $x \in [a, b]$, let $N(x) = \bracs{n \in \natp|x_n \in [a, x)}$. Let $\eps > 0$ and define
\[
S = \bracs{x \in [a, b] \bigg | f(y) - f(a) \le g(y) - g(a) + \eps\sum_{n \in N(x)}2^{-n} \quad \forall y \in [a, x]}
\]
then by continuity of $f$ and $g$, $S$ is closed.
Let $x \in S$ and suppose that $s < b$. If $x \in [a, b] \setminus N$, then since
\[
\lim_{t \downto 0}\frac{f(x + t) - f(x)}{t} < \lim_{t \downto 0}\frac{g(x + t) - f(x)}{t}
\]
there exists $\delta > 0$ such that $f(x + t) - f(x) < g(x + t) < g(x)$ for all $t \in [0, \delta)$. In which case, $[x, x + \delta) \subset S$.
If $x = x_n \in N$, then by continuity of $f$ and $g$, there exists $\delta > 0$ such that $f(x + t) - f(x) \le g(x + t) - g(x) + 2^{-n}$ for all $t \in (0, \delta)$. Hence $[x, x + \delta) \subset S$ as well.
Therefore $S = [a, b]$. As this holds for all $\eps > 0$, $f(b) - f(a) \le g(b) - g(a)$.
Finally, suppose that $D^+f(x) \le D^+g(x)$ for all $x \in [a, b] \setminus N$. Let $\eps > 0$ and $h(x) = g(x) + \eps(x - a)$. By the preceding case, $f(b) - f(a) \le g(b) - g(a) + \eps(b - a)$. As this holds for all $\eps > 0$, $f(x) - f(a) \le g(x) - g(a)$.
\end{proof}
\begin{theorem}[{{\cite[Theorem 4.6.1]{Bogachev}}}]
\label{theorem:right-differentiable-convex-form}
Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $f \in C([a, b]; E)$, $g \in C([a, b]; \real)$, $N \subset [a, b]$ be at most countable, and $B \subset E$ be a closed convex set such that:
\begin{enumerate}
\item $f, g$ are right-differentiable on $[a, b] \setminus N$.
\item For each $x \in [a, b] \setminus N$, $D^+f(x) \in D^+g(x)B$.
\item $g$ is non-decreasing.
\end{enumerate}
then
\[
f(b) - f(a) \in [g(b) - g(a)]B
\]
\end{theorem}
\begin{proof}
Let $\phi \in E^*$ and $x \in [a, b] \setminus N$, then since $g$ is non-decreasing,
\begin{align*}
\lim_{t \downto 0} \frac{\phi(f(x + t) - f(x))}{t} &\le \lim_{t \downto 0}\sup_{z \in B} \frac{\phi(g(x + t) - g(x)z)}{t} \\
D^+(\phi \circ f)(x) &\le \sup_{z \in B}\phi(z) \cdot \lim_{t \downto 0}\frac{\phi(g(x + t) - g(x))}{t} \\
D^+(\phi \circ f)(x) &\le D^+g(x) \cdot \sup_{z \in B}\phi(z)
\end{align*}
By \ref{lemma:right-differentiable-inequality},
\[
\phi(f(b) - f(a)) \le (g(b) - g(a)) \cdot \sup_{z \in B}\phi(z)
\]
Suppose that $f(b) - f(a) \not\in [g(b) - g(a)]B$. Given that $B$ is closed and convex, by the Hahn-Banach theorem (\ref{theorem:hahn-banach-geometric-2}), there exists $\phi \in E^*$ such that
\[
\phi(f(b) - f(a)) > \sup_{x \in B}\phi[(g(b) - g(a))b] = \phi(g(b) - g(a)) \cdot \sup_{x \in B}\phi(x)
\]
which is impossible. Therefore $f(b) - f(a) \in [g(b) - g(a)]B$.
\end{proof}
\begin{theorem}[Mean Value Theorem]
\label{theorem:mean-value-theorem}
Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $S \subset [a, b]$ be at most countable, and $f \in C([a, b]; E)$ be differentiable on $(a, b) \setminus N$, then
\[
f(b) - f(a) \in \overline{\text{Conv}\bracs{Df(x)(b - a)| x \in (a, b) \setminus N}}
\]
\end{theorem}
\begin{proof}
By \ref{proposition:derivative-sets-real}, $f$ is right-differentiable on $(a, b) \setminus N$ with
\[
D^+f(x) = Df(x) \in \bracs{Df(x)|x \in (a, b) \setminus N}
\]
for all $x \in (a, b)$. Let $g(x) = x$, then by \ref{theorem:right-differentiable-convex-form},
\[
f(b) - f(a) \in \overline{(b - a)\text{Conv}\bracs{Df(x)|x \in (a, b) \setminus N}}
\]
\end{proof}
\begin{proposition}
\label{proposition:zero-derivative-constant}
Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and connected, $f: V \to F$ be Gateaux-differentiable on $V$ such that $Df(x) = 0$ for all $x \in V$, then $f$ is constant.
\end{proposition}
\begin{proof}
Let $x, y \in V$ such that $\bracs{(1 - t)x + ty|t \in [0, 1]} \subset V$. Since $f$ is Gateaux-differentiable, $g: [0, 1] \to F$ defined by $g(t) = f((1 - t)x + ty)$ is differentiable with $Dg(t) = 0$ for all $t \in [0, 1]$. By the Mean Value Theorem (\ref{theorem:mean-value-theorem}), $f(y) - f(x) = 0$.
Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. Since $E$ is locally path-connected (\ref{proposition:tvs-locally-connected}), $W$ is open.
Let $y \in \overline{W}$. By \ref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_E(0)$ radial such that $U + y \subset V$. Since any point in $U$ is joined to $y$ via a line segment, $f$ is constant on $U + y$. As $y \in \overline{W}$, $U + y \cap W \ne \emptyset$, so $y \in W$.
Since $W$ is both open and closed, $W$ must coincide with $V$.
\end{proof}

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src/dg/derivative/sets.tex
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@@ -39,6 +39,12 @@
Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
\end{definition}
\begin{definition}]
\label{definition:derivative-garden}
Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b \subset 2^E$ be the collection of all finite, compact, and bounded subsets, respectively, then differentiability with respect to $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b$ correspond to \textbf{Gateaux}, \textbf{Hadamard}, and \textbf{Fréchet} differentiability.
\end{definition}
\begin{proposition}[{{\cite[Proposition 4.5.2]{Bogachev}}}]
\label{proposition:chain-rule-sets}
Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be upward-directed families that contain all finite sets. If:
@@ -130,3 +136,32 @@
\]
is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
\end{definition}
\begin{proposition}
\label{proposition:derivative-sets-real}
Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be an upward-directed system that contains finite sets, then
\begin{enumerate}
\item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{B(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
\item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is ($\sigma$-)differentiable at $x_0$ if and only if
\[
\lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
\]
exists. In which case, the above limit is identified with the ($\sigma$-)derivative of $f$ at $0$.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $r \in \mathcal{R}_\sigma(\real; E)$. For any $R > 0$ and $U \in \cn_E(0)$, there exists $\delta > 0$ such that $t^{-1}r(tR), t^{-1}r(-tR) \in U$ for all $t \in (0, \delta)$. Thus $t^{-1}r(tB(0, R)) \subset U$, and $r \in \mathcal{R}_{B(\real)}(\real; E)$.
(2): Suppose that $f$ is differentiable at $x_0$, then there exists $r \in \mathcal{R}_\sigma$ such that for any $t \in \real$ with $x_0 + t \in U$,
\begin{align*}
f(x_0 + t) - f(x_0) &= Df(x_0)(t) + r(t) \\
\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1) + t^{-1}r(t) \\
\lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1)
\end{align*}
Now suppose that $v = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$ exists. Let $T: \real \to E$ be defined by $t \mapsto tv$, then
\[
\lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
\]
and $Df(x_0) = T$.
\end{proof}

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src/fa/tvs/definition.tex
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@@ -202,3 +202,12 @@
(Uniqueness): Let $\mathcal{S} \subset 2^E$ be a topology on $E$ satisfying (1) and (2), then for each $x \in E$, $\cn_{(E, \mathcal{S})}(x) = \cn_{(E, \mathcal{T})}(x)$. By \ref{proposition:neighbourhoodcharacteristic}, $\mathcal{S} = \mathcal{T}$.
\end{proof}
\begin{proposition}
\label{proposition:tvs-locally-connected}
Let $E$ be a TVS over $K \in \RC$, then $E$ is locally connected.
\end{proposition}
\begin{proof}
Let $U \in \cn(0)$ be radial, then for any $y \in U$, the mapping $t \mapsto ty$ is a path from $0$ to $y$ contained in $U$. Thus $U$ is path-connected. By \ref{proposition:tvs-0-neighbourhood-base}, the radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods, so the path-connected neighbourhoods of $0$ forms a fundamental system as well.
\end{proof}

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src/fa/tvs/spaces-of-linear.tex
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@@ -39,7 +39,7 @@
\begin{definition}[Space of Bounded Linear Maps]
\label{definition:bounded-linear-map-space}
Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $L_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology.
Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $B_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology.
\end{definition}
\begin{proposition}
@@ -48,7 +48,7 @@
\begin{enumerate}
\item The map
\[
I: L_{\mathfrak{S}}^k(E; L_{\mathfrak{S}}(E; F)) \to L^{k+1}_{\mathfrak{S}}(E; F)
I: B_{\mathfrak{S}}^k(E; B_{\mathfrak{S}}(E; F)) \to B^{k+1}_{\mathfrak{S}}(E; F)
\]
defined by
\[
@@ -57,7 +57,7 @@
is an isomorphism.
\item The map
\[
I: \underbrace{L_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to L^k_{\mathfrak{S}}(E; F)
I: \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to B^k_{\mathfrak{S}}(E; F)
\]
defined by
\[
@@ -67,30 +67,30 @@
\end{enumerate}
which allows the identification
\[
\underbrace{L_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = L^k_{\mathfrak{S}}(E; F)
\underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = B^k_{\mathfrak{S}}(E; F)
\]
under the map $I$ in (2).
\end{proposition}
\begin{proof}
(1): To see that $I$ is surjective, let $T \in L_{\mathfrak{S}}^{k+1}(E; F)$ and
(1): To see that $I$ is surjective, let $T \in B_{\mathfrak{S}}^{k+1}(E; F)$ and
\[
I^{-1}T: E \to L_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot)
I^{-1}T: E \to B_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot)
\]
Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
\[
T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
\]
by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in L_{\mathfrak{S}}(E; F)$.
by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\mathfrak{S}}(E; F)$.
In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $L_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(L_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in L^k_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; F))$.
In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F))$.
It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \mathfrak{S}$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
On the other hand, let $S_1 \in \mathfrak{S}$ and $E(S_2, U)$ be an entourage of $L_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$. Let $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
On the other hand, let $S_1 \in \mathfrak{S}$ and $E(S_2, U)$ be an entourage of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$. Let $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
(2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
\[
\underbrace{L_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = L^k_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; F)) = L^{k+1}_{\mathfrak{S}}(E; F)
\underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F)) = B^{k+1}_{\mathfrak{S}}(E; F)
\]
Thus (2) holds for all $k \in \natp$.
\end{proof}

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@@ -0,0 +1,46 @@
\section{Connectedness}
\label{section:connected}
\begin{definition}[Connected]
\label{definition:connected}
Let $X$ be a topological space, then the following are equivalent:
\begin{enumerate}
\item For any $\emptyset \ne U, V \subset X$ open with $U \cup V = X$, $U \cap V \ne \emptyset$.
\item There exists no surjective $f \in C(X; \bracs{0, 1})$.
\item For any $U \subset X$ open and closed, either $U = \emptyset$ or $U = X$.
\end{enumerate}
If the above holds, then $X$ is \textbf{connected}.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Let $f \in C(X; \bracs{0, 1})$, then $X = f^{-1}(0) \cup f^{-1}(1)$. If $f$ is surjective, then $f^{-1}(0) \cap f^{-1}(1) \ne \emptyset$, which is impossible.
$\neg (1) \Rightarrow \neg (2)$: If there exists $\emptyset \ne U, V \subset X$ open with $X = U \sqcup V$, then $f = \one_U \in C(X; \bracs{0, 1})$ is surjective.
$\neg (3) \Rightarrow \neg (1)$: Let $\emptyset \ne U \subsetneq X$ be open and closed, then $X = U \sqcup U^c$ is a disjoint union of two non-empty open sets.
$\neg (1) \Rightarrow \neg (3)$: Let $\emptyset \ne U, V \subset X$. If $X = U \sqcup V$, then $\emptyset \ne U \subsetneq X$ is open and closed.
\end{proof}
\begin{proposition}
\label{proposition:connected-union}
Let $X$ be a topological space, $\seqi{A} \subset 2^X$ be connected with $\bigcap_{i \in I}A_i \ne \emptyset$, then $\bigcup_{i \in I}A_i$ is connected.
\end{proposition}
\begin{proof}
Let $x \in \bigcap_{i \in I}A_i = A$ and $U, V \subset X$ be open with $U \cap A, V \cap A \ne \emptyset$. Assume without loss of generality that $x \in U$. Let $i \in I$ such that $V \cap A_i \ne \emptyset$, then since $U \cap A_i \ne \emptyset$, $U \cap V \ne \emptyset$ by connectedness of $A_i$.
\end{proof}
\begin{proposition}
\label{proposition:connected-image}
Let $X$ be a connected space, $Y$ be a topological space, and $f \in C(X; Y)$, then $f(X)$ is connected.
\end{proposition}
\begin{proof}
Let $U, V \subset Y$ be open with $U \cap f(X), V \cap f(X) \ne \emptyset$. By continuity of $f$ and connectedness of $X$, $f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V) \ne \emptyset$. Hence $U \cap V \ne \emptyset$.
\end{proof}
\begin{definition}[Connected Component]
\label{definition:connected-component}
Let $X$ be a topological space and $A \subset X$ be connected, then there exists a unique connected set $C \supset A$ such that for any $C' \supset A$ connected, $C \supset C'$. The set $C$ is the \textbf{connected component} of $A$.
\end{definition}
\begin{proof}
Let $C$ be the union of all connected sets that contain $A$, then $C$ is connected by \ref{proposition:connected-union}, and is the maximum connected set containing $A$ by definition.
\end{proof}

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src/topology/main/index.tex
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@@ -12,6 +12,9 @@
\input{./src/topology/main/regular.tex}
\input{./src/topology/main/normal.tex}
\input{./src/topology/main/quotient.tex}
\input{./src/topology/main/connected.tex}
\input{./src/topology/main/path-connected.tex}
\input{./src/topology/main/local-path-connected.tex}
\input{./src/topology/main/unity.tex}
\input{./src/topology/main/compact.tex}
\input{./src/topology/main/sigma-compact.tex}

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@@ -0,0 +1,25 @@
\section{Local Path-Connectedness}
\label{section:local-path-connected}
\begin{definition}[Locally Path-Connected]
\label{definition:locally-path-connected}
Let $X$ be a topological space, then $X$ is \textbf{locally path-connected} if for every $x \in X$, there exists a fundamental system of neighbourhoods consisting of path-connected sets at $x$.
\end{definition}
\begin{proposition}
\label{proposition:locally-path-connected-properties}
Let $X$ be a locally path-connected space, then
\begin{enumerate}
\item The path components of $X$ are open.
\item The path components of $X$ are equal to its components.
\item $X$ is connected if and only if it is path-connected.
\item Every open subset of $X$ is locally path-connected.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $A \subset X$ be a path component and $x \in A$, then there exists $U \in \cn(x)$ connected. By \ref{proposition:path-connected-union}, $A \cup U \in \cn(x)$ is also connected. Since $A$ is a path-component, $A \cup U \subset A \in \cn(x)$. Thus $A$ is open by \ref{lemma:openneighbourhood}.
(2): Let $P$ be a path component in $X$ and $C \supset P$ be its connected components. If $C$ is not path-connected, then $P \subsetneq C$ there exists path-components $\seqi{P} \subset 2^C$ such that $C = \bigsqcup_{i \in I}P_i$. In which case, $C \setminus P$ is open by (1), and $C$ is not connected, which is impossible.
(3): By (2) applied to $X$.
\end{proof}

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@@ -0,0 +1,49 @@
\section{Path-Connectedness}
\label{section:path-connected}
\begin{definition}[Path]
\label{definition:path}
Let $X$ be a topological space, $x, y \in X$, then a \textbf{path} from $x$ to $y$ is a mapping $f \in C([0, 1]; X)$ such that $f(0) = x$ and $f(1) = y$.
\end{definition}
\begin{definition}[Path-Connected]
\label{definition:path-connected}
Let $X$ be a topological space, then $X$ is \textbf{path-connected} if for every $x, y \in X$, there exists a path from $x$ to $y$.
\end{definition}
\begin{proposition}
\label{proposition:path-connected-connected}
Let $X$ be a path-connected space, then $X$ is connected.
\end{proposition}
\begin{proof}
Let $U, V \subset [0, 1]$ be open with $[0, 1] = U \cup V$. If $\sup U = \sup V$, then $\sup U = \sup V = 1$ and $U, V \in \cn^o(1)$, so $U \cap V \ne \emptyset$. If $\sup U < \sup V \le 1$, then $x \not\in U$ and $x \in V$. In which case, $V \in \cn^o(x)$ and $V \cap U \ne \emptyset$. Therefore $[0, 1]$ is connected.
Fix $x \in X$. For any $y \in X$, let $f_y \in C([0, 1]; X)$ be a path from $x$ to $y$, then $f_y([0, 1])$ is connected with $x, y \in f_y([0, 1])$ by \ref{proposition:connected-image}. By \ref{proposition:connected-union},
\[
X = \bigcup_{y \in X}f_y([0, 1])
\]
is connected.
\end{proof}
\begin{proposition}
\label{proposition:path-connected-union}
Let $X$ be a topological space, $\seqi{A}$ be path-connected sets with $\bigcap_{i \in I}A_i \ne \emptyset$, then $\bigcup_{i \in I}A_i$ is connected.
\end{proposition}
\begin{proof}
Let $x \in \bigcap_{i \in I}A_i$, $i, j \in I$, $y \in A_i$, and $z \in A_j$. By connectedness of $A_i$ and $A_j$, there exists paths $f \in C([0, 1]; X)$ from $y$ to $x$, and $g \in C([0, 1]; X)$ from $x$ to $z$. Thus the concatenation
\[
g \cdot f: [0, 1] \to X \quad t \mapsto \begin{cases}
f(2t) &t \in [0, 1/2] \\
g(2(t - 1/2)) &t \in [1/2, 1]
\end{cases}
\]
is a path from $y$ to $z$.
\end{proof}
\begin{definition}[Path Component]
\label{definition:path-component}
Let $X$ be a topological space and $A \subset X$ be path-connected, then there exists a unique path-connected set $C \supset A$ such that for any $C' \supset A$ path-connected, $C \supset C'$. The set $C$ is the \textbf{path-component} of $A$.
\end{definition}
\begin{proof}
Let $C$ be the union of all path-connected sets containing $A$, then $C$ is path-connected by \ref{proposition:path-connected-union} and the maximum path-connected set containing $A$ by definition.
\end{proof}