Added the Mean Value Theorem.

src/topology/main/connected.tex

@@ -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}

src/topology/main/index.tex

@@ -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}

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

@@ -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}

src/topology/main/path-connected.tex

@@ -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}