\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 \autoref{proposition:connected-image}. By \autoref{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 \autoref{proposition:path-connected-union} and the maximum path-connected set containing $A$ by definition.
\end{proof}