Path-Components Partition the Space
# Statement
Suppose $X$ is a Topological Space. Then the collection $$\mathcal{P} := {P \subset X : P \text{ is a nonempty path-connected component}}$$ is a Partition of $X$.
# Proof
We must show
- $\emptyset \not\in \mathcal{P}$
- $P \cap P’ \neq \emptyset \Rightarrow P = P’$ for $P, P’ \in \mathcal{P}$
- $\bigcup\limits_{P \in \mathcal{P}} P = X$
(1) follows from our construction of $\mathcal{P}$. For (2), note that if $P \cap P’ \neq \emptyset$, then by If Path-Connected Sets share a point, then their Union is Path-Connected, $P \cup P’$ is Path-Connected. $P \cup P’ \supset P, P’$ so $P = P \cup P’ = P’$ because both $P$ and $P’$ are Path-Connected Components.
For (3), let $x \in X$ be arbitrary. Then consider $\mathcal{P}{x} = {D \subset X : x \in D, D \text{ is path-connected}}$. By If Path-Connected Sets share a point, then their Union is Path-Connected, $P{x} := \bigcup\limits_{D \in \mathcal{P}{x}} D$ is Path-Connected. If there were some Path-Connected $E \supset P{x}$, then $x \in E$ so $E \in \mathcal{P}{x}$ making $E \subset P{x}$, so $E = P_{x}$. Thus $P_{x}$ is a Path-Connected Component containing $x \in X$. Thus taking $\bigcup\limits_{x \in X} P_{x} = X$ shows (3) and completes the proof $\blacksquare$.