Path-Connected Components are equal to the Connected Components in a Locally Path-Connected Space

Last updated Nov 1, 2022

Let $X$ be a Topological Space that is Locally Path-Connected. Then the collection $\mathcal{P} := {P \subset X : P \text{ is a path-component of }X}$ is equal to $\mathcal{C} := {C \subset X : C \text{ is a connected component of }X}$.

If $X = \emptyset$, then $\mathcal{P} = \emptyset = \mathcal{C}$.

Otherwise, assume $X$ is Nonempty. We claim that for each $C \in \mathcal{C}$ there exists $P \in \mathcal{P}$ s.t. $C = P$. Then by Correspondence of Partition Sets means Partitions are the same, we have that $\mathcal{P} = \mathcal{C}$.

To verify the claim, let $C \in \mathcal{C}$. Because Path-Connected Components Partition the Space, for each $x \in C$, there is a $P_{x} \in \mathcal{P}$ so that $x \in P_{x}$. Because Path-Connected implies Connected, Every Connected Set is contained in a single Component, and Connected Components Partition the Space we know $P_{x} \subset C$ for all $x \in C$.

Pick some $y \in C$. Because Path-Connected implies Connected, Every Connected Set is contained in a single Component, and Connected Components Partition the Space we know $P_{y} \subset C$. Denote $R := \bigcup\limits_{P \in \mathcal{P} \setminus {P_{y}}} P$. $R = P_{y} ^{C}$ because Path-Connected Components Partition the Space. Since Path-Connected Components of Locally Path-Connected Spaces are Open, both $R$ and $P_{y}$ are Open. Thus $R$ and $P_{y}$ disconnect $C$. Since $C$ is Connected, and because $C \cap P_{y} \neq \emptyset$. we have that $P_{y} \subset C$. Therefore, $P_{y} = C$. $\blacksquare$