Path Component of Locally Path-Connected Space is Closed

Theorem

Let $T = \struct {S, \tau}$ be a locally path-connected topological space.

Let $G$ be a path component of $T$.

Then $G$ is open in $T$.

Proof

Let $x \in \partial G$, where $\partial G$ denotes the boundary of $G$.

By Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology, it follows that $S \in \tau$.

As $x \in S$, it follows that $S$ is a neighborhood of $x$.

By definition of locally path-connected space, it follows that there exists a path-connected neighborhood $N$ of $x$ such that $N \subseteq S$.

By definition of boundary, there exists $y \in N$ such that $y \in G$.

As $N$ is path-connected, there exists a path between $x$ and $y$.

By definition of path component , it follows that $x \in G$.

Thus, $\partial G \subseteq G$.

From Set is Closed iff it Contains its Boundary, it follows that $G$ is closed.

$\blacksquare$

Sources

• 2000: James R. Munkres: Topology (2nd ed.): $3$: Connectedness and Compactness: $\S 25$: Components and Local Connectedness