Path Component of Locally Path-Connected Space is Open

Theorem

Let $T = \left({S, \tau}\right)$ be a locally path-connected topological space.

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

Then $G$ is open in $T$.

Proof

By definition of locally path-connected, $T$ has a basis of path-connected set.

Thus $S$ is a union of open path-connected sets of $T$.

By Path Components are Open iff Union of Open Path-Connected Sets, the path components of $T$ are open in $T$.

$\blacksquare$

Also see

• Component of Locally Connected Space is Open, an analogous result for connected components
• Equivalence of Definitions of Path Component where it is shown that a space is locally path-connected iff the path components of open subspaces are open.