Component of Locally Connected Space is Open

Theorem

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

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

Then $G$ is open.

Proof

By definition of locally connected space, $T$ has a basis of connected sets in $T$.

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

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

$\blacksquare$

Also see

• Path Component of Locally Path-Connected Space is Open, an analogous result for path components
• Equivalence of Definitions of Locally Connected