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.

Also see

 * Path Component of Locally Path-Connected Space is Open, an analogous result for path components
 * Locally Connected iff Components of Open Subsets are Open