Component of Locally Connected Space is Open
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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