Components are Open iff Union of Open Connected Sets

Theorem
Let $T = \struct{S, \tau}$ be a topological space.


 * (1) $\quad$ The components of $T$ are open.


 * (2) $\quad S$ is a union of open connected sets of $T$.

(1) implies (2)
Let the components of $T$ be open.

By definition, the components of $T$ are a partition of $S$.

Hence $S$ is the union of the open components of $T$.

Since a component is a maximal connected set by definition, then $S$ is a union of open connected sets of $T$

(2) implies (1)
Let $S = \bigcup \{ U \subseteq S : U \in \tau \text { and } U \text { is connected} \}$.

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

Then:

Hence $C$ is the union of open sets.

By definition of a topology then $C$ is an open set.

The result follows.

Also see

 * Path Components are Open iff Union of Open Path-Connected Sets, an analogous result for path components