Lemma 5.12.5.label Let $X$ be a topological space and $A \subset X$ be both open and closed, then $A$ is a union of connected components of $X$.
Proof. Let $C \subset X$ be a connected component, then $C \cap A$ and $C \setminus A$ are both open. Since $C$ is connected, either $C \cap A = \emptyset$ and $C \subset A^{c}$, or $C \setminus A = \emptyset$ and $C \subset A$.$\square$
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