Lemma 5.22.5. Let $X$ be a topological space. If $X$ is quasi-compact and every connected component of $X$ is the intersection of the open and closed subsets containing it, then $\pi _0(X)$ is a profinite space.
Proof. We will use Lemma 5.22.2 to prove this. Since $\pi _0(X)$ is the image of a quasi-compact space it is quasi-compact (Lemma 5.12.7). It is totally disconnected by construction (Lemma 5.7.9). Let $C, D \subset X$ be distinct connected components of $X$. Write $C = \bigcap U_\alpha $ as the intersection of the open and closed subsets of $X$ containing $C$. Any finite intersection of $U_\alpha $'s is another. Since $\bigcap U_\alpha \cap D = \emptyset $ we conclude that $U_\alpha \cap D = \emptyset $ for some $\alpha $ (use Lemmas 5.7.3, 5.12.3 and 5.12.6) Since $U_\alpha $ is open and closed, it is the union of the connected components it contains, i.e., $U_\alpha $ is the inverse image of some open and closed subset $V_\alpha \subset \pi _0(X)$. This proves that the points corresponding to $C$ and $D$ are contained in disjoint open subsets, i.e., $\pi _0(X)$ is Hausdorff. $\square$
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