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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