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The Stacks project

Lemma 5.22.2. Let X be a topological space. The following are equivalent

  1. X is a profinite space, and

  2. X is Hausdorff, quasi-compact, and totally disconnected.

If this is true, then X is a cofiltered limit of finite discrete spaces.

Proof. Assume (1). Choose a diagram i \mapsto X_ i of finite discrete spaces such that X = \mathop{\mathrm{lim}}\nolimits X_ i. As each X_ i is Hausdorff and quasi-compact we find that X is quasi-compact by Lemma 5.14.5. If x, x' \in X are distinct points, then x and x' map to distinct points in some X_ i. Hence x and x' have disjoint open neighbourhoods, i.e., X is Hausdorff. In exactly the same way we see that X is totally disconnected.

Assume (2). Let \mathcal{I} be the set of finite disjoint union decompositions X = \coprod _{i \in I} U_ i with U_ i nonempty open (and closed) for all i \in I. For each I \in \mathcal{I} there is a continuous map X \to I sending a point of U_ i to i. We define a partial ordering: I \leq I' for I, I' \in \mathcal{I} if and only if the covering corresponding to I' refines the covering corresponding to I. In this case we obtain a canonical map I' \to I. In other words we obtain an inverse system of finite discrete spaces over \mathcal{I}. The maps X \to I fit together and we obtain a continuous map

X \longrightarrow \mathop{\mathrm{lim}}\nolimits _{I \in \mathcal{I}} I

We claim this map is a homeomorphism, which finishes the proof. (The final assertion follows too as the partially ordered set \mathcal{I} is directed: given two disjoint union decompositions of X we can find a third refining both.) Namely, the map is injective as X is totally disconnected and hence \{ x\} is the intersection of all open and closed subsets of X containing x (Lemma 5.12.11) and the map is surjective by Lemma 5.12.6. By Lemma 5.17.8 the map is a homeomorphism. \square


Comments (2)

Comment #3566 by Laurent Moret-Bailly on

In the proof that (2) implies (1), assume each nonempty.

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