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. $\square$

Comment #3566 by Laurent Moret-Bailly on

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

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