
## 5.22 Profinite spaces

Here is the definition.

Definition 5.22.1. A topological space is profinite if it is homeomorphic to a limit of a diagram of finite discrete spaces.

This is not the most convenient characterization of a profinite space.

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. $\square$

Lemma 5.22.3. A cofiltered limit of profinite spaces is profinite.

Proof. Let us use the characterization of profinite spaces in Lemma 5.22.2. By Lemma 5.14.1 the limit exists. By Theorem 5.14.4 the limit is quasi-compact. A cofiltered limit of totally disconnected spaces is totally disconnected (details omitted). A cofiltered limit of Hausdorff spaces is Hausdorff (details omitted). This finishes the proof. $\square$

Lemma 5.22.4. Let $X$ be a profinite space. Every open covering of $X$ has a refinement by a finite covering $X = \coprod U_ i$ with $U_ i$ open and closed.

Proof. Write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as a limit of an inverse system of finite discrete spaces over a directed set $I$ (Lemma 5.22.2). Denote $f_ i : X \to X_ i$ the projection. For every point $x = (x_ i) \in X$ a fundamental system of open neighbourhoods is the collection $f_ i^{-1}(\{ x_ i\} )$. Thus, as $X$ is quasi-compact, we may assume we have an open covering

$X = f_{i_1}^{-1}(\{ x_{i_1}\} ) \cup \ldots \cup f_{i_ n}^{-1}(\{ x_{i_ n}\} )$

Choose $i \in I$ with $i \geq i_ j$ for $j = 1, \ldots , n$ (this is possible as $I$ is a directed set). Then we see that the covering

$X = \coprod \nolimits _{t \in X_ i} f_ i^{-1}(\{ t\} )$

refines the given covering and is of the desired form. $\square$

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 5.7.8). 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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