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

## 5.22 Profinite spaces

Here is the definition.

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

$X$ is a profinite space, and

$X$ 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

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 limit of profinite spaces is profinite.

**Proof.**
Let $i \mapsto X_ i$ be a diagram of profinite spaces over the index category $\mathcal{I}$. Let us use the characterization of profinite spaces in Lemma 5.22.2. In particular each $X_ i$ is Hausdorff, quasi-compact, and totally disconnected. By Lemma 5.14.1 the limit $X = \mathop{\mathrm{lim}}\nolimits X_ i$ exists. By Lemma 5.14.5 the limit $X$ is quasi-compact. Let $x, x' \in X$ be distinct points. Then there exists an $i$ such that $x$ and $x'$ have distinct images $x_ i$ and $x'_ i$ in $X_ i$ under the projection $X \to X_ i$. Then $x_ i$ and $x'_ i$ have disjoint open neighbourhoods in $X_ i$. Taking the inverse images of these opens we conclude that $X$ is Hausdorff. Similarly, $x_ i$ and $x'_ i$ are in distinct connected components of $X_ i$ whence necessarily $x$ and $x'$ must be in distinct connected components of $X$. Hence $X$ is totally disconnected. 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

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

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 5.7.9). 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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (4)

Comment #5552 by Curious dilettante on

Comment #5736 by Johan on

Comment #6209 by Amnon Yekutieli on

Comment #6210 by Johan on