The Stacks project

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. $\square$

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. $\square$

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. $\square$

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. $\square$