Lemma 5.12.12 (04PL)—The Stacks project

Lemma 5.12.12. Let $X$ be a topological space. Assume

$X$ is quasi-compact,
$X$ has a basis for the topology consisting of quasi-compact opens, and
the intersection of two quasi-compact opens is quasi-compact.

For a subset $T \subset X$ the following are equivalent:

$T$ is an intersection of open and closed subsets of $X$, and
$T$ is closed in $X$ and is a union of connected components of $X$.

Proof. It is clear that (a) implies (b). Assume (b). Let $x \in X$, $x \not\in T$. Let $x \in C \subset X$ be the connected component of $X$ containing $x$. By Lemma 5.12.10 we see that $C = \bigcap V_\alpha $ is the intersection of all open and closed subsets $V_\alpha $ of $X$ which contain $C$. In particular, any pairwise intersection $V_\alpha \cap V_\beta $ occurs as a $V_\alpha $. As $T$ is a union of connected components of $X$ we see that $C \cap T = \emptyset $. Hence $T \cap \bigcap V_\alpha = \emptyset $. Since $T$ is quasi-compact as a closed subset of a quasi-compact space (see Lemma 5.12.3) we deduce that $T \cap V_\alpha = \emptyset $ for some $\alpha $, see Lemma 5.12.6. For this $\alpha $ we see that $U_\alpha = X \setminus V_\alpha $ is an open and closed subset of $X$ which contains $T$ and not $x$. The lemma follows. $\square$

Comments (2)

Comment #8514 by Elías Guisado on June 20, 2023 at 10:06

Instead of "any pairwise intersection $V_\alpha\cap V_\beta$ occurs as a $V_\alpha$ ," one could write "any finite intersection $V_{\alpha_1}\cap\cdots\cap V_{\alpha_n}$ equals $V_\beta$ , for some $\beta$ " (the latter is what one actually uses to deduce $T\cap V_\alpha=\emptyset$ for some $\alpha$ ).

Comment #8533 by Johan on June 26, 2023 at 14:31

Agreed. Going to leave as is until more people chime in.

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