The Stacks project

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

  1. $X$ is quasi-compact,

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

  3. the intersection of two quasi-compact opens is quasi-compact.

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

  1. $T$ is an intersection of open and closed subsets of $X$, and

  2. $T$ 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 on

Instead of "any pairwise intersection occurs as a ," one could write "any finite intersection equals , for some " (the latter is what one actually uses to deduce for some ).

Comment #8533 by on

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

There are also:

  • 2 comment(s) on Section 5.12: Quasi-compact spaces and maps

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