
Lemma 5.12.10. 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 any $x \in X$ the connected component of $X$ containing $x$ is the intersection of all open and closed subsets of $X$ containing $x$.

Proof. Let $T$ be the connected component containing $x$. Let $S = \bigcap _{\alpha \in A} Z_\alpha$ be the intersection of all open and closed subsets $Z_\alpha$ of $X$ containing $x$. Note that $S$ is closed in $X$. Note that any finite intersection of $Z_\alpha$'s is a $Z_\alpha$. Because $T$ is connected and $x \in T$ we have $T \subset S$. It suffices to show that $S$ is connected. If not, then there exists a disjoint union decomposition $S = B \amalg C$ with $B$ and $C$ open and closed in $S$. In particular, $B$ and $C$ are closed in $X$, and so quasi-compact by Lemma 5.12.3 and assumption (1). By assumption (2) there exist quasi-compact opens $U, V \subset X$ with $B = S \cap U$ and $C = S \cap V$ (details omitted). Then $U \cap V \cap S = \emptyset$. Hence $\bigcap _\alpha U \cap V \cap Z_\alpha = \emptyset$. By assumption (3) the intersection $U \cap V$ is quasi-compact. By Lemma 5.12.6 for some $\alpha ' \in A$ we have $U \cap V \cap Z_{\alpha '} = \emptyset$. Since $X \setminus (U \cup V)$ is disjoint from $S$ and closed in $X$ hence quasi-compact, we can use the same lemma to see that $Z_{\alpha ''} \subset U \cup V$ for some $\alpha '' \in A$. Then $Z_\alpha = Z_{\alpha '} \cap Z_{\alpha ''}$ is contained in $U \cup V$ and disjoint from $U \cap V$. Hence $Z_\alpha = U \cap Z_\alpha \amalg V \cap Z_\alpha$ is a decomposition into two open pieces, hence $U \cap Z_\alpha$ and $V \cap Z_\alpha$ are open and closed in $X$. Thus, if $x \in B$ say, then we see that $S \subset U \cap Z_\alpha$ and we conclude that $C = \emptyset$. $\square$

## Comments (2)

Comment #636 by Wei Xu on

Dear stacks project,

There is a possible small gap between "for some $\alpha \in A$ we have $U \cap V \cap Z_\alpha = \emptyset$." and "Hence $Z_\alpha = U \cap Z_\alpha \coprod V \cap Z_\alpha$". Possibly we might need to add words like "(with some argurements) we may aslo assume this $Z_{\alpha}\subset (U\cup V)$" before the sentense "Hence ..."

Comment #644 by on

Yes, I agree one needs an argument there. I added something here. Thanks!

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