Lemma 5.12.11. Let $X$ be a topological space. Assume $X$ is quasi-compact and Hausdorff. 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. By Lemma 5.12.4 there exist disjoint opens $U, V \subset X$ with $B \subset U$ and $C \subset V$. Then $X \setminus U \cup V$ is closed in $X$ hence quasi-compact (Lemma 5.12.3). It follows that $(X \setminus U \cup V) \cap Z_\alpha = \emptyset$ for some $\alpha$ by Lemma 5.12.6. In other words, $Z_\alpha \subset U \cup V$. Thus $Z_\alpha = Z_\alpha \cap V \amalg Z_\alpha \cap U$ 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 $C = \emptyset$. $\square$

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