Lemma 5.12.9. Let $X$ be a quasi-compact Kolmogorov space. Then the set $X_0$ of closed points of $X$ is quasi-compact.

**Proof.**
Let $X_0 = \bigcup U_{i, 0}$ be an open covering. Write $U_{i, 0} = X_0 \cap U_ i$ for some open $U_ i \subset X$. Consider the complement $Z$ of $\bigcup U_ i$. This is a closed subset of $X$, hence quasi-compact (Lemma 5.12.3) and Kolmogorov. By Lemma 5.12.8 if $Z$ is nonempty it would have a closed point which contradicts the fact that $X_0 \subset \bigcup U_ i$. Hence $Z = \emptyset $ and $X = \bigcup U_ i$. Since $X$ is quasi-compact this covering has a finite subcover and we conclude.
$\square$

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