Lemma 5.12.8. Let $X$ be a topological space. Assume that

$X$ is nonempty,

$X$ is quasi-compact, and

$X$ is Kolmogorov.

Then $X$ has a closed point.

Lemma 5.12.8. Let $X$ be a topological space. Assume that

$X$ is nonempty,

$X$ is quasi-compact, and

$X$ is Kolmogorov.

Then $X$ has a closed point.

**Proof.**
Consider the set

\[ \mathcal{T} = \{ Z \subset X \mid Z = \overline{\{ x\} } \text{ for some }x \in X\} \]

of all closures of singletons in $X$. It is nonempty since $X$ is nonempty. Make $\mathcal{T}$ into a partially ordered set using the relation of inclusion. Suppose $Z_\alpha $, $\alpha \in A$ is a totally ordered subset of $\mathcal{T}$. By Lemma 5.12.6 we see that $\bigcap _{\alpha \in A} Z_\alpha \not= \emptyset $. Hence there exists some $x \in \bigcap _{\alpha \in A} Z_\alpha $ and we see that $Z = \overline{\{ x\} }\in \mathcal{T}$ is a lower bound for the family. By Zorn's lemma there exists a minimal element $Z \in \mathcal{T}$. As $X$ is Kolmogorov we conclude that $Z = \{ x\} $ for some $x$ and $x \in X$ is a closed point. $\square$

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