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
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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