Lemma 5.12.15 (Alexander subbase theorem). Let $X$ be a topological space. Let $\mathcal{B}$ be a subbase for $X$. If every covering of $X$ by elements of $\mathcal{B}$ has a finite refinement, then $X$ is quasi-compact.

Proof. Assume there is an open covering of $X$ which does not have a finite refinement. Using Zorn's lemma we can choose a maximal open covering $X = \bigcup _{i \in I} U_ i$ which does not have a finite refinement (details omitted). In other words, if $U \subset X$ is any open which does not occur as one of the $U_ i$, then the covering $X = U \cup \bigcup _{i \in I} U_ i$ does have a finite refinement. Let $I' \subset I$ be the set of indices such that $U_ i \in \mathcal{B}$. Then $\bigcup _{i \in I'} U_ i \not= X$, since otherwise we would get a finite refinement covering $X$ by our assumption on $\mathcal{B}$. Pick $x \in X$, $x \not\in \bigcup _{i \in I'} U_ i$. Pick $i \in I$ with $x \in U_ i$. Pick $V_1, \ldots , V_ n \in \mathcal{B}$ such that $x \in V_1 \cap \ldots \cap V_ n \subset U_ i$. This is possible as $\mathcal{B}$ is a subbasis for $X$. Note that $V_ j$ does not occur as a $U_ i$. By maximality of the chosen covering we see that for each $j$ there exist $i_{j, 1}, \ldots , i_{j, n_ j} \in I$ such that $X = V_ j \cup U_{i_{j, 1}} \cup \ldots \cup U_{i_{j, n_ j}}$. Since $V_1 \cap \ldots \cap V_ n \subset U_ i$ we conclude that $X = U_ i \cup \bigcup U_{i_{j, l}}$ a contradiction. $\square$

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