Alexander's Compactness Theorem

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Theorem

If $T$ has a sub-basis $\mathcal S$ such that from every cover of $X$ by elements of $\mathcal S$, a finite subcover of $X$ can be selected.

Suppose that $T$ is compact.

By definition of sub-basis, $\mathcal S$ is an open cover.

Let $\mathcal S$ be a sub-basis of $X$.

Then from every cover of $X$ by elements of $\mathcal S$, a finite subcover can be selected.

$\Box$

Now, with a view to obtain a contradiction, suppose that the space $X$ is not compact, yet every cover of $X$ by elements of $\mathcal S$ has a finite subcover.

Use Zorn's Lemma to find an open cover $\mathcal C$ which has no finite subcover that is maximal among such covers.

That means that if $V \notin \mathcal C$, then $\mathcal C \cup V$ has a finite subcover, necessarily of the form $\mathcal C_0 \cup V$.

Consider $\mathcal C \cap \mathcal S$, that is, the subbasic subfamily of $\mathcal C$.

If it covered $X$, then by hypothesis, it would have a finite subcover.

But $\mathcal C$ does not have such, so $\mathcal C \cap \mathcal S$ does not cover $X$.

Let $x \in X$ that is not covered.

$\mathcal C$ covers $X$, so $\exists U \in \mathcal C: x \in U$.

$\mathcal S$ is a subbasis, so for some $S_1, \ldots, S_n \in \mathcal S$, $x \in S_1 \cap \cdots \cap S_n \subseteq U$.

Since $x$ is uncovered, $S_i \notin \mathcal C$.

As noted above, this means that for each $i$, $S_i$ along with a finite subfamily $\mathcal C_i$ of $\mathcal C$, covers $X$.

But then $U$ and all the $\mathcal C_i$ cover $X$, so $\mathcal C$ has a finite subcover after all.

$\blacksquare$

Although this proof makes use of Zorn's Lemma, the proof does not need the full strength of the Axiom of Choice.

Instead, it relies on the intermediate Ultrafilter Principle.