Alexander's Compactness Theorem

Theorem
Let $T = \left({X, \tau}\right)$ be a topological space.

Then $T$ is compact :


 * If $\tau$ 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.

Proof
Suppose that $T$ is compact.

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

By definition of a compact space, from every cover of $X$ by elements of $\mathcal S$, a finite subcover can be selected.

that the space $T$ 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 open covers.

So if:
 * $V$ is an open set

and:
 * $V \notin \mathcal C$,

then $\mathcal C \cup \left\{ {V}\right\}$ has a finite subcover, necessarily of the form:
 * $\mathcal C_0 \cup \left\{ {V}\right\}$

for some finite subset $\mathcal C_0$ of $\mathcal C$.

Consider $\mathcal C \cap \mathcal S$, that is, the sub-basic subset of $\mathcal C$.

Suppose $\mathcal C \cap \mathcal S$ covers $X$.

Then, by hypothesis, $\mathcal C \cap \mathcal S$ would have a finite subcover.

But $\mathcal C$ does not have a finite subcover.

So $\mathcal C \cap \mathcal S$ does not cover $X$.

Let $x \in X$ that is not covered by $\mathcal C \cap \mathcal S$.

We have that $\mathcal C$ covers $X$, so:
 * $\exists U \in \mathcal C: x \in U$

We have that $\mathcal S$ is a sub-basis.

So for some $S_1, \ldots, S_n \in \mathcal S$, we have that:
 * $x \in S_1 \cap \cdots \cap S_n \subseteq U$

Since $x$ is not covered, $S_i \notin \mathcal C$.

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

But then $U$ and all the $\mathcal C_i$ cover $X$.

Hence $\mathcal C$ has a finite subcover.

This contradicts our supposition that we can construct $\mathcal C$ so as to have no finite subcover.

It follows that we cannot construct an open cover $\mathcal C$ of $X$ which has no finite subcover.

That is, $T$ is compact.