# Alexander's Compactness Theorem

## Theorem

The following definitions of the concept of Compact Space in the context of Topology are equivalent:

### Definition by Open Covers

A topological space $T = \left({S, \tau}\right)$ is compact if and only if every open cover for $S$ has a finite subcover.

### Definition Elements of Sub-Basis

A topological space $T = \left({S, \tau}\right)$ is compact if and only if $\tau$ has a sub-basis $\mathcal B$ such that:

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

## Proof

Let every open cover of $S$ have a finite subcover.

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

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

$\Box$

Let the space $T$ have a sub-basis $\mathcal B$ such that every cover of $S$ by elements of $\mathcal B$ has a finite subcover.

Aiming for a contradiction, suppose $T$ is not such that every open cover of $S$ 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 $\mathcal C \cap \mathcal B$, that is, the sub-basic subset of $\mathcal C$.

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

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

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

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

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

We have that $\mathcal C$ covers $S$, so:

$\exists U \in \mathcal C: x \in U$

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

So for some $B_1, \ldots, B_n \in \mathcal B$, we have that:

$x \in B_1 \cap \cdots \cap B_n \subseteq U$

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

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

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

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 $S$ which has no finite subcover.

$\blacksquare$

## Axiom of Choice

This theorem depends on the Axiom of Choice, by way of Zorn's Lemma.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.