Equivalent Definitions of Compactness

Theorem

Let $T = \left({X, \tau}\right)$ be a topological space. The following are equivalent:

$(1): \quad$ $T$ is compact, i.e. every open cover of $X$ has a finite subcover.

$(2): \quad$ In every set $\mathcal A$ of closed subsets of $X$ satisfying $\displaystyle \bigcap \mathcal A = \varnothing$ exists a finite subset $\tilde{\mathcal A}$ such that $\displaystyle \bigcap \tilde{\mathcal A} = \varnothing$. That is, $X$ satisfies the Finite Intersection Axiom.

$(3): \quad$ Each filter on $X$ has a limit point in $X$.

$(4): \quad$ Each ultrafilter on $X$ converges.

Proof

$(1) \iff (2)$: Compact Space satisfies Finite Intersection Axiom

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

Let $\mathcal A$ be any set of closed subsets of $X$ satisfying $\bigcap \mathcal A = \varnothing$.

We define the set:

$\mathcal V := \left\{{X \setminus A : A \in \mathcal A}\right\}$

which is clearly an open cover of $X$.

From De Morgan's Laws:

$\displaystyle X \setminus \bigcup \mathcal V = \bigcap \left\{{X \setminus V : V \in \mathcal V}\right\} = \bigcap \left\{{A : A \in \mathcal A}\right\} = \varnothing$

and therefore $X = \bigcup \mathcal V$.

By definition, there exists a finite subcover $\tilde{\mathcal V} \subseteq \mathcal V$.

We define:

$\tilde{\mathcal A} := \left\{{X \setminus V : V \in \tilde{\mathcal V}}\right\}$

then $\tilde{\mathcal A} \subseteq \mathcal A$ by definition of $\mathcal V$.

Because $\tilde{\mathcal V}$ covers $X$, it follows directly that:

$\displaystyle \bigcap \tilde{\mathcal A} = \bigcap \left\{{X \setminus V : V \in \tilde{\mathcal V}}\right\} = X \setminus \bigcup \tilde{\mathcal V} = \varnothing$

Thus, in every set $\mathcal A$ of closed subsets of $X$ satisfying $\displaystyle \bigcap \mathcal A = \varnothing$ exists a finite subset $\tilde{\mathcal A}$ such that $\displaystyle \bigcap \tilde{\mathcal A} = \varnothing$.

That is, $X$ satisfies the Finite Intersection Axiom.

$\Box$

The converse works exactly as the previous, but with the roles of the open cover and $\mathcal A$ reversed.

$\blacksquare$

(3) $\implies$ (4)

Let $\mathcal F$ be an ultrafilter on $X$.

By $(3)$ $\mathcal F$ has a limit point $x \in X$.

Thus there exists a filter $\mathcal F'$ on $X$ which converges to $x$ satisfying $\mathcal F \subseteq \mathcal F'$.

Because $\mathcal F$ is an ultrafilter, $\mathcal F = \mathcal F'$.

Thus $\mathcal F$ converges to $x$.

(4) $\implies$ (3)

Let $\mathcal F$ be a filter on $X$.

As we have that Every Filter is Contained in an Ultrafilter, there exists an ultrafilter $\mathcal F'$ such that $\mathcal F \subseteq \mathcal F'$.

By $(4)$ we know that $\mathcal F'$ converges to a certain $x \in X$.

This implies that $x$ is a limit point of $\mathcal F$.

Axiom of Choice

This proof depends on the Axiom of Choice, by way of Every Filter is Contained in an Ultrafilter.

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.

(2) $\implies$ (3)

Let $\mathcal F$ be a filter on $X$.

Assume that $\mathcal F$ has no limit point.

This would imply that $\bigcap \left\{{\overline F : F \in \mathcal F}\right\} = \varnothing$.

By $(2)$ there are therefore sets $F_1, \ldots, F_n \in \mathcal F$ such that $\overline F_1 \cap \ldots \cap \overline F_n = \varnothing$.

Because for any set $M$ we have $M \subseteq \overline M$, we know that $\overline F_1, \ldots, \overline F_n \in \mathcal F$.

This contradicts the fact that $\mathcal F$ is a filter, because filters are closed under finite intersections and must not contain the empty set.

Thus $\mathcal F$ has a limit point.

(3) $\implies$ (2)

Let $\mathcal A \subset \mathcal P \left({X}\right)$ be a set of closed subsets of $X$.

Assume that $\bigcap \tilde{\mathcal A} \ne \varnothing$ for all finite subsets $\tilde{\mathcal A}$ of $\mathcal A$.

We show that this implies $\bigcap \mathcal A \ne \varnothing$.

Because of our assumption, $\mathcal B := \left\{{\bigcap \tilde{\mathcal A} : \tilde{\mathcal A} \subseteq \mathcal A \text{ finite}}\right\}$ is a filter basis.

Let $\mathcal F$ be the corresponding generated filter.

Then $\mathcal F$ has a limit point by $(3)$ and thus $\varnothing \ne \bigcup \left\{{\overline F : F \in \mathcal F}\right\} \subseteq \bigcap \mathcal B \subseteq \bigcap \mathcal A$.

Thus $\bigcap \mathcal A \ne \varnothing$.

Therefore $(2)$ follows.

$\blacksquare$