# Equivalence of Definitions of Compact Topological Space

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## Theorem

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

### Definition 1

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

### Definition 2

A topological space $T = \struct {S, \tau}$ is compact if and only if it satisfies the Finite Intersection Axiom.

### Definition 3

A topological space $T = \struct {S, \tau}$ is compact if and only if $\tau$ has a sub-basis $\BB$ such that:

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

### Definition 4

A topological space $T = \left({S, \tau}\right)$ is compact if and only if every filter on $S$ has a limit point in $S$.

### Definition 5

A topological space $T = \left({S, \tau}\right)$ is compact if and only if every ultrafilter on $S$ converges.

## Proof

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

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

Let $\AA$ be any set of closed subsets of $S$ satisfying $\displaystyle \bigcap \AA = \O$.

We define the set:

$\VV := \set {S \setminus A : A \in \AA}$

which is trivially an open cover of $S$.

$\displaystyle S \setminus \bigcup \VV = \bigcap \set {S \setminus V : V \in \VV} = \bigcap \set {A : A \in \AA} = \O$

and therefore:

$S = \displaystyle \bigcup \VV$

By definition, there exists a finite subcover $\tilde \VV \subseteq \VV$.

We define:

$\tilde \AA := \set {S \setminus V : V \in \tilde \VV}$

then $\tilde \AA \subseteq \AA$ by definition of $\VV$.

Because $\tilde \VV$ covers $S$, it follows directly that:

$\displaystyle \bigcap \tilde \AA = \bigcap \set {S \setminus V : V \in \tilde \VV} = S \setminus \bigcup \tilde \VV = \O$

Thus, in every set $\AA$ of closed subsets of $S$ satisfying $\displaystyle \bigcap \AA = \O$ exists a finite subset $\tilde \AA$ such that $\displaystyle \bigcap \tilde \AA = \O$.

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

$\Box$

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

$\blacksquare$

### $(1) \iff (3)$: Alexander's Compactness Theorem

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

Let $\BB$ be a sub-basis of $\tau$.

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

$\Box$

Let the space $T$ have a sub-basis $\BB$ such that every cover of $S$ by elements of $\BB$ 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 $\CC$ which has no finite subcover that is maximal among such open covers.

So if:

$V$ is an open set

and:

$V \notin \CC$

then $\CC \cup \set V$ has a finite subcover, necessarily of the form:

$\CC_0 \cup \set V$

for some finite subset $\CC_0$ of $\CC$.

Consider $\CC \cap \BB$, that is, the sub-basic subset of $\CC$.

Suppose $\CC \cap \BB$ covers $S$.

Then, by hypothesis, $\CC \cap \BB$ would have a finite subcover.

But $\CC$ does not have a finite subcover.

So $\CC \cap \BB$ does not cover $S$.

Let $x \in S$ that is not covered by $\CC \cap \BB$.

We have that $\CC$ covers $S$, so:

$\exists U \in \CC: x \in U$

We have that $\BB$ is a sub-basis.

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

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

Since $x$ is not covered, $B_i \notin \CC$.

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

But then $U$ and all the $\CC_i$ cover $S$.

Hence $\CC$ has a finite subcover.

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

It follows that we cannot construct an open cover $\CC$ of $S$ which has no finite subcover.

$\blacksquare$

### $(4) \implies (5)$: Every Filter has Limit Point implies Every Ultrafilter Converges

Let $T = \left({S, \tau}\right)$ be such that each filter on $S$ has a limit point in $S$.

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

By hypothesis, $\mathcal F$ has a limit point $x \in S$.

By Limit Point iff Superfilter Converges, there exists a filter $\mathcal F'$ on $S$ 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 $x$.

### $(5) \implies (4)$: Every Ultrafilter Converges implies Every Filter has Limit Point

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

By the Ultrafilter Lemma, there exists an ultrafilter $\mathcal F'$ such that $\mathcal F \subseteq \mathcal F'$.

By hypothesis, $\mathcal F'$ converges to some $x \in S$.

This, by Limit Point iff Superfilter Converges, implies that $x$ is a limit point of $\mathcal F$.

$\blacksquare$

### $(2) \implies (4)$

Let $T = \left({S, \tau}\right)$ satisfy the Finite Intersection Axiom.

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

Aiming for a contradiction, suppose that $\mathcal F$ has no limit point.

Thus:

$\bigcap \left\{{\overline F : F \in \mathcal F}\right\} = \varnothing$

By hypothesis 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 $M \subseteq \overline M$:

$\overline F_1, \ldots, \overline F_n \in \mathcal F$

But by definition of a filter, $\mathcal F$ must not contain the empty set.

Thus $\mathcal F$ has a limit point.

$\Box$

### $(4) \implies (2)$

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

Let:

$\bigcap \tilde{\mathcal A} \ne \varnothing$

for all finite subsets $\tilde{\mathcal A}$ of $\mathcal A$.

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

From 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.

By hypothesis $\mathcal F$ has a limit point.

Thus:

$\varnothing \ne \bigcap \left\{{\overline F : F \in \mathcal F}\right\} \subseteq \bigcap \mathcal B \subseteq \bigcap \mathcal A$.

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

Hence $T = \left({S, \tau}\right)$ satisfies the Finite Intersection Axiom.

$\blacksquare$