# Equivalence of Definitions of Compact Topological Space

## Contents

- 1 Theorem
- 2 Proof
- 2.1 $(1) \iff (2)$: Compact Space satisfies Finite Intersection Axiom
- 2.2 $(1) \iff (3)$: Alexander's Compactness Theorem
- 2.3 $(4) \implies (5)$: Every Filter has Limit Point implies Every Ultrafilter Converges
- 2.4 $(5) \implies (4)$: Every Ultrafilter Converges implies Every Filter has Limit Point
- 2.5 $(2) \implies (4)$
- 2.6 $(4) \implies (2)$

## Theorem

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

### Definition 1

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

### Definition 2

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

### Definition 3

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.

### 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 $\mathcal A$ be any set of closed subsets of $S$ satisfying $\displaystyle \bigcap \mathcal A = \varnothing$.

We define the set:

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

which is clearly an open cover of $S$.

From De Morgan's Laws: Difference with Union:

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

and therefore:

- $S = \displaystyle \bigcup \mathcal V$

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

We define:

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

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

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

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

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

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 $\mathcal A$ reversed.

$\blacksquare$

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

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.

$\blacksquare$

### $(4) \implies (5)$: 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:

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

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

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

$\blacksquare$