# Compact Space satisfies Finite Intersection Axiom

## Contents

## Theorem

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

### Definition by Open Covers

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

### Definition by Finite Intersection Axiom

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

## Proof

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 $S$.

From De Morgan's Laws: Difference with Union:

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

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.

$\blacksquare$

## Also see

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties