# 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.

### Definition 6

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

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

## 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 $\ds \bigcap \AA = \O$.

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:

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

and therefore:

- $S = \ds \bigcup \VV$

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:

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

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

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:

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

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

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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.

$\blacksquare$

### $(1) \iff (6)$: Compactness from Basis

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

Let $\BB$ be a basis $\BB$.

Then every cover of $S$ by elements of $\BB$ is an open cover for $S$.

So:

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

$\Box$

Let $\tau$ have a basis $\BB$ such that:

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

Let $\AA$ be an open cover of $S$.

Let $f: \AA \to \powerset B$ be the mapping defined by:

- $\forall A \in \AA : \map f A = \set{B \in \BB : B \subseteq A}$

Since each element of $\AA$ is open:

- $A = \bigcup \map f A$ for each $A \in \AA$.

Let $\AA' = \bigcup f \sqbrk {\AA}$.

Then $\AA'$ is an cover of $S$ by elements of $\BB$

By the premise, $\AA'$ has a finite subset $\FF'$ that covers $S$.

Let $g: \FF' \to \AA$ be the map each element of $\FF'$ to an element of $\AA$ that contains it.

Note that since $\FF'$ is finite, this does not require the Axiom of Choice.

Let $\FF = g \sqbrk {\FF'}$.

Then $\FF$ is a finite subcover of $\AA$.

It follows that every open cover of $S$ has a finite subcover of $S$.

$\blacksquare$

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

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

Let $\FF$ be an ultrafilter on $S$.

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

By Limit Point iff Superfilter Converges, there exists a filter $\FF'$ on $S$ which converges to $x$ satisfying $\FF \subseteq \FF'$.

By Definition of Ultrafilter on Set:

- $\FF$ is an ultrafilter if and only if whenever $\GG$ is a filter on $S$ and $\FF \subseteq \GG$ holds, then $\FF = \GG$

Therefore:

- $\FF = \FF'$.

Thus $\FF$ converges to $x$.

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

Let $\FF$ be a filter on $S$.

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

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

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

$\blacksquare$

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

Let $T = \struct {S, \tau}$ satisfy the Finite Intersection Axiom.

Let $\FF$ be a filter on $X$.

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

Thus:

- $\ds \bigcap_{F \mathop \in \FF} \overline F = \O$

By hypothesis there are therefore sets $F_1, \ldots, F_n \in \FF$ such that:

- $\overline F_1 \cap \ldots \cap \overline F_n = \O$

Because for any set $M$ we have $M \subseteq \overline M$:

- $\overline F_1, \ldots, \overline F_n \in \FF$

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

Thus $\FF$ has a limit point.

$\Box$

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

Let $\AA \subset \powerset S$ be a set of closed subsets of $S$.

Let:

- $\bigcap \tilde \AA \ne \O$

for all finite subsets $\tilde \AA$ of $\AA$.

We show that this implies $\bigcap \AA \ne \O$.

From our assumption, $\BB := \set {\bigcap \tilde \AA : \tilde \AA \subseteq \AA \text{ finite} }$ is a filter basis.

Let $\FF$ be the corresponding generated filter.

By hypothesis $\FF$ has a limit point.

Thus:

- $\ds \O \ne \bigcap_{F \mathop \in \FF} \overline F \subseteq \bigcap \BB \subseteq \bigcap \AA$.

Thus $\bigcap \AA \ne \O$.

Hence $T = \struct {S, \tau}$ satisfies the Finite Intersection Axiom.

$\blacksquare$