# Compactness

(Redirected from Compact (topology))

See Notes:Compactness and sequences - I think there's a different definition for metric spaces, I have not seen a proof that the metric one [ilmath]\implies[/ilmath] this one

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## Definition

There are 2 distinct definitions of compactness, however they are equivalent:

1. We may only say a topological space is compact, we may not speak of the compactness of subsets. Compactness is strictly a property of topological spaces.
2. Sure talk about the compactness of subsets of a space.

For 1) we may talk about the compactness of subsets if we consider them as topological subspaces

### Definition 1

A topological space, [ilmath](X,\mathcal{J})[/ilmath] is compact if:

• Every open covering of [ilmath]X[/ilmath], [ilmath]\{U_\alpha\}_{\alpha\in I}\subseteq\mathcal{J} [/ilmath] contains a finite sub-cover

Note that in this definition we'll actually have (if [ilmath]\{U_\alpha\}_{\alpha\in I} [/ilmath] is actually a covering) [ilmath]X=\bigcup_{\alpha\in I}U_\alpha[/ilmath] (notice equality rather than [ilmath]\subseteq[/ilmath], this is because the union on the right cannot contain more than [ilmath]X[/ilmath] itself, The elements of [ilmath]\mathcal{J} [/ilmath] are subsets of [ilmath]X[/ilmath] and the [ilmath]U_\alpha[/ilmath] are elements of [ilmath]\mathcal{J} [/ilmath] after all.

#### Compactness of a subset

A subset, [ilmath]S\subseteq X[/ilmath] of a topological space [ilmath](X,\mathcal{J})[/ilmath] is compact if:

• The topology [ilmath](S,\mathcal{J}_\text{subspace})[/ilmath] is compact (the subspace topology on [ilmath]S[/ilmath] inherited from [ilmath]X[/ilmath]) as per the definition above.

This maintains compactness as a strictly topological property.

### Definition 2

A subset, [ilmath]S\subseteq X[/ilmath] of a topological space [ilmath](X,\mathcal{J})[/ilmath] is compact if:

TODO: Find reference

Note that: when [ilmath]S=X[/ilmath] we get definition 1.

### Claim 1: The definitions are equivalent

These 2 definitions are the same, that is:

• Claim 1: A subspace [ilmath]Y\subseteq X[/ilmath] is a compact (def 1) in [ilmath](X,\mathcal{J})[/ilmath] [ilmath]\iff[/ilmath] every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering (def 2).

## Compactness of a metric space

Given a metric space [ilmath](X,d)[/ilmath], the following are equivalent[Note 1]:

1. [ilmath]X[/ilmath] is compact
2. Every sequence in [ilmath]X[/ilmath] has a subsequence that converges (AKA: having a convergent subsequence)
3. [ilmath]X[/ilmath] is totally bounded and complete

(see Equivalent statements to compactness of a metric space for proof)

## Proof of claims

Claim 1: A subspace [ilmath]Y\subseteq X[/ilmath] is a compact (def 1) in [ilmath](X,\mathcal{J})[/ilmath] [ilmath]\iff[/ilmath] every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering (def 2).

Suppose that [ilmath](Y,\mathcal{J}_\text{subspace})[/ilmath] is compact [ilmath]\implies[/ilmath] every covering consisting of open sets of [ilmath](X,\mathcal{J})[/ilmath] contains a finite subcover.

Let [ilmath]\{A_\alpha\}_{\alpha\in I}\subseteq\mathcal{J}[/ilmath] be a family of open sets in [ilmath]X[/ilmath] with [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha[/ilmath]
Take [ilmath]B_\alpha=A_\alpha\cap Y[/ilmath], then [ilmath]\{B_\alpha\}_{\alpha\in I} [/ilmath] is an open (in [ilmath]Y[/ilmath]) covering of [ilmath]Y[/ilmath], that is [ilmath]Y\subseteq\cup_{\alpha\in I}B_\alpha[/ilmath] (infact we have [ilmath]Y=\cup_{\alpha\in I}B_\alpha[/ilmath])
Proof of [ilmath]Y\subseteq\cup_{\alpha\in I}B_\alpha[/ilmath] (we actually have [ilmath]Y=\cup_{\alpha\in I}B_\alpha[/ilmath])
We wish to show that [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha\implies Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath], using the Implies-subset relation we actually just want to show that:
• Given [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha[/ilmath] that [ilmath]y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] - which is what we'll do.
• Note additionally that [ilmath]y\in\cup_{\alpha\in I}(A_i\cap Y)\iff \exists\beta\in I[y\in A_\beta\wedge y\in Y][/ilmath]
Let [ilmath]y\in Y[/ilmath], then by hypothesis [ilmath]y\in\cup_{\alpha\in I}A_\alpha\iff\exists \beta\in I[y\in A_\beta][/ilmath]
It is easily seen that [ilmath]y\in Y\wedge\exists\beta\in I[y\in A_\beta]\implies\exists\gamma\in I[y\in A_\gamma\wedge y\in Y][/ilmath] simply by choosing [ilmath]\gamma:=\beta[/ilmath].
Lastly, note that [ilmath]\exists\gamma\in I[y\in A_\gamma\wedge y\in Y]\iff y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath]
• We have shown that [ilmath]y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] and by the Implies-subset relation we see
• [ilmath]Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] - as required.
I earlier claimed that actually [ilmath]Y=\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] - this isn't important to the proof but it shows something else.
This shows that considering an open covering as a union of sets open in [ilmath]Y[/ilmath] whose union is exactly [ilmath]Y[/ilmath] is the same as a covering by open sets in [ilmath]X[/ilmath] whose union contains (but need not be exactly equal to) [ilmath]Y[/ilmath]. So we have shown so far that:
• Compact in the subspace with equality for an open covering [ilmath]\implies[/ilmath] compact with the open cover of sets in [ilmath]X[/ilmath] whose union contains [ilmath]Y[/ilmath]
Claim: [ilmath]\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y[/ilmath]
Let [ilmath]y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] then:
• [ilmath]\exists\beta\in I[y\in A_\beta\wedge y\in Y]\iff \exists\beta\in I[y\in(A_\beta\cap Y)][/ilmath]
As the intersection of sets is a subset of each set we see that (it's trivial to show without this result too, but this uses a general result)
[ilmath]y\in(A_\beta\cap Y)\implies y\in Y[/ilmath]
Thus we have shown that [ilmath]y\in\cup_{\alpha\in I}(A_\alpha\cap Y)\implies y\in Y[/ilmath] and finally this means:
• [ilmath]\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y[/ilmath]
Combining this with [ilmath]Y\subseteq \cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] above we see that:
• [ilmath]Y=\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath]
This completes the proof
By hypothesis, [ilmath]Y[/ilmath] is compact, this means that [ilmath]\{B_\alpha\}_{\alpha\in I} [/ilmath] contains a finite subcover
• call this subcover [ilmath]\{B'_i\}_{i=1}^n[/ilmath] where each [ilmath]B'_i\in\{B_\alpha\}_{\alpha\in I} [/ilmath], now we have [ilmath]Y\subseteq\cup_{i=1}^n B'_i[/ilmath] (we actually have equality, see the blue box in the yellow note box above)
As each [ilmath]B'_i=A'_i\cap Y[/ilmath] (where [ilmath]A'_i[/ilmath] is the corresponding [ilmath]A_\alpha[/ilmath] for the [ilmath]B_\alpha[/ilmath] that [ilmath]B'_i[/ilmath] represents) we see that [ilmath]\{A_i\}_{i=1}^n[/ilmath] is a finite subcover by sets open in [ilmath]X[/ilmath]
Proof of: [ilmath]Y\subseteq\cup_{i=1}^nB'_i\implies[/ilmath] [ilmath]Y\subseteq\cup_{i=1}^nA'_i[/ilmath] (proving that [ilmath]\{A'_i\}_{i=1}^n[/ilmath] is an open cover)
For each [ilmath]i[/ilmath] we have [ilmath]B'_i:=A'_i\cap Y[/ilmath], by invoking the intersection of sets is a subset of each set we note that:
• [ilmath]B'_i\subseteq A'_i[/ilmath]
We now invoke Union of subsets is a subset of the union

This theorem states that given two families of sets, [ilmath]\{A_\alpha\}_{\alpha\in I} [/ilmath] and [ilmath]\{B_\alpha\}_{\alpha\in I} [/ilmath] with [ilmath]\forall\alpha\in I[B_\alpha\subseteq A_\alpha][/ilmath] we have [ilmath]\cup_{\alpha\in I}B_\alpha\subseteq\cup_{\alpha\in I}A_\alpha[/ilmath]

It follows that [ilmath]Y\subseteq\cup_{i=1}^nB'_i\subseteq\cup_{i=1}^nA'_i[/ilmath], in particular:
• [ilmath]Y\subseteq\cup_{i=1}^nA'_i[/ilmath]
This confirms that [ilmath]\{A'_i\}_{i=1}[/ilmath] is an open cover by sets in [ilmath]X[/ilmath]
This completes this half of the proof.

[ilmath](Y,\mathcal{J}_\text{subspace})[/ilmath] is compact $\impliedby$ every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering

Suppose that every covering of $Y$ by sets open in $X$ contains a finite subcollection covering $Y$. We need to show $Y$ is compact.
Suppose we have a covering, $\mathcal{A}'=\{A'_\alpha\}_{\alpha\in I}$ of $Y$ by sets open in $Y$
For each $\alpha$ choose an open set $A_\alpha$ open in $X$ such that: $A'_\alpha=A_\alpha\cap Y$
Then the collection $\mathcal{A}=\{A_\alpha\}_{\alpha\in I}$ covers $Y$
By hypothesis we have a finite sub-collection from [ilmath]\mathcal{A} [/ilmath] of things open in $X$ that cover $Y$
Thus the corresponding finite subcollection of $\mathcal{A}'$ covers $Y$