Compactness
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
Contents
Definition
There are 2 distinct definitions of compactness, however they are equivalent:
- 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.
- 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^{[1]}^{[2]}:
- 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^{[1]}^{[2]}:
- 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^{[1]}^{[Note 1]}:
- [ilmath]X[/ilmath] is compact
- Every sequence in [ilmath]X[/ilmath] has a subsequence that converges (AKA: having a convergent subsequence)
- [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.
- 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:
- 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](A_\beta\cap Y)\subseteq Y)[/ilmath] by the implies-subset relation we see immediately that:
- [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
- Let [ilmath]y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] then:
- 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:
- 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 [math]\impliedby[/math] every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering
- Suppose that every covering of [math]Y[/math] by sets open in [math]X[/math] contains a finite subcollection covering [math]Y[/math]. We need to show [math]Y[/math] is compact.
- Suppose we have a covering, [math]\mathcal{A}'=\{A'_\alpha\}_{\alpha\in I}[/math] of [math]Y[/math] by sets open in [math]Y[/math]
- For each [math]\alpha[/math] choose an open set [math]A_\alpha[/math] open in [math]X[/math] such that: [math]A'_\alpha=A_\alpha\cap Y[/math]
- Then the collection [math]\mathcal{A}=\{A_\alpha\}_{\alpha\in I}[/math] covers [math]Y[/math]
- By hypothesis we have a finite sub-collection from [ilmath]\mathcal{A} [/ilmath] of things open in [math]X[/math] that cover [math]Y[/math]
- Thus the corresponding finite subcollection of [math]\mathcal{A}'[/math] covers [math]Y[/math]
References
- ↑ ^{1.0} ^{1.1} ^{1.2} Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
- ↑ ^{2.0} ^{2.1} Introduction to Topology - Bert Mendelson
OLD PAGE
Not to be confused with Sequential compactness
There are two views here.
- Compactness is a topological property and we cannot say a set is compact, we say it is compact and implicitly consider it with the subspace topology
- We can say "sure that set is compact".
The difference comes into play when we cover a set (take the interval [ilmath][0,5]\subset\mathbb{R} [/ilmath]) with open sets. Suppose we have the covering [ilmath]\{(-1,3),(2,6)\} [/ilmath] this is already finite and covers the interval. The corresponding sets in the subspace topology are [ilmath]\{[0,3),(2,5]\} [/ilmath] which are both open in the subspace topology.
Definition
- A topological space is compact^{[1]} if every open cover of [math]X[/math] contains a finite sub-covering that also covers [math]X[/math].
That is to say that given an arbitrary collection of sets:
- [ilmath]\mathcal{A}=\{A_\alpha\}_{\alpha\in I}[/ilmath] such that each [ilmath]A_\alpha[/ilmath] is open in [ilmath]X[/ilmath] and
- [math]X=\bigcup_{\alpha\in I}A_\alpha[/math]^{[Note 2]}
The following is true:
- [ilmath]\exists \{i_1,\cdots,i_n\}\subset I[/ilmath] such that [math]X=\bigcup_{\alpha\in\{i_1,\cdots,i_n\} }A_\alpha[/math]
Then [ilmath]X[/ilmath] is compact^{[1]}
Lemma for a set being compact
Take a set [math]Y\subset X[/math] in a topological space [math](X,\mathcal{J})[/math]. Then to say:
- [math]Y[/math] is compact
Means [math]Y[/math] satisfies the definition of compactness when considered as a subspace of [math](X,\mathcal{J})[/math]
Theorem: A set [ilmath]Y\subseteq X[/ilmath] is a compact in [ilmath](X,\mathcal{J})[/ilmath] if and only if every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering.
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.
- 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:
- 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](A_\beta\cap Y)\subseteq Y)[/ilmath] by the implies-subset relation we see immediately that:
- [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
- Let [ilmath]y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] then:
- 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:
- 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 [math]\impliedby[/math] every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering
- Suppose that every covering of [math]Y[/math] by sets open in [math]X[/math] contains a finite subcollection covering [math]Y[/math]. We need to show [math]Y[/math] is compact.
- Suppose we have a covering, [math]\mathcal{A}'=\{A'_\alpha\}_{\alpha\in I}[/math] of [math]Y[/math] by sets open in [math]Y[/math]
- For each [math]\alpha[/math] choose an open set [math]A_\alpha[/math] open in [math]X[/math] such that: [math]A'_\alpha=A_\alpha\cap Y[/math]
- Then the collection [math]\mathcal{A}=\{A_\alpha\}_{\alpha\in I}[/math] covers [math]Y[/math]
- By hypothesis we have a finite sub-collection from [ilmath]\mathcal{A} [/ilmath] of things open in [math]X[/math] that cover [math]Y[/math]
- Thus the corresponding finite subcollection of [math]\mathcal{A}'[/math] covers [math]Y[/math]
See also
Notes
- ↑ To say statements are equivalent means we have one [ilmath]\iff[/ilmath] one of the other(s)
- ↑ Note that we actually have [ilmath]X\subseteq\bigcup_{\alpha\in I}A_\alpha[/ilmath] but as topologies are closed under arbitrary union and contain the set the open sets are subsets of we cannot "exceed [ilmath]X[/ilmath]", so we must have [ilmath]X=\bigcup_{\alpha\in I}A_\alpha[/ilmath]
References
- ↑ ^{1.0} ^{1.1} Topology - James R. Munkres - Second Edition