Compactness

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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 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, (X,J) is compact if[1]

  • Every open covering of X, {Uα}αIJ contains a finite sub-cover

Note that in this definition we'll actually have (if {Uα}αI is actually a covering) X=αIUα (notice equality rather than , this is because the union on the right cannot contain more than X itself, The elements of J are subsets of X and the Uα are elements of J after all.

Compactness of a subset

A subset, SX of a topological space (X,J) is compact if[1]:

  • The topology (S,Jsubspace) is compact (the subspace topology on S inherited from X) as per the definition above.

This maintains compactness as a strictly topological property.

Definition 2

A subset, SX of a topological space (X,J) is compact if:


TODO: Find reference


Note that: when S=X we get definition 1.

Claim 1: The definitions are equivalent

These 2 definitions are the same, that is:

Claim 1: A subspace YX is a compact (def 1) in (X,J) every covering of Y by sets open in X contains a finite subcovering (def 2).



Suppose that (Y,Jsubspace) is compact every covering consisting of open sets of (X,J) contains a finite subcover.

Let {Aα}αIJ be a family of open sets in X with YαIAα
Take Bα=AαY, then {Bα}αI is an open (in Y) covering of Y, that is YαIBα (infact we have Y=αIBα)
Proof of YαIBα (we actually have Y=αIBα)
We wish to show that YαIAαYαI(AαY), using the Implies-subset relation we actually just want to show that:
  • Given YαIAα that yYyαI(AαY) - which is what we'll do.
  • Note additionally that yαI(AiY)βI[yAβyY]
Let yY, then by hypothesis yαIAαβI[yAβ]
It is easily seen that yYβI[yAβ]γI[yAγyY] simply by choosing γ:=β.
Lastly, note that γI[yAγyY]yαI(AαY)
  • We have shown that yYyαI(AαY) and by the Implies-subset relation we see
  • YαI(AαY) - as required.
I earlier claimed that actually Y=αI(AαY) - 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 Y whose union is exactly Y is the same as a covering by open sets in X whose union contains (but need not be exactly equal to) Y. So we have shown so far that:
  • Compact in the subspace with equality for an open covering compact with the open cover of sets in X whose union contains Y
Claim: αI(AαY)Y
Let yαI(AαY) then:
  • βI[yAβyY]βI[y(AβY)]
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)
y(AβY)yY
Thus we have shown that yαI(AαY)yY and finally this means:
  • αI(AαY)Y
Combining this with YαI(AαY) above we see that:
  • Y=αI(AαY)
This completes the proof
By hypothesis, Y is compact, this means that {Bα}αI contains a finite subcover
  • call this subcover {Bi}ni=1 where each Bi{Bα}αI, now we have Yni=1Bi (we actually have equality, see the blue box in the yellow note box above)
As each Bi=AiY (where Ai is the corresponding Aα for the Bα that Bi represents) we see that {Ai}ni=1 is a finite subcover by sets open in X
Proof of: Yni=1Bi Yni=1Ai (proving that {Ai}ni=1 is an open cover)
For each i we have Bi:=AiY, by invoking the intersection of sets is a subset of each set we note that:
  • BiAi
We now invoke Union of subsets is a subset of the union

This theorem states that given two families of sets, {Aα}αI and {Bα}αI with αI[BαAα] we have αIBααIAα

It follows that Yni=1Bini=1Ai, in particular:
  • Yni=1Ai
This confirms that {Ai}i=1 is an open cover by sets in X
This completes this half of the proof.


(Y,Jsubspace) is compact

every covering of Y by sets open in X 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, A={Aα}αI
of Y
by sets open in Y
For each α
choose an open set Aα
open in X
such that: Aα=AαY
Then the collection A={Aα}αI
covers Y
By hypothesis we have a finite sub-collection from A of things open in X
that cover Y
Thus the corresponding finite subcollection of A
covers Y


References

  1. 1.0 1.1 Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene

OLD PAGE

Not to be confused with Sequential compactness


There are two views here.

  1. 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
  2. We can say "sure that set is compact".

The difference comes into play when we cover a set (take the interval [0,5]R) with open sets. Suppose we have the covering {(1,3),(2,6)} this is already finite and covers the interval. The corresponding sets in the subspace topology are {[0,3),(2,5]} which are both open in the subspace topology.

Definition

That is to say that given an arbitrary collection of sets:

  • A={Aα}αI such that each Aα is open in X and
  • X=αIAα
    [Note 1]

The following is true:

  • {i1,,in}I such that X=α{i1,,in}Aα

Then X is compact[1]

Lemma for a set being compact

Take a set YX

in a topological space (X,J)
. Then to say:

  • Y
    is compact

Means Y

satisfies the definition of compactness when considered as a subspace of (X,J)

Theorem: A set YX is a compact in (X,J) if and only if every covering of Y by sets open in X contains a finite subcovering.



Suppose that (Y,Jsubspace) is compact every covering consisting of open sets of (X,J) contains a finite subcover.

Let {Aα}αIJ be a family of open sets in X with YαIAα
Take Bα=AαY, then {Bα}αI is an open (in Y) covering of Y, that is YαIBα (infact we have Y=αIBα)
Proof of YαIBα (we actually have Y=αIBα)
We wish to show that YαIAαYαI(AαY), using the Implies-subset relation we actually just want to show that:
  • Given YαIAα that yYyαI(AαY) - which is what we'll do.
  • Note additionally that yαI(AiY)βI[yAβyY]
Let yY, then by hypothesis yαIAαβI[yAβ]
It is easily seen that yYβI[yAβ]γI[yAγyY] simply by choosing γ:=β.
Lastly, note that γI[yAγyY]yαI(AαY)
  • We have shown that yYyαI(AαY) and by the Implies-subset relation we see
  • YαI(AαY) - as required.
I earlier claimed that actually Y=αI(AαY) - 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 Y whose union is exactly Y is the same as a covering by open sets in X whose union contains (but need not be exactly equal to) Y. So we have shown so far that:
  • Compact in the subspace with equality for an open covering compact with the open cover of sets in X whose union contains Y
Claim: αI(AαY)Y
Let yαI(AαY) then:
  • βI[yAβyY]βI[y(AβY)]
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)
y(AβY)yY
Thus we have shown that yαI(AαY)yY and finally this means:
  • αI(AαY)Y
Combining this with YαI(AαY) above we see that:
  • Y=αI(AαY)
This completes the proof
By hypothesis, Y is compact, this means that {Bα}αI contains a finite subcover
  • call this subcover {Bi}ni=1 where each Bi{Bα}αI, now we have Yni=1Bi (we actually have equality, see the blue box in the yellow note box above)
As each Bi=AiY (where Ai is the corresponding Aα for the Bα that Bi represents) we see that {Ai}ni=1 is a finite subcover by sets open in X
Proof of: Yni=1Bi Yni=1Ai (proving that {Ai}ni=1 is an open cover)
For each i we have Bi:=AiY, by invoking the intersection of sets is a subset of each set we note that:
  • BiAi
We now invoke Union of subsets is a subset of the union

This theorem states that given two families of sets, {Aα}αI and {Bα}αI with αI[BαAα] we have αIBααIAα

It follows that Yni=1Bini=1Ai, in particular:
  • Yni=1Ai
This confirms that {Ai}i=1 is an open cover by sets in X
This completes this half of the proof.


(Y,Jsubspace) is compact

every covering of Y by sets open in X 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, A={Aα}αI
of Y
by sets open in Y
For each α
choose an open set Aα
open in X
such that: Aα=AαY
Then the collection A={Aα}αI
covers Y
By hypothesis we have a finite sub-collection from A of things open in X
that cover Y
Thus the corresponding finite subcollection of A
covers Y


See also

Notes

  1. Note that we actually have XαIAα but as topologies are closed under arbitrary union and contain the set the open sets are subsets of we cannot "exceed X", so we must have X=αIAα

References

  1. 1.0 1.1 Topology - James R. Munkres - Second Edition