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 ⟹ 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, (X,J) is compact if[1]
- Every open covering of X, {Uα}α∈I⊆J contains a finite sub-cover
Note that in this definition we'll actually have (if {Uα}α∈I is actually a covering) X=⋃α∈IUα (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, S⊆X 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, S⊆X 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 Y⊆X 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}α∈I⊆J be a family of open sets in X with Y⊆∪α∈IAα
- Take Bα=Aα∩Y, then {Bα}α∈I is an open (in Y) covering of Y, that is Y⊆∪α∈IBα (infact we have Y=∪α∈IBα)
- Proof of Y⊆∪α∈IBα (we actually have Y=∪α∈IBα)
- We wish to show that Y⊆∪α∈IAα⟹Y⊆∪α∈I(Aα∩Y), using the Implies-subset relation we actually just want to show that:
- Given Y⊆∪α∈IAα that y∈Y⟹y∈∪α∈I(Aα∩Y) - which is what we'll do.
- Note additionally that y∈∪α∈I(Ai∩Y)⟺∃β∈I[y∈Aβ∧y∈Y]
- Let y∈Y, then by hypothesis y∈∪α∈IAα⟺∃β∈I[y∈Aβ]
- It is easily seen that y∈Y∧∃β∈I[y∈Aβ]⟹∃γ∈I[y∈Aγ∧y∈Y] simply by choosing γ:=β.
- Lastly, note that ∃γ∈I[y∈Aγ∧y∈Y]⟺y∈∪α∈I(Aα∩Y)
- We have shown that y∈Y⟹y∈∪α∈I(Aα∩Y) and by the Implies-subset relation we see
- Y⊆∪α∈I(Aα∩Y) - as required.
- We wish to show that Y⊆∪α∈IAα⟹Y⊆∪α∈I(Aα∩Y), using the Implies-subset relation we actually just want to show that:
- I earlier claimed that actually Y=∪α∈I(Aα∩Y) - 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: ∪α∈I(Aα∩Y)⊆Y
- Let y∈∪α∈I(Aα∩Y) then:
- ∃β∈I[y∈Aβ∧y∈Y]⟺∃β∈I[y∈(Aβ∩Y)]
- 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)
- (Aβ∩Y)⊆Y) by the implies-subset relation we see immediately that:
- y∈(Aβ∩Y)⟹y∈Y
- Thus we have shown that y∈∪α∈I(Aα∩Y)⟹y∈Y and finally this means:
- ∪α∈I(Aα∩Y)⊆Y
- Combining this with Y⊆∪α∈I(Aα∩Y) above we see that:
- Y=∪α∈I(Aα∩Y)
- This completes the proof
- Let y∈∪α∈I(Aα∩Y) then:
- 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:
- By hypothesis, Y is compact, this means that {Bα}α∈I contains a finite subcover
- call this subcover {B′i}ni=1 where each B′i∈{Bα}α∈I, now we have Y⊆∪ni=1B′i (we actually have equality, see the blue box in the yellow note box above)
- As each B′i=A′i∩Y (where A′i is the corresponding Aα for the Bα that B′i represents) we see that {Ai}ni=1 is a finite subcover by sets open in X
- Proof of: Y⊆∪ni=1B′i⟹ Y⊆∪ni=1A′i (proving that {A′i}ni=1 is an open cover)
- For each i we have B′i:=A′i∩Y, by invoking the intersection of sets is a subset of each set we note that:
- B′i⊆A′i
- We now invoke Union of subsets is a subset of the union
This theorem states that given two families of sets, {Aα}α∈I and {Bα}α∈I with ∀α∈I[Bα⊆Aα] we have ∪α∈IBα⊆∪α∈IAα
- It follows that Y⊆∪ni=1B′i⊆∪ni=1A′i, in particular:
- Y⊆∪ni=1A′i
- This confirms that {A′i}i=1 is an open cover by sets in X
- This completes this half of the proof.
(Y,Jsubspace) is compact ⟸
- Suppose that every covering of Yby sets open in Xcontains a finite subcollection covering Y. We need to show Yis compact.
- Suppose we have a covering, A′={A′α}α∈Iof Yby sets open in Y
- For each αchoose an open set Aαopen in Xsuch that: A′α=Aα∩Y
- Then the collection A={Aα}α∈Icovers Y
- By hypothesis we have a finite sub-collection from A of things open in Xthat cover Y
- Thus the corresponding finite subcollection of A′covers Y
References
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 [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
- A topological space is compact[1] if every open cover of Xcontains a finite sub-covering that also covers X.
That is to say that given an arbitrary collection of sets:
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 Y⊂X
- Yis compact
Means Y
Theorem: A set Y⊆X 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}α∈I⊆J be a family of open sets in X with Y⊆∪α∈IAα
- Take Bα=Aα∩Y, then {Bα}α∈I is an open (in Y) covering of Y, that is Y⊆∪α∈IBα (infact we have Y=∪α∈IBα)
- Proof of Y⊆∪α∈IBα (we actually have Y=∪α∈IBα)
- We wish to show that Y⊆∪α∈IAα⟹Y⊆∪α∈I(Aα∩Y), using the Implies-subset relation we actually just want to show that:
- Given Y⊆∪α∈IAα that y∈Y⟹y∈∪α∈I(Aα∩Y) - which is what we'll do.
- Note additionally that y∈∪α∈I(Ai∩Y)⟺∃β∈I[y∈Aβ∧y∈Y]
- Let y∈Y, then by hypothesis y∈∪α∈IAα⟺∃β∈I[y∈Aβ]
- It is easily seen that y∈Y∧∃β∈I[y∈Aβ]⟹∃γ∈I[y∈Aγ∧y∈Y] simply by choosing γ:=β.
- Lastly, note that ∃γ∈I[y∈Aγ∧y∈Y]⟺y∈∪α∈I(Aα∩Y)
- We have shown that y∈Y⟹y∈∪α∈I(Aα∩Y) and by the Implies-subset relation we see
- Y⊆∪α∈I(Aα∩Y) - as required.
- We wish to show that Y⊆∪α∈IAα⟹Y⊆∪α∈I(Aα∩Y), using the Implies-subset relation we actually just want to show that:
- I earlier claimed that actually Y=∪α∈I(Aα∩Y) - 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: ∪α∈I(Aα∩Y)⊆Y
- Let y∈∪α∈I(Aα∩Y) then:
- ∃β∈I[y∈Aβ∧y∈Y]⟺∃β∈I[y∈(Aβ∩Y)]
- 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)
- (Aβ∩Y)⊆Y) by the implies-subset relation we see immediately that:
- y∈(Aβ∩Y)⟹y∈Y
- Thus we have shown that y∈∪α∈I(Aα∩Y)⟹y∈Y and finally this means:
- ∪α∈I(Aα∩Y)⊆Y
- Combining this with Y⊆∪α∈I(Aα∩Y) above we see that:
- Y=∪α∈I(Aα∩Y)
- This completes the proof
- Let y∈∪α∈I(Aα∩Y) then:
- 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:
- By hypothesis, Y is compact, this means that {Bα}α∈I contains a finite subcover
- call this subcover {B′i}ni=1 where each B′i∈{Bα}α∈I, now we have Y⊆∪ni=1B′i (we actually have equality, see the blue box in the yellow note box above)
- As each B′i=A′i∩Y (where A′i is the corresponding Aα for the Bα that B′i represents) we see that {Ai}ni=1 is a finite subcover by sets open in X
- Proof of: Y⊆∪ni=1B′i⟹ Y⊆∪ni=1A′i (proving that {A′i}ni=1 is an open cover)
- For each i we have B′i:=A′i∩Y, by invoking the intersection of sets is a subset of each set we note that:
- B′i⊆A′i
- We now invoke Union of subsets is a subset of the union
This theorem states that given two families of sets, {Aα}α∈I and {Bα}α∈I with ∀α∈I[Bα⊆Aα] we have ∪α∈IBα⊆∪α∈IAα
- It follows that Y⊆∪ni=1B′i⊆∪ni=1A′i, in particular:
- Y⊆∪ni=1A′i
- This confirms that {A′i}i=1 is an open cover by sets in X
- This completes this half of the proof.
(Y,Jsubspace) is compact ⟸
- Suppose that every covering of Yby sets open in Xcontains a finite subcollection covering Y. We need to show Yis compact.
- Suppose we have a covering, A′={A′α}α∈Iof Yby sets open in Y
- For each αchoose an open set Aαopen in Xsuch that: A′α=Aα∩Y
- Then the collection A={Aα}α∈Icovers Y
- By hypothesis we have a finite sub-collection from A of things open in Xthat cover Y
- Thus the corresponding finite subcollection of A′covers Y
See also
Notes
- ↑ Note that we actually have X⊆⋃α∈IAα 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=⋃α∈IAα