Compact Sets, Countably/Sequentially Compact

Countably Compact

A space or subspace is countably compact if every countable open cover has a finite subcover.

Assume a space S is second countable, and let U be countably compact in S. We will show that U is compact (in the traditional sense).

Given an open cover for U, perhaps an uncountable cover, replace each open set with a union of base sets. Thus a collection of base sets covers U, and there are countably many of these. A finite subcover will do. Let U be covered by n base open sets. Call them B₁ through B_n for convenience. Find an open set, from the original cover, that contains B₁ and bring it in. Then find an open set from the original cover that includes B₂. Repeate this process for B₃B₄ etc, up to B_n, and build a finite subcover for U. Thus U is compact. If a space is second countable, compact and countably compact are the same thing.

Review the intro page, and extend those concepts to countably compact spaces. Verify the following four assertions.

The finite union of countably compact sets is countably compact.
A closed subspace of a countably compact set is countably compact.
The continuous image of a countably compact set is countably compact.
Countable compactness is equivalent to the statement that every countable collection of closed sets with the finite intersection property has a nonempty intersection.

Cluster Point

Assume S has the bolzano weierstrass property, i.e. every sequence has a cluster point. Take a countable family of closed sets with the finite intersection property and let x_n be a point in the intersection of the first n sets. Let x be a cluster point of this sequence. Let O be an open set containing x. since x is cluster, O contains infinitely many points in the sequence. Thus O intersects every closed set in the collection. This holds for every open set O containing x. Thus x is in the closure of every closed set in our collection. This means x is in every closed set. The closed sets intersect, and S is countably compact.

Conversely, let S be countably compact. Start with a sequence x₁ x₂ x₃ etc. Let B₁ B₂ B₃ etc be a sequence of sets where B_n contains x_k for k ≥ n. Thus B₃ is the set containing x₃ x₄ x₅ etc.

Let C_n be the closure of B_n. Note that every finite subcollection of these sets intersects. In other words, C has the finite intersection property. Since S is countably compact, there is a point z in the intersection. We will prove z is a cluster point of our sequence x.

Let O be an arbitrary open set containing z, and suppose O has only a finite number of points from the x sequence. Beyond x_n, it has none. This means z is not in C_n+1, which is a contradiction. Therefore z is a cluster point of x.

In summary, countably compact is equivalent to bolzano weierstrass.

Sequentially Compact

A space is sequentially compact if every sequence has a convergent subsequence.

If the subsequence converges to p, then p is a cluster point for the original sequence. Therefore sequentially compact implies countably compact.

Next assume S is countably compact and first countable. Let a sequence x₁ x₂ x₃ etc have a cluster point p. Let B₁ B₂ B₃ etc be the base open sets assigned to p. Start with B₁, which contains an infinite subsequence of x. Perhaps B₁ contains x₃ x₇ x₃₅ x₂₂₇₉ and so on. The intersection of B₁ and B₂ extracts an infinite subsequence from this subsequence. Now B₂ may include some additional points from x, outside of the intersection, but we're not going to worry about that. Similarly, B₃∩B₂∩B₁ pulls out another infinite subset. Continue this process, building a descending chain of infinite subsets from the sequence x.

Build the subsequence v as follows. Let v₁ be any point of x that is in B₁. In the above example, we might set v₁ = x₇. Let v₂ be any point from x in B₁∩ B₂, that has an index beyond 7. We always want to select a point farther out in the x sequence. Let v₃ be any point from x in B₁∩B₂∩B₃, beyond the previous two points. Continue this process, building a subsequence of the original sequence x.

consider any open set containing p. This includes a base open set about p, say B_j. Now B_j includes v_j, and v_j+1, and v_j+2, and so on - the tail of the sequence v. This holds for every open set about p, hence v is a convergent subsequence of x, converging to p. Since x was an arbitrary sequence, countably compact and first countable implies sequentially compact.