Compactness of Topological Space

Cover and Subcover

A cover of a set is collection of sets whose union contains : A subcover of a cover is a subset of whose elements still cover . A cover is open if all its elements are open.

e.g. is an open cover of ; is a subcover.

Compactness

A topological space is compact if every open cover of has a finite subcover. A subset of is compact if every open cover of by subsets of has a finite subcover. This is the same as being compact with the subspace topology.

e.g.

• is not compact, is an open cover with no finite subcover;
• is not compact, has no finite subcover;
• with the box topology is not compact. In fact, any discrete infinite space is not compact, since the cover of singletons has no finite subcover.
  

Lemma

If is a topological space and then is compact in the if and only if is compact in the subspace topology on .

Continuous Image of a Compact Set is Compact

Let and be topological spaces. If is compact and is continuous, then is compact.

Extreme Value Theorem

Let be a compact topological space and continuous, where is endowed with the standard topology. Then achieves its maximum and minimum value on .

Heine–Borel Theorem

Any closed interval is a compact subset of . More generally, a subset of is compact if and only if, it is bounded and closed.

Proof

Theorem

Let be a compact topological space and a closed subspace of . Then is compact.

Proof Let be closed and let be an open covering of . Then is an open covering of . Since is compact, there exist such that It follows that , therefore is compact.

Corollary

Any intersection of a compact set with a closed set is compact.

Proof Suppose is closed and is compact. Then is closed in (w.r.t subspace topology), and hence compact by the previous theorem and lemma.

Tube Lemma

Let be any topological space and be a compact space. If and is an open set containing , then there is an open neighborhood of so that .

Proof is an open neighborhood of any , so contains some with an open neighborhood of in and an open neighborhood of in . Then forms an open cover of , since is compact, we can take a finite subcover of . Then is an open neighborhood of in and

Corollary

Any finite disjoint union, quotient, or finite product of compact spaces is compact.

Proof Finite disjoint unions of compact spaces are compact since any open cover of the union is an open cover of each component; The quotient of a compact space is compact since the continuous image of any compact space is compact; To show finite product of compact spaces is compact, it suffices to only consider the product of two compact spaces, say . Suppose is an open cover of , then for any , there is a subcover that covers , which is compact. So we can pick a finite subcover, where is a finite index set. By the tube lemma, we can enlarge to a tube that is also covered by . Now is an open cover of , so we can pick a finite subcover , then is a finite subcover of .

Tychonoff’s Theorem

We have just showed that any finite product of compact spaces is compact. However, this is true even for infinite products if we assume the axiom of choice. This is known as Tychonoff’s theorem.

Tychonoff's Theorem

The product of any collection of compact topological spaces is compact with respect to the product topology.

To prove this theorem, we need an alternative characterization of compactness in terms of closed sets:

Finite Intersection Property

A collection of subsets of a set is said to have the finite intersection property (FIP) if the intersection of any finite subcollection is nonempty.

Proposition

A topological space is compact if and only if every collection of closed subsets of with the finite intersection property has a nonempty intersection.

Proof We prove that is compact iff any collection of closed sets with empty intersection has a finite subcollection with empty intersection. Note that if is an open cover of , then it has a finite subcover if and only if has a finite subcollection with empty intersection. This finishes the proof.

Lindelöf Spaces

There is a slightly weaker notion of compactness called Lindelöf spaces, which is defined as follows:

Lindelöf Space

A topological space is called a Lindelöf space if every open cover of has a countable subcover.

Proposition

A second-countable space is Lindelöf.

Proof Let be a countable basis for the topology of . Let be an open cover of . For each , there exists such that . Since is a basis, there exists such that . The collection is an open cover of consisting of elements from the countable basis . Since is countable, is countable. Now, for each , we can choose the corresponding such that . The collection is a countable subcollection of that covers . Therefore, is Lindelöf.

Remark

The proof above utilized countable axiom of choice.

Sequential and Limit Point Compactness

Limit Point Compactness

A topological space is called limit point compact if every infinite subset of has a limit point in .

Lemma

Every compact space is limit point compact.

Proof Suppose is compact, and is infinite, and has no limit points. Then for each , there exists an open neighborhood of such that , and for each , there exists an open neighborhood of such that . Then is an open cover of , but it does not have a finite subcover because every for is needed to cover . This contradicts the compactness of .

Sequential Compactness

Let be a topological space and . We say that is sequentially compact if every sequence in has a subsequence converges to a point in .

Lemma

Suppose is limit point compact, first-countable and Hausdorff, then is sequentially compact.

Proof Suppose is a sequence in . If is eventually constant, then it converges to the constant value. Otherwise, it is an infinite set in , and it has a limit point . Since is first-countable, we can pick a nested countable neighborhood basis at , say, . Then we can define a convergent subsequence recursively as follows: pick as the point in ; assume all have been chosen, then by the super Hausdorff lemma, there exists an open neighbourhood of not intersecting . As is a neighborhood basis, for some , then we pick as the point in . Then by our construction, we have as .

Remark

Here, we utilized Hausdorffness to construct a subsequence (i.e., we need to ensure that the subindex for all ), and first-countability to ensure the subsequence converges.

Lemma

If is sequentially compact, and second-countable, then is compact.

Proof Since is second-countable, it is Lindelöf, so it suffices to show that any countable open cover of admits a finite cover. For the sake of contradiction, suppose is not compact, then any finite subcollection does not cover the whole space, so we can pick for each , and is a sequence in , which has convergent subsequence, say, as . Suppose for some , then there exists an integer such that for all . However, for sufficiently large such that , we have , which is a contradiction.

In summary, the above lemmas tell us immediately that

Theorem

Compactness, limit point compactness, and sequential compactness are equivalent for metric spaces and second-countable Hausdorff spaces.

Local Compactness

Precompactness

A subset of a topological space is precompact if its closure is compact.

Local Compactness

A topological space is locally compact if every point has a compact neighbourhood.

References and Other Resources

• Morphocular, The Concept So Much of Modern Math is Built On