## Thoughts about compactness (Pt. IV (cont.) compactness and metric spaces)

Before we get to the main part of the post we prove one nice little theorem:

**Theorem:** A metric space is compact if and only if it’s complete and totally bounded.

**Proof:**

: This is relatively simple. We know that since every compact metric space is sequentially compact and every sequentially compact metric space is totally bounded we already have the second part. But, for didactic purposes we can prove the second part without the use of sequential compactness (in fact it’s easier).

**Lemma:** Every compact metric space is totally bounded.

**Proof:** Let be arbitrary and consider . This is clearly an open cover for and by assumption there exists a finite subcover . Clearly then will serve as a perfectly fine -net. The conclusion follows.

So it remains to show that is complete. But, we know that given any in which is Cauchy that it has some convergent subsequence (since every compact metric space is sequentially compact). So, consider the following lemma

**Lemma:** Let be a Cauchy sequence in a metric space then is convergent if and only if it has a convergent subsequence.

**Proof:** Clearly if is convergent it will serve as it’s own convergent subsequence.

Conversely, let be the guaranteed convergent subsequence of with limit . Since there exists some such that and since is Cauchy it is relatively easy to see there exists some such that . Taking clearly gives that . The conclusion follows.

With this it’s clear that since any Cauchy sequence has a convergent subsequence (by sequential compactness) that every Cauchy sequence is convergent.

: Conversely, by previous argument we must only show that is sequentially compact. But, under the assumption of completeness it suffices to show that every sequence has a Cauchy subsequence. So let be sequence in it is clear that we could relabel this sequence as . Since is totally bounded, there exists a finite class of open balls with radius which cover . From this we see that has a subsequence all of which lie in some open ball of radius one-half. Applying the total boundedness of shows there exists a subsequence of which lies entirely in an open ball of radius . Continuing in this way we see that we may define which is Cauchy. The conclusion follows

Now, I’ve looked through a bunch of books and found what I believe to be the most useful/interesting results of compactness.

The following is actually a GREATLY generalized result that has a compactness related corollary.

**Quasi-totally bounded: **Call a metric space *quasi-totally bounded* if for every there exists some countable set such that is an open cover for .

**Theorem: **Every quasi-totally bounded metric space is separable.

**Proof:** Let be the guaranteed countable discusssed above with and let . Clearly we have is countable since it is the countable union of countable sets. We must merely prove that is dense in . So, let be arbitrary and be given. We must find some such that . But, by the Archimedean principle we have that there exists some such that . So considering we see that there exists some such that and so and so . The conclusion follows.

**Corollary:** Every compact metric space being totally bounded and thus quasi-totally bounded is separable.

**Theorem:** Let be a metric space and Let and be disjoint subsets such that is compact and closed, then .

**Proof:** Define by . We see that and similarly and thus and thus is Lipschitz and thus continuous. Since is compact we know then that for some . But, since and disjoint we have that . The conclusion follows.

**Corollary:** Let be disjoint compact subsets of then .

**Corollary:** Let be compact and latex then

We can now prove the following

**Theorem:** Let be compact metric space and let be continuous. Suppose that is expanding in the sense that . Then** a) ** is injective and is continuous and **b)**

**Proof:**

**a) ** Injectivity follows immediately since if there existed such that but then .

Also, we see that is automatically uniformly continuous since it is Lipschitz with Lipschitz constant one since given any we have that

**b)** Suppose there exists some such that . Since is compact and continuous we know that is compact and thus by our previous theorem we know that .

So, consider the sequence where . Clearly this is a sequence in and by ‘s compactness we know that there exists some subsequence which converges, and is thus Cauchy. Consequently, there exists some such that . But

which is clearly a contradiction.

**Corollary:** Every isometry where is compact is a homeomorphism.

The next theorem is surprisingly powerful.

**Theorem: **Let be a metric space and a convergent sequence in with . Then, is compact.

**Proof:** Let be an open cover for , then there exists some such that and since is open there exists some . But, since we have that all but finitely many points of are in and consequently . Thus, taking and an open set containing each of the finitely many points not in procures the necessary finite subcover of . The conclusion follow.

**Corollary:** Let and be as above, then is closed.

Before we get to the next theorem we point something out that is true in any topological space.

T**heorem: **Let be a topological space and let and be a compact and a closed subspace of respectively. Then, is a compact subspace.

**Proof:** This follows since is closed subspace of .

**Theorem:** Let and be metric spaces and be continuous. Also, let be a decreasing sequence of non-empty compact subsets of . Then .

**Prove:** It is well-known that . So it remains to prove the reverse inclusion.

Firstly, it is easily proved that is a decreasing sequence of compact sets with the F.I.P, and thus we know that . Thus, let since every metric space is Hausdorff (thus ) we know that is closed and since is continuous we know that is closed as well. But, by the above theorem we see then that is compact for every . Also, and thus is decreasing sequence of non-empty compact sets. Thus, we have that and thus . The conclusion follows.

**Theorem:** Let be a compact metric space and . Prove that if for every which is continuous that attains a maximum then is compact.

**Proof:** Since is compact it suffices to show that is closed. If this is trivial, so assume otherwise and let . Define by . It is trivial to prove that this is a continuous mapping and so attains a maximum on , let’s say at . Thus, . But, since and so and the conclusion follows.

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