## Thoughts about compactness (Pt. IV compactness in metric spaces)

Up until this point we’ve talked about compactness mainly in the context of general topological spaces, with only brief allusions to metric spaces. Now, while topologically spaces may be a more fruitful study than metric spaces it is admittedly easier to apply facts about metric spaces. That said, we devote a few sections specifically to the study of compactness in metric spaces. Since metric spaces are in general much nicer than general topological spaces it isn’t surprising that there are other, more applicable, forms of compactness. So, we need to start with some definitions:

**Sequential Compactness:** A metric space is said to be *sequentially compact* if every sequence has a convergent subsequence.

**Bolzano-Weierstrass Property (B.W.P): **A metric space is said to have the *Bolzano-Weierstrass Property (B.W.P)* if every infinite subset has a limit point.

We begin by showing that these are equivalent statements.

**Theorem:** A metric space is sequentially compact if and only if it has the B.W.P.

**Proof:**

: Suppose that is sequentially and let is an infinite subset of . It is clear that we can extract an infinite sequence of distinct points from and by assumption there exists some subsequence which is convergent. The conclusion follows by considering the following lemma

**Lemma:** Let where has an infinite number of points, then is a limit point of .

**Proof:** Let be arbitrary. Since we know that contains all but finitely many points of and in particular it contains more than two distinct points of . Since both of these can’t be equal to it follows that there exist some point of in distinct from . The conclusion follows

: Conversely, let be a sequence in . If has a value infinitely repeated then that constitutes a convergent subsequence. Otherwise, the set is infinite and by assumption has a limit point. It is easy to find a convergent subsequence from this. (consider taking to be any point in where is the limit point afforded by the B.W.P).

The conclusion follows.

Our ultimate goal is to prove that a metric space being sequentially compact, having the B.W.P. , and being compact are all equivalent. We have laid the ground-work for this above. The rest will actually uncover some equally fruitful concepts.

**Theorem:** A compact metric space has the B.W.P.

**Proof:** Suppose not. Then there exists some infinite subset which has no limit point in . Thus, for every there exists some such that . Clearly then is an open cover for and by assumption there exists a finite subcover. But, since is contained in the set of centers of these open balls it follows that is finite. This is clearly a contradiction, and the conclusion follows.

To complete our goal we must now prove that sequential compactness implies compactness. This, fortunately, is not as simple as the previous theorem. It is fortunate because the sequence of theorems needed to ultimately prove this fact ends up being equally, if not more so, useful than the theorem itself. We begin by discussing the concept of a Lebesgue number.

**Lebesgue number:** Let be a metric space and let be an open cover for . Then, we call a *Lebesgue number* for in if whenever there exists some such that .

In other words, an open cover has a Lebesgue number if whenever the diameter of a set gets small enough it is contained in at least one of the sets in the open cover.

We now prove that every sequentially compact set has a Lebesgue number.

**Theorem:** Let be a sequentially compact metric space. Then has a Lebesgue number for any open cover .

**Proof:** Call a subset of * big* if it is not contained in any and define . If there are no big sets, then and thus any real number will serve as a Lebesgue number. So, assume that there are big sets. Clearly we must have that . Once again, if any real number will do and if then will work. So, it is only problematic if . So, for a contradiction assume that it does.

Clearly we must have a sequence such that . So let be an arbitrary but fixed sequence such that . By assumption we have that has a convergent subsequence which converges to a point . Since covers there exists some such that and since is open we have that there exists some such that . Consider , since we have that contains all but finitely many points of , in particular there exists some such that . Since and it can be easily seen that which contradicts that is big. The conclusion follows.

To discuss the next topic we need to define an -net

**-net: **Let be a metric space. We see that a finite point set is an -net for if is an open cover for .

**Totally bounded:** A metric space is totally bounded if it has an -net for ever

**Theorem:** Every sequentially compact metric space is totally bounded.

**Proof:** Let and be arbitrary. If covers then is an -net and we’re done, so assume not. Let . If cover we are done, so assume not. Continuing on this way we get a sequence . If is infinite then it clearly has no convergent subsequence which contradicts the sequentially compactness of . It follows that is finite and thus an -net for . The conclusion follows.

Using this we can prove what we’re really after.

**Theorem:** If is sequentially compact then it is compact.

**Proof:** Let be an arbitrary open cover of . Since is sequentially compact it has a Lebesgue number . Let . Once again, by virtue of ‘s sequential compactness, we may find an -net of . Since we know that for some . Doing this for produces a finite subclass of such that . The conclusion follows.

We can summarize this by saying that

From this we can prove one of the most useful theorems in real analysis.

**Theorem:** Let be metric spaces and be continuous, then is uniformly continuous.

**Proof:** Let be arbitrary and consider , clearly then is an open cover for . And since is compact we know that this open cover has a Lebesgue number . In particular if and thus and thus . The conclusion follows

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