## Thoughts about compactness (Pt. II)

In the previous post we looked at the real number motivation for the concept of compactness and generalized it to topological spaces. We also gave some useful alternative formulations of compactness. We continue briefly here:

Basic open cover: If is a topological space and an open base for then is a basic open cover for if it is an open cover for and composed entirely of basic sets.

**Theorem:** A set is compact if and only if every basic open cover has a finite subcover.

**Proof:**

: Let be an open cover for , clearly we have that each where is some class of open basic sets. Clearly then is a basic open cover for and by assumption there exists some set which covers clearly taking one of the ‘s that contain for finishes the argument.

: This is obvious since a basic open cover is an open cover.

Subbasic open cover: Similarly to before, a subbasic open cover is an open cover consisting entirely of sets from a given open subbase.

**Theorem**: A topological space is compact if and only if every subbasic open cover has a finite subcover.

**Proof:** It is clear from the last post that it is sufficient to show that is compact if and only if every class of closed subbasic sets with the F.I.P. have non-empty intersection.

So, let be the subbase in question. By the previous theorem it suffices to show that every set of closed basic sets with the F.I.P has non-empty intersection.

So, let be the open base generated by and let have the F.I.P. We first show that is contained in some maximal collection of open basic sets in the sense that if then does not have the F.I.P.

To see this we use Zorn’s lemma. Let be the collection of all superset of which have the F.I.P. It is clear that this class along with set inclusion induced a partially ordered set. Now, it is realtively easy to show that if is a chain, then is a class with the finite intersection property, and thus an upper bound. It follows by Zorn’s lemma that has a maximal element, call it and since it only remains to show that

Since each is a basic open set we know that where each is a subbasic closed set. And so if we can show that we are done because .

It merely remains to show that at least one of is in for each . So now assume that none of the aforementioned sets are in . Since is a subbasic open set is is necessarily a basic closed set and so to assume that would mean that and by the maximality of we see that does not have the finite intersection property. This then implies that for some . Doing this for each of gives us a class whose union is disjoint from which contradicts having the F.I.P. Doing this for each finishes the proof.

These will be extremely useful later in the proof of some very powerful theorems. Here is one of them.

**Theorem (Heine-Borel)**: A subset is compact if is closed and bounded.

**Proof:** Since is bounded we know that for some finite closed interval and if we can prove that is compact it our theorem will follow since will be a closed subspace of a compact space.

So, now if this is obvious, so assume . It is relatively easy to prove that the set is an open subbase for so the set is a closed subbase, we must merely show that any subclass of with the F.I.P. has non-empty intersection. So let be non-empty and posses the F.I.P. If contains only intervals of one of the two types then their intersection clearly contiains or , so assume that contains intervals of both types. Let . We finish the argument by showin that for ever . Suppose not, and there existed some . By definition then there exists some and so contradicting ‘s F.I.P. The conclusion follows.

One immediately corollary of this is the following:

**Theorem:** Let be sequence of decreasing closed intervals. Then, .

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To continue from here we need to define the following:

Compact subspace: If and is compact with the relative topology we say that is a compact subspace of .

It is clear that not only is this definition necessary for many reasons one would be we might want to consider whether is compact implies that , but as one can see we run into the problem of under what topology? As we will see later, in a generalization, if are compact subspaces of some ambient space then is a compact subspace. But for now:

**Theorem:** If is a compact topological space and is a closed subset of then is a compact subspace.

**Proof:** Let be an open cover for , by definition we have that for some open in . Clearly covers and thus is an open cover for . By assumption then there exists some finite subcover of , if we may discard it and be left with some finite set which covers . Clearly then taking the corresponding in produces the necessary finite subcover.

Of course any student worth their weight in sawdust is asking “Is the converse necessarily true? Do compact subspaces have to be closed?”. The answer is, no. Consider for example the topological space with and . Clearly is a compact subspace since it’s finite but it is clearly not closed since . Fortunately, that’s not all she wrote. There is a partial converse:

**Theorem:** Let be a Hausdorff topological space and a compact subspace, then is closed in .

**Proof:** We prove that is closed by proving that is open. Let . For each let be the open sets containing respectively such that . Clearly, if we let then is clearly an open cover for and so there exists some such that . Clearly then we have that for the corresponding set . So let which is clearly an open set disjoint from . The conclusion follows.

We now proved what was promised earlier.

**Theorem:** Let be a topological space and let be a finite class of compact subspaces of . Then is a compact subspace of .

**Proof:** Let be an open cover for , then for some open set in . Clearly then covers and thus each . Clearly then are open covers for respectively which by assumption means that for each there exists some which is a finite subcover of . Let . Then is a class of open sets in that covers . Clearly then intersecting each element of will result in the desired subclass of . The conclusion follows.

Using this we can prove a nice little theorem. Suppose that you were a naive analyst and you encounter the following dilemma. You have a function where but is not compact. You wonder, could I possibly make compact by removing a few points? The answer is no.

Theorem: Let be a topological space and let be a non-compact subspace of , then

is non-compact.

Proof: Suppose that was compact, then since each is compact (since they’re finite) it follows that is compact. Boom.

Also, one might ask “what about the infinite case?” The answer is…well no. It should be obvious intuitively that since the union of an infinite number of closed sets need not be closed and any compact subspace of a Hausdorff space is closed that this probably isn’t true. Just to sate your doubtlessly eager minds though, consider with the discrete topology and consider that is an infinite class of compact spaces who’s union is the full space and by previous discussion, this is not compact.

Now, what about the intersection of compact subspaces?

**Theorem:** Let be a non-empty class of compact closed subspaces of a topological space then if (we don’t ever want to talk about empty topological spaces) then is compact.

**Proof:** Clearly we have that given any that since is a closed subset of that is a closed subset of . And since is compact it follows that is a compact subspace of and thus a compact subspace of .

[…] of metric spaces are closed and bounded the converse is true in (this can be proven first for and then extended by Tychonoff’s theorem to the general […]

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