Thoughts about compactness (Pt. V compactness misc.)
In this post we will discuss some points that may be put into one of the other categories, and it is just happenstance that I “forgot” to put it there. Topics that are indirectly related to compactness will also be included here. I also feel as though I should mention that a lot of things aren’t going to be discussed. It just isn’t feasible. Ascoli’s theorem for example, local compactness (at least not yet…maybe when we deal with the one-point compactification).
We begin with a few definitions.
Perfect set: We call a subset of a topological space perfect if where is the derived set or the set of all limit points of .
Condensation point: If is a topological space is a subset of then is a condensation point of if for every neighborhood of there are uncountably many points of .
Theorem: Let be a second countable topological space and latex an uncountable subset of . a) Then the set of all condensation points of , denoted , is perfect and b) all but countably many points of lie in
Proof: a) First suppose that but . Then, there exists some neighborhood latex of such that . In particular, for each there exists some such that is countable. We therefore see that is an open cover of and since is second countable we see that Lindelof’s theorem we are guaranteed a countable subcover . Thus, . But, by assumption we have that is countable and thus is the countable union of countable sets thus countable. This contradicts that is a condensation point of , and so
Conversely, let then then for every neighborhood of we have that there exists some such that . But, since is open there exists some neighborhood of such that , but by assumption every neighborhood of contains uncountably many points of and thus so does , but since it follows that contains uncountably many points of . Since was arbitrary we see that . The conclusion follows.
b) Let the guaranteed countable base of . Define and let . We claim that .
Let then for there exists some of such that is countable, but since is an open base for there exists some such that and clearly then since we see that
Conversely, let then there exists some such that and is countable. Thus, is a neighborhood of which has only countably many points of and so
It follows that . Thus, and since the right hand side is the countable union of countable sets the conclusion follows.
From this we can prove a nice little theorem.
Theorem: Let be a separable metric space, then every closed subset of may be written as where is perfect and is countable.
Proof: Since is closed we have that but clearly and so . Thus, we have that . But from the above we see that is perfect and countable. The conclusion follows. .
Corollary: Every compact subspace of a separable space mEay be written as the union of a perfect set and a closed set.
A theorem which has a cute little solution is the following.
Theorem: Let be a compact metric space and closed subsets of . Then for some and .
Proof: Notice that and since are closed subsets of a compact space we know that they are compact. And, by Tychonoff’s theorem we know that is compact and so given by is trivially continuous and so is continuous and by previous comments assumes its minimum on . The conclusion follows.
We can weaken the prerequisites of the sets involved slightly when dealing with
Theorem: Let be closed subsets of such that is compact. Then for some and some .
Proof: The idea here is that we could revert back to the previous theorem if was also compact. But, since every closed and bounded subset of is compact and since is already closed we must merely find someone to bound it. But, this is pretty obvious in the sense that since is bounded that a lot of contributes nothing to the distance between the two sets. More formally, let . Since is bounded we know that for some . So, consider the set . Clearly this is a closed and bounded subset of . We pause for a lemma:
Lemma: Let be a topological space, and a class of closed subspaces of such that at least one of them, call it , is compact. Then, is a compact subspace of .
Proof: Clearly we have that and since they are all closed we see they are a closed subspace of a compact space. Thus, compact. Since is the same as a subspace of as it is as subspace of the conclusion follows. .
With this in mind we see that is a compact subspace of . Now, the conclusion follows from the last theorem if we can prove that but this is a relatively simple task.
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