Abstract Nonsense

Crushing one theorem at a time

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 E of a topological space X perfect if D(E)=E where D(E) is the derived set or the set of all limit points of E.

Condensation point: If X is a topological space E is a subset of X then p is a condensation point of E if for every neighborhood of p there are uncountably many points of E.

Theorem: Let X be a second countable topological space and latex E an uncountable subset of X. a) Then the set of all condensation points of E, denoted \mathfrak{C}, is perfect and b) all but countably many points of E lie in \mathfrak{C}

Proof: a) First suppose that c\in\mathfrak{C} but c\notin D\left(\mathfrak{C}\right). Then, there exists some neighborhood latex N of c such that N\cap D\left(\mathfrak{C}\right)=\{c\}. In particular, for each x\in N there exists some N_x such that N_x\cap E is countable. We therefore see that \left\{N_x\right\}_{x\in N} is an open cover of N and since X is second countable we see that Lindelof’s theorem we are guaranteed a countable subcover \Omega. Thus, \displaystyle\bigcup_{\omega\in \Omega}\omega\implies N\cap E=\bigcup_{\omega\in\Omega}\left\{\omega\cap E\right\}. But, by assumption we have that \omega\cap E is countable and thus N\cap E is the countable union of countable sets thus countable. This contradicts that c is a condensation point of E, and so c\in D\left(\mathfrak{C}\right)\implies \mathfrak{C}\subseteq D\left(\mathfrak{C}\right)

Conversely, let c\in\mathcal{D}\left(\mathcal{C}\right) then then for every neighborhood N of c we have that there exists some  k\in N\cap\mathcal{C}\ne\varnothing such that k\ne c. But, since N is open there exists some neighborhood M of k such that M\subseteq N, but by assumption every neighborhood of k contains uncountably many points of E and thus so does M, but since M\subseteq N it follows that N contains uncountably many points of E. Since N was arbitrary we see that c\in\mathfrak{C}\implies D\left(\mathfrak{C}\right)\subseteq\mathfrak{C}. The conclusion follows.

b) Let \mathfrak{B} the guaranteed countable base of X. Define \mathcal{B}=\left\{B\in\mathfrak{B}:B\cap E\text{ is countable}\right\} and let \displaystyle \mathfrak{U}=\bigcup_{B\in\mathcal{B}}B. We claim that \displaystyle \mathfrak{C}=\mathfrak{U}'=\bigcap_{B\in\mathcal{B}}B'\Leftrightarrow \mathfrak{C}'=\mathfrak{U}.

Let x\in \mathfrak{C}' then for there exists some N of X such that N\cap E is countable, but since \mathfrak{B} is an open base for X there exists some B\in\mathfrak{B} such that x\in B\subseteq N and clearly then since B\cap E\subseteq N\cap E we see that x\in\mathfrak{U}\implies \mathfrak{C}'\subseteq\mathfrak{U}

Conversely, let x\in \mathfrak{U} then there exists some B\in \mathfrak{B} such that B\in x and B\cap E is countable. Thus, B is a neighborhood of x which has only countably many points of E and so x\notin\mathfrak{C}\implies x\in\mathfrak{C}'\implies \mathfrak{U}\subseteq\mathfrak{C}'

It follows that \mathfrak{C}'=\mathfrak{U}. Thus, \displaystyle \mathfrak{C}'\cap E=\bigcup_{U\in\mathfrak{U}}\left\{U\cap E\right\} and since the right hand side is the countable union of countable sets the conclusion follows. \blacksquare

From this we can prove a nice little theorem.

Theorem: Let X be a separable metric space, then every closed subset C of X may be written as P\cup N where P is perfect and N is countable.

Proof: Since C is closed we have that D(C)\subseteq C but clearly P\subseteq D(C) and so P\subseteq C. Thus, we have that C=P\cup \left(C-P\right). But from the above we see that P is perfect and C-P countable. The conclusion follows. \blacksquare.

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 X be a compact metric space and E,G closed subsets of X. Then d(E,G)=d(e,g) for some e\in E and g\in G.

Proof: Notice that \displaystyle d(E,G)=\inf_{e\in E}\inf_{g\in G}d(e,g)=\inf_{(e,g)\in E\times G}d(e,g) and since E,G are closed subsets of a compact space we know that they are compact. And, by Tychonoff’s theorem we know that E\times G is compact and so d:X\times X\mapsto\mathbb{R} given by (x,y)\mapsto d(x,y)  is trivially continuous and so d:E\times G\mapsto \mathbb{R} is continuous and by previous comments assumes its minimum on E\times G. The conclusion follows. \blacksquare

We can weaken the prerequisites of the sets involved slightly when dealing with \mathbb{R}^n

Theorem: Let E,G be closed subsets of \mathbb{R}^n such that E is compact. Then d(E,G)=d(e,g) for some e\in E and some g\in G.

Proof: The idea here is that we could revert back to the previous theorem if G was also compact. But, since every closed and bounded subset of \mathbb{R}^n is compact and since G is already closed we must merely find someone to bound it. But, this is pretty obvious in the sense that since E is bounded that a lot of G contributes nothing to the distance between the two sets. More formally, let \delta=d(E,G). Since E is bounded we know that E\subseteq\left\{\bold{x}:\|\bold{x}\|\leqslant M\right\} for some M\in\mathbb{R}. So, consider the set A=\left\{\bold{x}:\|\bold{x}\|\leqslant M+\delta+1\right\}. Clearly this is a closed and bounded subset of \mathbb{R}^n. We pause for a lemma:

Lemma: Let X be a topological space, and \left\{C_j\right\}_{j\in\mathcal{J}} a class of closed subspaces of X such that at least one of them, call it C_\ell, is compact. Then, \displaystyle\bigcap_{j\in\mathcal{J}}C_j is a compact subspace of X.

Proof: Clearly we have that \displaystyle \bigcap_{j\in\mathcal{J}}C_j\subseteq C_\ell and since they are all closed we see they are a closed subspace of a compact space. Thus, compact. Since \displaystyle \bigcap_{j\in\mathcal{J}}C_j is the same as a subspace of C_\ell as it is as  subspace of X the conclusion follows. \blacksquare.

With this in mind we see that G\cap A=K is a compact subspace of \mathbb{R}^n. Now, the conclusion follows from the last theorem if we can prove that d(E,K)=d(E,G) but this is a relatively simple task. \blacksquare


February 27, 2010 - Posted by | General Topology, Topology | ,

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