# Abstract Nonsense

## 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. $\blacksquare$