# Abstract Nonsense

## Thoughts about compactness (Pt. IV (cont.) compactness and metric spaces)

Before we get to the main part of the post we prove one nice little theorem:

Theorem: A metric space $X$ is compact if and only if it’s complete and totally bounded.

Proof:

$\implies$: This is relatively simple. We know that since every compact metric space is sequentially compact and every sequentially compact metric space is totally bounded we already have the second part. But, for didactic purposes we can prove the second part without the use of sequential compactness (in fact it’s easier).

Lemma: Every compact metric space is totally bounded.

Proof: Let $\varepsilon>0$ be arbitrary and consider $\left\{B_{\varepsilon}(x)\right\}_{x\in X}$. This is clearly an open cover for $X$ and by assumption there exists a finite subcover $\left\{B_{\varepsilon}(x_1),\cdots,B_{\varepsilon}(x_n)\right\}$. Clearly then $\left\{x_1,\cdots,x_n\right\}$ will serve as a perfectly fine $\varepsilon$-net. The conclusion follows. $\blacksquare$

So it remains to show that $X$ is complete. But, we know that given any $\{x_n\}_{n\in\mathbb{N}}$ in $X$ which is Cauchy that it has some convergent subsequence (since every compact metric space is sequentially compact). So, consider the following lemma

Lemma: Let $\{c_n\}_{n\in\mathbb{N}}$ be a Cauchy sequence in a metric space $\left(C,d\right)$ then $\{c_n\}_{n\in\mathbb{N}}$ is convergent if and only if it has a convergent subsequence.

Proof: Clearly if $\{c_n\}_{n\in\mathbb{N}}$ is convergent it will serve as it’s own convergent subsequence.

Conversely, let $\{c_{\varphi(n)}\}_{n\in\mathbb{N}}$ be the guaranteed convergent subsequence of $\{c_n\}_{n\in\mathbb{N}}$ with limit $c$.  Since $c_{\varphi(n)}\to c$ there exists some $N_1\in\mathbb{N}$ such that $N_1\leqslant n\implies d(c_{\varphi(n)},c)<\frac{\varepsilon}{2}$ and since $\{c_n\}$ is Cauchy it is relatively easy to see there exists some $N_2\in\mathbb{N}$ such that $N_2\leqslant n\implies (d(c_{\varphi(n)},c_n)<\frac{\varepsilon}{2}$. Taking $N=\max\{N_1,N_2\}$ clearly gives that $n\leqslant N\implies d(c,c_n)\leqslant d(c_n,c_{\varphi(n)})+d(c,c_{\varphi(n)})<\varepsilon$. The conclusion follows. $\blacksquare$

With this it’s clear that since any Cauchy sequence has a convergent subsequence (by sequential compactness) that every Cauchy sequence is convergent.

$\Leftarrow$: Conversely, by previous argument we must only show that $X$ is sequentially compact. But, under the assumption of completeness it suffices to show that every sequence has a Cauchy subsequence. So let $\{x_n\}$ be sequence in $X$ it is clear that we could relabel this sequence as $\{x_{11},x_{12},x_{13},\cdots\}$. Since $X$ is totally bounded, there exists a finite class of open balls with radius $\frac{1}{2}$ which cover $X$. From this we see that $\{x_{11},x_{12},\cdots\}$ has a subsequence $\{x_{21},x_{22},\cdots\}$ all of which lie in some open ball of radius one-half. Applying the total boundedness of $X$ shows there exists a subsequence $\{x_{31},x_{32},\cdots\}$ of $\{x_{21},x_{22},\cdots\}$ which lies entirely in an open ball of radius $\frac{1}{3}$. Continuing in this way we see that we may define $\Delta=\left\{x_{11},x_{22},x_{33},\cdots\right\}\sqsubseteq\{x_n\}$  which is Cauchy. The conclusion follows $\blacksquare$

Now, I’ve looked through a bunch of books and found what I believe to be the most useful/interesting results of compactness.

The following is actually a GREATLY generalized result that has a compactness related corollary.

Quasi-totally bounded: Call a metric space $X$ quasi-totally bounded if for every $\varepsilon>0$ there exists some countable set $M_{\varepsilon}$ such that $\left\{B_{\varepsilon}(m)\right\}_{m\in M_{\varepsilon}}$ is an open cover for $X$.

Theorem: Every quasi-totally bounded metric space $X$ is separable.

Proof: Let $E_n$ be the guaranteed countable $M_{\varepsilon}$ discusssed above with $\displaystyle \varepsilon=\frac{1}{n}$ and let $\displaystyle E=\bigcup_{n=1}^{\infty}E_n$. Clearly we have $E$ is countable since it is the countable union of countable sets. We must merely prove that $E$ is dense in $X$. So, let $x\in X$ be arbitrary and $\varepsilon>0$ be given. We must find some $e\in E$ such that $e\in B_{\varepsilon}(x)$. But, by the Archimedean principle we have that there exists some $n\in\mathbb{N}$ such that $\frac{1}{n}<\varepsilon$. So considering $E_n$ we see that there exists some $e\in E_n$ such that $x\in B_{\frac{1}{n}}(e)$ and so $\displaystyle d(x,e)<\frac{1}{n}<\varepsilon$ and so $m\in B_{\varepsilon}(x)$. The conclusion follows. $\blacksquare$

Corollary: Every compact metric space being totally bounded and thus quasi-totally bounded is separable.

Theorem: Let $X$ be a metric space and Let $E$ and $G$ be disjoint subsets such that $E$ is compact and $G$ closed, then $d(E,G)=\delta>0$.

Proof: Define $\rho:E\mapsto\mathbb{R}$ by $\displaystyle \rho(e)=\inf_{g\in G}d(e,g)$. We see that $\displaystyle \rho(e)-\rho(e')=\inf_{g\in G}d(e,g)-\inf_{g\in G}d(e',g)\leqslant\inf_{g\in G}\left\{d(e,g)-d(e',g)\right\}\leqslant d(e,e')$ and similarly $\rho(e')-\rho(e)\leqslant d(e,e')$ and thus $\left|\rho(e)-\rho(e')\right|\leqslant d(e,e')$ and thus $\rho$ is Lipschitz and thus continuous. Since $E$ is compact we know then that $\displaystyle \inf_{e\in E}\rho(e)=\inf_{e\in E}\inf_{g\in G}d(e,g)=\inf_{g\in G}d(e',g)$ for some $e'\in E$. But, since $G$ and disjoint we have that $\displaystyle \inf_{g\in G}d(e',g)=\delta>0$. The conclusion follows. $\blacksquare$

Corollary: Let $E,G$ be disjoint compact subsets of $X$ then $d(E,G)=\delta>0$.

Corollary: Let $E$ be compact and latex $\{x\}\in X-E$ then $d(E,x)=\delta>0$

We can now prove the following

Theorem: Let $K$ be compact metric space and let $f:K\mapsto K$ be continuous. Suppose that $f$ is expanding in the sense that $d\left(f(x),f(y)\right)\geqslant d(x,y),\text{ }\forall x,y\in K$. Then a) $f$ is injective and $f^{-1}:f(K)\mapsto K$ is continuous and b) $f(K)=K$

Proof:

a) Injectivity follows immediately since if there existed $x,y\in K$ such that $x\ne y$ but $f(x)=f(y)$ then $0=d\left(f(x),f(y)\right)\geqslant d(x,y)>0$.

Also, we see that $f^{-1}$ is automatically uniformly continuous since it is Lipschitz with Lipschitz constant one since given any $f(x),f(y)\in f(K)$ we have that $d\left(f^{-1}(f(x))-f^{-1}(f(y))\right)=d\left(x-y\right)\leqslant d\left (f(x)-f(y)\right)$

b) Suppose there exists some $k\in K$ such that $k\notin f(K)$. Since $K$ is compact and $f$ continuous we know that $f(K)$ is compact and thus by our previous theorem we know that $d(f(K),k)=\delta>0$.

So, consider the sequence $\left\{f^n(k)\right\}_{n\in\mathbb{N}}$ where $f^n(k)=\underbrace{f(f(\cdots(f(k))\cdots)}_{n\text{ times}}$. Clearly this is a sequence in $K$ and by $K$‘s compactness we know that there exists some subsequence $\left\{f^{n_m}\right\}_{m\in\mathbb{N}}$ which converges, and is thus Cauchy. Consequently, there exists some $m\in\mathbb{N}$ such that $d\left(f^{n_{m+1}}(k),f^{n_m}(k)\right)<\frac{\delta}{2}$. But

\displaystyle \begin{aligned} d\left(f^{n_{m+1}}(k),f^{n_m}(k)\right) &\geqslant d\left(f^{n_m}(k),f^{n_{m-1}}(k)\right) \\& \vdots \\& \geqslant d(f(k),k)\geqslant \delta \end{aligned}

which is clearly a contradiction. $\blacksquare$

Corollary: Every isometry $f:X\mapsto X$ where $X$ is compact is a homeomorphism.

The next theorem is surprisingly powerful.

Theorem: Let $X$ be a metric space and $\{x_n\}_{n\in\mathbb{N}}$ a convergent sequence in $X$ with $x_n\to x$. Then, $E=\left\{x\right\}\cup\left\{x_n:n\in\mathbb{N}\right\}$ is compact.

Proof: Let $\Omega$ be an open cover for $E$, then there exists some $\omega\in \Omega$ such that $x\in\omega$ and since $\omega$ is open there exists some $B_{\delta}(x)\subseteq\omega$. But, since $x_n\to x$ we have that all but finitely many points of $\left\{x_n:n \in\mathbb{N}\right\}$ are in $B_{\delta}(x)$ and consequently $\omega$. Thus, taking $\omega$ and an open set containing each of the finitely many points not in $\omega$ procures the necessary finite subcover of $\Omega$. The conclusion follow. $\blacksquare$

Corollary: Let $X$ and $E$ be as above, then $E$ is closed.

Before we get to the next theorem we point something out that is true in any topological space.

Theorem: Let $X$ be a topological space and let $E$ and $F$ be a compact and a closed subspace of $X$ respectively. Then, $E\cap F$ is a compact subspace.

Proof: This follows since $E\cap F$ is closed subspace of $E$. $\blacksquare$

Theorem: Let $X$ and $Y$ be metric spaces and $f:X\mapsto Y$ be continuous. Also, let $\left\{K_n\right\}_{n\in\mathbb{N}}$ be a decreasing sequence of non-empty compact subsets of $X$. Then $\displaystyle f\left(\bigcap_{n\in\mathbb{N}}K_n\right)=\bigcap_{n\in\mathbb{N}}f(K_n)$.

Prove: It is well-known that $\displaystyle f\left(\bigcap_{n\in\mathbb{N}}K_n\right)\subseteq \bigcap_{n\in\mathbb{N}}f(K_n)$. So it remains to prove the reverse inclusion.

Firstly, it is easily proved that $\left\{f(K_n)\right\}_{n\in\mathbb{N}}$ is a decreasing sequence of compact sets with the F.I.P, and thus we know that $\displaystyle \bigcap_{n\in\mathbb{N}}f(K_n)\ne\varnothing$. Thus, let $\displaystyle k\in\bigcap_{n\in\mathbb{N}}f(K_n)$ since every metric space is Hausdorff (thus $T_1$) we know that $\{k\}$ is closed and since $f$ is continuous we know that $f^{-1}(\{k\})$ is closed as well. But, by the above theorem we see then that $f^{-1}(\{k\})\cap K_n$ is compact for every $n$. Also, $f^{-1}(\{k\})\cap K_{n+1}\subseteq f^{-1}(\{k\})\cap K_{n}$ and thus $\left\{f^{-1}(\{k\})\cap K_n\right\}_{n\in\mathbb{N}}$ is decreasing sequence of non-empty compact sets. Thus, we have that $\displaystyle \bigcap_{n\in\mathbb{N}}\left(f^{-1}(\{k\})\cap K_n\right)=f^{-1}(\{k\})\cap\bigcap_{n\in\mathbb{N}}K_n\ne\varnothing$ and thus $\displaystyle k\in f\left(\bigcap_{n\in\mathbb{N}}K_n\right)$. The conclusion follows. $\blacksquare$

Theorem: Let $X$ be a compact metric space and $E\subseteq X$. Prove that if for every $f:X\mapsto\mathbb{R}$ which is continuous that $f\mid E:E\mapsto\mathbb{R}$ attains a maximum then $E$ is compact.

Proof: Since $X$ is compact it suffices to show that $E$ is closed. If $E=X$ this is trivial, so assume otherwise and let $c\in X-E$. Define $\rho:X\mapsto\mathbb{R}$ by $\rho(x)=-d(x,c)$. It is trivial to prove that this is a continuous mapping and so $\rho\mid E$ attains a maximum on $E$, let’s say at $e$. Thus, $-d(x,e')\leqslant -d(x,e),\text{ }\forall e\in E$. But, $-d(x,e)<0$ since $d(x,e)>0$ and so $0 and the conclusion follows.$\blacksquare$