# Abstract Nonsense

## Thoughts about compactness (Pt. IV compactness in metric spaces)

Up until this point we’ve talked about compactness mainly in the context of general topological spaces, with only brief allusions to metric spaces. Now, while topologically spaces may be a more fruitful study than metric spaces it is admittedly easier to apply facts about metric spaces. That said, we devote a few sections specifically to the study of compactness in metric spaces. Since metric spaces are in general much nicer than general topological spaces it isn’t surprising that there are other, more applicable, forms of compactness. So, we need to start with some definitions:

Sequential Compactness: A metric space $\left(X,d\right)$ is said to be sequentially compact if every sequence has a convergent subsequence.

Bolzano-Weierstrass Property (B.W.P): A metric space $(X,d)$ is said to have the Bolzano-Weierstrass Property (B.W.P) if every infinite subset has a limit point.

We begin by showing that these are equivalent statements.

Theorem: A metric space $(X,d)$ is sequentially compact if and only if it has the B.W.P.

Proof:

$\implies$: Suppose that $X$ is sequentially and let $E$ is an infinite subset of $X$. It is clear that we can extract an infinite sequence $\{e_n\}_{n\in\mathbb{N}}$ of distinct points from $E$ and by assumption there exists some subsequence $\{e_k\}_{k\in\mathcal{K}}\sqsubseteq \{e_n\}_{n\in\mathbb{N}}$ which is convergent. The conclusion follows by considering the following lemma

Lemma: Let $x_n\to x$ where $\{x_n\}_{n\in\mathbb{N}}$ has an infinite number of points, then $x$ is a limit point of $\left\{x_n:n\in\mathbb{N}\right\}$.

Proof: Let $\varepsilon>0$ be arbitrary. Since $x_n\to x$ we know that $B_{\varepsilon}(x)$ contains all but finitely many points of $\{x_n\}_{n\in\mathbb{N}}$ and in particular it contains more than two distinct points of $\{x_n:n\in\mathbb{N}\}$. Since both of these can’t be equal to $x$ it follows that there exist some point of $\{x_n:n\in\mathbb{N}\}$ in $B_{\varepsilon}(x)$ distinct from $x$. The conclusion follows $\blacksquare$

$\Leftarrow$: Conversely, let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence in $X$. If $\{x_n\}_{n\in\mathbb{N}}$ has a value infinitely repeated then that constitutes a convergent subsequence. Otherwise, the set $\{x_n:n\in\mathbb{N}\}$ is infinite and by assumption has a limit point. It is easy to find a convergent subsequence from this. (consider taking $y_n$ to be any point in $B_{\frac{1}{n}}(x)$ where $x$ is the limit point afforded by the B.W.P).

The conclusion follows. $\blacksquare$

Our ultimate goal is to prove that a metric space being sequentially compact, having the B.W.P. , and being compact are all equivalent. We have laid the ground-work for this above.  The rest will actually uncover some equally fruitful concepts.

Theorem: A compact metric space $X$ has the B.W.P.

Proof: Suppose not. Then there exists some infinite subset $E$ which has no limit point in $X$. Thus, for every $x\in X$ there exists some $\varepsilon_x$ such that $B_{\varepsilon_x}(x)\cap E\subseteq \{x\}$. Clearly then $\left\{B_{\varepsilon_x}(x)\right\}_{x\in X}$ is an open cover for $X$ and by assumption there exists a finite subcover. But, since $E$ is contained in the set of centers of these open balls it follows that $E$ is finite. This is clearly a contradiction, and the conclusion follows. $\blacksquare$

To complete our goal we must now prove that sequential compactness implies compactness. This, fortunately, is not as simple as the previous theorem. It is fortunate because the sequence of theorems needed to ultimately prove this fact ends up being equally, if not more so, useful than the theorem itself. We begin by discussing the concept of a Lebesgue number.

Lebesgue number: Let $X$ be a metric space and let $\Omega$ be an open cover for $X$. Then, we call $\mathfrak{l}>0$ a Lebesgue number for $\Omega$ in $X$ if whenever $\text{diam }E\leqslant \mathfrak{l}$ there exists some $\omega\in\Omega$ such that $E\subseteq\omega$.

In other words, an open cover has a Lebesgue number if whenever the diameter of a set gets small enough it is contained in at least one of the sets in the open cover.

We now prove that every sequentially compact set has a Lebesgue number.

Theorem: Let $X$ be a sequentially compact metric space. Then $\Omega$ has a Lebesgue number for any open cover $\Omega$.

Proof: Call a subset of $X$ big if it is not contained in any $\omega\in \Omega$ and define $\mathfrak{l}=\inf\left\{\text{diam }B:B\text{ is a big set}\right\}$. If there are no big sets, then $\mathfrak{l}=\infty$ and thus any real number will serve as a Lebesgue number. So, assume that there are big sets. Clearly we must have that $0\leqslant\mathfrak{l}\leqslant\infty$. Once again, if $\mathfrak{l}=\infty$ any real number will do and if $\mathfrak{l}>0$ then $\displaystyle \frac{\mathfrak{l}}{2}$ will work. So, it is only problematic if $\mathfrak{l}=0$. So, for a contradiction assume that it does.

Clearly we must have a sequence $\left\{B_n\right\}_{n\in\mathbb{N}}$ such that $\text{diam }B_n<\frac{1}{n}$. So let $\{b_n\}_{n\in\mathbb{N}}$ be an arbitrary but fixed sequence such that $b_n\in B_n$. By assumption we have that $\{b_n\}_{n\in\mathbb{N}}$ has a convergent subsequence which converges to a point $b\in X$. Since $\Omega$ covers $X$ there exists some $\omega\in\Omega$ such that $b\in\omega$ and since $\omega$ is open we have that there exists some $\delta>0$ such that $B_{\delta}(x)\subseteq\omega$. Consider $B_{\frac{\delta}{2}}(b)$, since $b_n\to b$ we have that $B_{\frac{\delta}{2}}(b)$ contains all but finitely many points of $\left\{b_n\right\}_{n\in\mathbb{N}}$, in particular there exists some $N\in\mathbb{N}$ such that $\displaystyle \frac{1}{N}<\frac{\delta}{2}$. Since $\displaystyle \text{diam }B_N<\frac{\delta}{2}$ and $B_N\cap B_{\frac{\delta}{2}}(b)\ne\varnothing$ it can be easily seen that $B_N\subseteq B_{\delta}(b)$ which contradicts that $B_N$ is big. The conclusion follows. $\blacksquare$

To discuss the next topic we need to define an $\varepsilon$-net

$\varepsilon$-net: Let $X$ be a metric space. We see that a finite point set $E=\{e_1,\cdots,e_n\}$ is an $\varepsilon$-net for $X$ if $\left\{B_{\varepsilon}(e_k)\right\}_{k=1}^n$ is an open cover for $X$.

Totally bounded: A metric space $X$ is totally bounded if it has an $\varepsilon$-net for ever $\varepsilon>0$

Theorem: Every sequentially compact metric space $X$ is totally bounded.

Proof: Let $\varepsilon>0$ and $x_1\in X$ be arbitrary. If $B_{\varepsilon}(x)$ covers $X$ then $\{x_1\}$ is an $\varepsilon$-net and we’re done, so assume not. Let $x_2\in X-B_{\varepsilon}(x_1)$. If $\left\{B_\varepsilon(x_1),B_{\varepsilon}(x_2)\right\}$ cover $X$ we are done, so assume not. Continuing on this way we get a sequence $\{x_n\}$. If $\{x_n\}$ is infinite then it clearly has no convergent subsequence which contradicts the sequentially compactness of $X$. It follows that $\{x_n\}$ is finite and thus an $\varepsilon$-net for $X$. The conclusion follows. $\blacksquare$

Using this we can prove what we’re really after.

Theorem: If $X$ is sequentially compact then it is compact.

Proof: Let $\Omega$ be an arbitrary open cover of $X$. Since $X$ is sequentially compact it has a Lebesgue number $\mathfrak{l}$. Let $\varepsilon=\frac{\mathfrak{l}}{3}$. Once again, by virtue of $X$‘s sequential compactness, we may find an $\varepsilon$-net $\{e_1,\cdots,e_n\}$ of $X$. Since $\text{diam }B_{\varepsilon}(e_k)=\frac{2\mathfrak{l}}{3}<\mathfrak{l}$ we know that $B_{\varepsilon}(e_k)\subseteq \omega_k$ for some $\omega\in \Omega$. Doing this for $1\leqslant k\leqslant n$ produces a finite subclass of $\Omega$ such that $\omega_1\cup\cdots\cup\omega_n\supseteq B_\varepsilon(e_1)\cup\cdots\cup B_\varepsilon(e_n)=X$. The conclusion follows. $\blacksquare$

We can summarize this by saying that $X\text{ is compact }\Leftrightarrow\text{ }X\text{ is sequentially compact }\Leftrightarrow \text{ }X\text{ has the B.W.P}$

From this we can prove one of the most useful theorems in real analysis.

Theorem: Let $\left(X,d\right),\left(Y,d'\right)$ be metric spaces and $f:X\to Y$ be continuous, then $f$ is uniformly continuous.

Proof: Let $\varepsilon>0$ be arbitrary and consider $\left\{B_{\frac{\varepsilon}{2}}(f(x))\right\}_{x\in X}$, clearly then $\left\{f^{-1}\left(B_{\frac{\varepsilon}{2}}(f(x))\right)\right\}_{x\in X}$ is an open cover for $X$. And since $X$ is compact we know that this open cover has a Lebesgue number $\delta$.  In particular if $d(x,y)<\delta\implies \text{diam }\{x,y\}<\delta$ and thus $\{x,y\}\subseteq f^{-1}\left(B_{\frac{\varepsilon}{2}}(f(x))\right)\implies f(\{x,y\})=\{f(x),f(y)\}\subseteq B_{\frac{\varepsilon}{2}}(f(x))$ and thus $|f(x)-f(y)|<\text{diam }B_{\frac{\varepsilon}{2}}(f(x))=\varepsilon$. The conclusion follows $\blacksquare$