LEC ANA2 6

Reviewed second countability. A metric space is separable iff it is second countable.

Example 166.1(Example).

Bounded sets need not be totally bounded. In $l_1$, let ${\mathbf{e}}_n=\{e_n^k{\}}_{k=1}^{\infty }$ be the sequence defined by $e_n^k={\delta }_{n,k}$. $\Vert {\mathbf{e}}_n-{\mathbf{e}}_k{\Vert }_1=2$ for all $n\ne m$. Now, the set $B=\{\mathbf{x}\in l_1|\Vert \mathbf{x}{\Vert }_1\le 1\}\supset \{{\mathbf{e}}_n\}$. Any open ball of radius $1/2$ can contain at most one ${\mathbf{e}}_n$. Thus, $B$ is not totally bounded.

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Compactness

Compact $⟹$ totally bounded.
Proof is trivial.

Any closed and bounded set in ${\mathbb{R}}^n$ is sequentially compact (Cf Heine Borel Theorem)

Closed subsets of compact sets are compact.

Compact $⟹$ sequentially compact (Cf Theorem 176.5).
Vasanth’s proof: Suppose $X$ is compact and not sequentially compact, that is, there exists a sequence $\{x_n\}\subseteq X$ such that $\{x_n\}$ does not have any convergent subsequence. We claim that $K=\{x_n\}$ is closed; this should be easy to see. Since $X$ is compact, this implies that $K$ is compact too. Since $\{x_n\}$ does not have any converging subsequence, there exist ${\delta }_n$ such that $B_{{\delta }_n}(x_n)\cap K=\varnothing$ for each $n$. This yields an open cover $\{B_{{\delta }_n}(x_n){\}}_{n=1}^{\infty }$ of $K$ which does not have a finite subcover, contradicting our earlier conclusion that $K$ must be compact.

If $X$ is sequentially compact, then for any open cover $X={\bigcup }_{\alpha \in I}{U}_{\alpha }$ there exists $\delta >0$ such that for any $x\in X$, there exists $\alpha \in I$ such that $B_{\delta }(x)\subseteq U_{\alpha }$.
This is the Lebesgue covering lemma with ‘compact’ replaced with ‘sequentially compact’ in the hypothesis (note that we haven’t shown these to be equal yet).
Vasanth’s proof: Suppose there does not exist such a $\delta$. Then for each $n\in \mathbb{N}$, there exists $x_n\in X$ such that $B_{1/n}(x_n)\cap U_{\alpha }^c\ne \varnothing$ for all $\alpha \in I$. WLOG, assume $\{x_n\}\to x\in X$. $x$ must be contained in $B_{ϵ}(x)\subseteq U_{\alpha }$ for some $\alpha \in I$ and $ϵ>0$. Choose $n$ large enough such that $x_n\in B_{ϵ/2}(x)$ and $1/n<ϵ/2$. Then, we have $B_{1/n}(x_n)\subseteq B_{ϵ}(x)\subseteq U_{\alpha }$, a contradiction.

Compact $⟹$ complete.
Let $X$ be compact. We have shown that this implies $X$ is sequentially compact. If $\{x_n\}$ is a Cauchy sequence in $X$, then it has a convergent subsequence, and hence must itself converge.

TFAE:

1. $X$ is compact;
2. $X$ is complete and totally bounded.
3. $X$ is sequentially compact;

We have shown: $1⟹2$, $1⟹3$,

$2⟹3$: Let $\{x_n\}\subseteq X$. If $\{x_n\}$ has finitely many distinct elements, then we are done. Otherwise, $X$ is the union of finitely many $1$-balls, and at least one such ball must contain infinitely many elements of $\{x_n\}$. Take the closure of this ball. It is totally bounded (since $X$ is totally bounded), so we can cover it with finitely many balls of radius $1/2$, once of which must again contain infinitely many points of $\{x_n\}$. Keep going to obtain a contracting sequence of closed sets, whose intersection must be a singleton $\{x\}$ since $X$ is complete. It is now easy to obtain a subsequence of $\{x_n\}$ converging to $x$.

$3⟹1$: Let $\{U_{\alpha }{\}}_{\alpha \in I}$ be an open cover of $X$. Since $X$ is sequentially compact, there exists $\delta >0$ such that for every $x\in X$, there is $\alpha \in I$ such that $B_{\delta }(x)\subseteq U_{\alpha }$. We now claim that there exist $x_1,\dots ,x_n$ such that $X={\bigcup }_{i=1}^n{B}_{\delta }(x_i)$. Indeed, if this were not true, we would be able to construct a sequence $\{y_n\}$ such that $d(y_n,y_m)\ge \delta$ for any $n,m$, contradicting sequential compactness.

Here, we have shown the implications in the opposite directions.