A metric space is compact if and only if its complete and totally bounded.

Question

I just need help with one direction. In particular suppose that $(X,d)$ is complete and totally bounded. Suppose that $X$ is infinite. Take any infinite subset $Y$ in $X$ and then just show that $Y$ has a limit point in $X$. Construct finite $1$ net, $\frac{1}{2}$ net, $\frac{1}{4}$ net, etc for $X$. Consider the balls comprising these nets now. Therefore $\exists$ a ball of radius $1$ such that the set $Y_{1} = Y_{0} \cap B_{0}$ is infinite. Then $\exists$ a ball $B_{1}$ of radius $\frac{1}{2}$ such that $Y_{2} = Y_{1} \cap B_{1}$ is infinite. And so forth

Now quadruple the radius of each ball $B_{k}$ to obtain a larger ball $\tilde{B_{k}}$. Then $\tilde{B_{0}} \supseteq \tilde{B_{1}} \supseteq ...$.

My goal is to obtain a sequence of nested balls that has empty intersection and to figure out why we quadrupled the radius. In particular quadrupling the radius was an idea Hausdorff had but I dont know why he did it in his proof. The help would be appreciated!

Answer 1

Assume the space is not compact, so it has an open cover $\{U_i\}_{i\in I}$ with no finite subcober. Since $X$ is totally bounded, you can cover it by finitely many balls of radius $1$, and by assumption, at least one of these balls cannot be covered by finitely many of the $U_i$. Let's say that $B(x_1,1)$ is a ball that cannot be covered by finitely many of the $U_i$.

Now, continue by covering the ball $B(x_1,1)$ by finitely many balls of radius $1/2$, and choose a ball $B(x_2,1/2)$ with $x_2\in B(x_1,1)$, and $B(x_2,1/2)$ cannot be covered by finitely many of the $U_i$.

Inductively, for each $n$ find $x_n\in B(x_{n-1},1/2^{n-2})$ and a ball $B(x_n,1/2^{n-1})$ that cannot be covered by finitely many of the $U_i$.

Note that $x_{m+1}\in B(x_m,1/2^{m-1})$ for all $m$, therefore $$d(x_{m+1},x_m)\leq \textrm{diam}(B(x_m,1/2^{m-1}))=1/2^{m-2}$$ and for $n>m$ we have $$d(x_{n},x_m)\leq d(x_n,x_{n-1})+\dots+d(x_{m+1},x_m)\leq \sum_{k=m}^{n-1} \frac{1}{2^{k-2}}$$ $$\leq \sum_{k=m}^\infty \frac{1}{2^{k-2}}=\frac{1}{2^{m-3}}$$ which shows that the sequence $x_n$ is Cauchy, so it converges to some $x\in X$ by completeness.

In particular $x\in U_i$ for some $i\in I$. Also, $x\in B(x_n,1/2^{n-1})$ for all sufficiently large $n$ and since the radii go to $0$, and $U_i$ is open, you can find some $n$ such that: $$x\in B(x_n,1/2^{n-1})\subset U_i$$ But this contradicts the assumption that the balls we chose cannot be covered by finitely many of the $U_i$.

Answer 2

I particularly like this line of argument. I hope you don't mind if I do not address your question directly. I will try to leave some things as interesting exercises.

Proof First, suppose $(M,d)$ is compact. Then any Cauchy sequence has a convergent subsequence by compactness iff sequential compactness, so it must converge altogether (why?). It is immediate that $M$ is totally bounded: $\varepsilon$-nets are open covers.

Now suppose $(M,d)$ is complete, and totally bounded. I show first that

Claim $(M,d)$ is totally bounded iff every sequence has a Cauchy subsequence.

Proof Suppose $(M,d)$ is not totally bounded. Then there exists $\varepsilon_0$ such that no $\varepsilon_0$-net has a finite subnet. Thus, pick a point $x_1\in X$. Then $X\smallsetminus\{x_1\}$ also fails to be covered, since we removed a finite subset, that is, $\{x_1\}$ is not a $\varepsilon_0$-net. This means there exists $x_2\in X$ that is $\varepsilon_0$-away from $x_1$. Again $X\smallsetminus \{x_1,x_2\}$ fails to be covered, so there is $x_3$ that is $\varepsilon_0$-away from both $x_1,x_2$. Continuing, what can we say about $d(x_n,x_m)$ if $m\neq n$? What does this tell us about the subequences of $(x_n)$?

Supose now $(M,d)$ is totally bounded. We may assume $M$ is infinite, else things are trivial. Pick any $x_1\in M$ Cover $M$ by finitely many $1$-balls. Then there exists a ball, $B_1$; that contains infinitely many points of $M$. We may pick thus some $x_2\neq x_1$. Evidently $B_1$ is also totally bounded, so we may cover it by finitely many $1/2$-balls. Since $B_1$ is infinite, there is a $B_2$ in this cover that is also infinite. Continuing, we find a sequence of distinct points $(x_n)$ and a sequence of nested balls $B_1\supseteq B_2\supseteq B_3\supseteq\cdots$ of radius $1,1/2,1/3,\ldots$. I claim $(x_n)$ is Cauchy. Why? Hint If $m>n$, $x_m,x_n\in A_n\supseteq A_m$. What can $d(x_n,x_m)$ be, at most?

Continuing, we know that if $(M,d)$ is complete and totally bounded, it is complete and every sequence has a Cauchy subsequence. But since $M$ is complete, every sequence has a convergent subsequence. Since we know compact iff sequentially compact for second countable spaces, we're done. $\blacktriangleleft$