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Show that complete and totally bounded metric space $X$ is compact.

Can anyone check my attempt?

Attempt: Let $(x_n)_n$ be a sequence in $X$ having no convergent subsequence. By the completeness of $X$, it has no Cauchy subsequence. Say $F=:\cup_{n=1}^{\infty} \{x_n\}$. So there exists a number $\epsilon >0$ such that $d(x_n,x_m)>\epsilon$ for all distinct $n,m$. Then for each $n$, the open ball $B(x_n, \epsilon)$ centered at $x_n$ with radius $\epsilon$ does not contain any member of $F-\{x_n\}$. Such open balls form an infinite open cover of $F$ having no finite cover. So this is a contradiction totally boundedness.

I needed to check my attempt because my textbook uses Cantor diagonal arguement. Thanks

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  • $\begingroup$ How do you get from "there is no Cauchy subsequence" to "there exists $\epsilon>0$ such that $d(x_n,x_m)>\epsilon$ for all distinct $m,n$"? $\endgroup$ – Lee Mosher Jun 14 '16 at 15:51
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As Lee Mosher implied in the comments, the fact that $\langle x_n:n\in\Bbb N\rangle$ has no Cauchy subsequence does not imply that there is an $\epsilon>0$ such that $d(x_m,x_n)>\epsilon$ for all distinct $m,n\in\Bbb N$.

For a counterexample we can take $x_0=0$ and $x_n=\sum_{k=1}^n\frac1k$ for $n\in\Bbb Z^+$. (In other words, $x_n$ is the harmonic number $H_n$.) The sequence $\langle x_n:n\in\Bbb N\rangle$ has no Cauchy subsequence, but for any $\epsilon>0$ there is an $n\in\Bbb Z^+$ such that $\frac1n<\epsilon$, and clearly $|x_n-x_{n-1}|=\frac1n<\epsilon$.

The fact that $\sigma=\langle x_n:n\in\Bbb N\rangle$ has no Cauchy subsequence means that if $\langle x_{n_k}:k\in\Bbb N\rangle$ is any subsequence of $\sigma$, there is an $\epsilon>0$ such that for each $m\in\Bbb N$ there are $k,\ell\ge m$ such that $d(x_{n_k},x_{n_\ell})\ge\epsilon$. Note that $\epsilon$ may depend on exactly which subsequence of $\sigma$ we’re considering at the moment.

In particular, $\sigma$ is a subsequence of itself, so there is some $\epsilon>0$ such that for each $m\in\Bbb N$ there are $k,\ell\ge m$ such that $d(x_k,x_\ell)\ge\epsilon$. Thus, we could pick $n_0$ and $n_1$ so that $n_0<n_1$ and $d(x_{n_0},x_{n_1})\ge\epsilon$. Then we could pick $n_2$ and $n_3$ so that $n_1<n_2<n_3$ and $d(x_{n_2},x_{n_3})\ge\epsilon$. Continuing in this fashion, we could pick $n_4$ and $n_5$ so that $n_3<n_4<n_5$ and $d(x_{n_4},x_{n_5})\ge\epsilon$, and keep going to get a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ such that $d(x_{n_{2k}},x_{n_{2k+1}})\ge\epsilon$ for each $k\in\Bbb N$. Unfortunately, nothing in this construction constrains $d(x_0,x_2)$, for instance, or $d(x_1,x_3)$: these distances may be much smaller than $\epsilon$.

In fact, it says very little about the subsequence $\langle x_{n_{2k}}:k\in\Bbb N\rangle$ of even-indexed terms, or the subsequence of odd-indexed terms. Of course these are both subsequences of $\sigma$, so we could play the same game with them, but then we’d have the same problem all over again with the resulting subsequences. If we use this general approach, we need to find some way to deal with all of these subsequences at once, and I suspect that this is what the diagonal argument in your text is designed to do.

You should also be worried about the fact that you did not use the hypothesis of total boundedness: since $\Bbb R$ with the usual metric is complete and not compact, it’s clear that total boundedness must be used somewhere in the proof.

The approach in your text may be pædagogically useful, but in my opinion it’s a good example of doing something the hard way. It’s much easier simply to show that total boundedness implies that every sequence has a Cauchy subsequence; once you’ve done this, you’re home free: your space is complete, so every Cauchy sequence has a convergent subsequence. You can find a proof here.

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Maybe the following proof may be more clear than the one in your textbook.

Let $(X,d)$ be metric space with metric d, complete and totally bounded and let $S = \{s_n\}_{n\in\mathbb{N}}\subseteq X$ be a sequence. Since $X$ is totally bounded, there exists an infinite subsequence $S_1\subseteq S$ which lies in some set $B_1 = \{y\in X \mid d(x_1,y)<1\}$. Since $B_1$ is totally bounded, there exists an infinite subsequence $S_2\subseteq S_1$ which lies in some set $B_2 = \{y\in B_1 \mid d(x_2,y)<\frac{1}{2}\}$. Continuing in this manner, we have for each $n$ an infinite sequence $S_{n+1}\subseteq S_n$ lying in $B_{n+1}=\{y\in B_n \mid d(x_{n+1},y)<\frac{1}{n+1}\}$. Let $i_1<i_2<\dots$ be such that $s_{i_{n}}\in S_n$. Then $\{s_{i_n}\}_{n\in\mathbb{N}}$ is a Cauchy sequence. Since $X$ is complete, this subsequence is convergent. Therefore, $S$ has an accumulation point, thus, $(X,d)$ is compact.

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  • $\begingroup$ why $B_{1}$ is totally bounded? because totally bounded + complete space is $\bar{B_{1}}$ compact, but it's not true in infinite banach space $\endgroup$ – user89940 Oct 19 '18 at 4:11

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