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My book says "A metric space $(X,d)$ is totally bounded if and only if every sequence in $X$ contains a cauchy subsequence."

Let us take a sequence $\{x_i\}$ such that for every prime number $p$, when $i$ is of the form $p^k$, $x_i$ or $x_{p^k}$ is part of the subsequence converging to $p$. For example, for $p=3$, $\{x_3,x_9,x_{27},\dots\}$ forms a cauchy subsequence converging to $3$.

Clearly, this sequence contains infinite cauchy subsequences. But is it totally bounded? Any help with proving this sequence is totally bounded would be greaty appreciated. For a given $\epsilon\in\Bbb{R}$, selecting the primes as part of the finite set isn't an option, as there are an infinite number of them.

Thanks in advance!

  • If the sequence contains subsequences converging to $p$ for all primes $p$, then the sequence is not even bounded, hence it is a fortiori not totally bounded. The implication goes only one way, totally bounded implies Cauchy subsequences. – Daniel Fischer Aug 19 '13 at 12:28
  • The problem is that it has to be every sequence contains a Cauchy subsequence. The sequence $x_i=i$ doesn't contain a Cauchy subsequence, so of course it isn't totally bounded... – Clayton Aug 19 '13 at 12:44
  • @DanielFischer- Doesn't the if and only if part imply the argument runs both ways? Actually the text I have is not a standard text by any stretch of imagination, so it might be a misprint or something of the sort. –  Aug 19 '13 at 13:37
  • But look at what the if and only if stands between: (1) $X$ is totally bounded, and (2) every sequence in $X$ has a Cauchy subsequence. Both (1) and (2) give properties of the whole space $X$, and the property in (2) has to do with all sequences in $X$. It’s not hard to construct, as you’ve done, an example of a sequence with a Cauchy subsequence in a space that is not totally bounded, but in every such case you’ll find that there’s some other sequence in the space that has no Cauchy subsequence, so that the space does not in fact have the property (2). – Brian M. Scott Aug 19 '13 at 19:52

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In principle, the problem was already pointed out in the comments - let me just clarify this a little more:

The term "totally bounded" refers, as stated, to the metric space $(X,d)$, i.e. it is a property of the metric space and not of a single sequence in the metric space. Now the statement reads that if a metric space is totally bounded, then every sequence contains a Cauchy-subsequence and conversely, if every (not only one!) sequence contains a Cauchy subsequence, then the space (not the sequence) is totally bounded. Intuitively, totally bounded means that you cannot diverge with your sequences. Either you converge (within the topological closure), or you somehow oscillate (meaning that eventually every point of your sequence is close to some point in the space and the set of these points is bounded). But this behaviour cannot be decided upon looking at a single sequence.

For your example (you have not really specified the metric space), if you take the subsequence $\{x_{i_j}\}_j$ where $i_j=p_j^{k_j}$ with $p_j$ the j-th prime number and $k_j$ sufficiently large, such that $|x_{i_j}-p_j|<\varepsilon$ for some prechosen $\varepsilon>0$, which is certainly a subsequence of your sequence and hence a sequence of the metric space, then this sequence does not contain a Cauchy-subsequence - therefore your space cannot be totally bounded.

Martin
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This is a very useful statement, since it readily follows that a metric space is compact iff it is complete and totally bounded. So I'll give a proof, even if that wasn't explicitly asked for.

Suppose $E$ is a totally bounded metric space. Let $\{p_n\}$ be a sequence in $E$ and $\varepsilon>0$. Then there exists a finite set of points $\{x_1, \ldots, x_m\}\subset E$ such that $$E=\bigcup_{j=1}^N B\left(x_j,\frac\varepsilon2\right).$$ Then one of these balls must contain $p_n$ for infinitely many $n$, which gives us a subsequence $\{p_{n_k}\}$ with $$d(p_{n_l}, p_{n_m}) \leqslant d(p_{n_l},x_j) + d(x_j, p_{n_m}) < \frac\varepsilon2 + \frac\varepsilon2=\varepsilon $$ for all $l, m$.

Conversely, if $E$ is not totally bounded, then there exists $\varepsilon>0$ such that for any finite subset $\{x_1,\ldots, x_N\}\subset E$, there exists $p\in E$ with $$p\notin\bigcup_{j=1}^N B(x_j,\varepsilon). $$ Let $p_1\in E$. Given $p_1,\ldots, p_N\in E$, choose $p_{N+1}\in E$ with $$p_{N+1}\notin\bigcup_{n=1}^N B(p_n,\varepsilon). $$ Then for all $N$, $d(p_n, p_N)\geqslant\varepsilon$ for $n>N$, and so $\{p_n\}$ cannot have a Cauchy subsequence.

Math1000
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    Just leaving a note for future reference. The first part of this proof is incomplete. You have proven that for every $\varepsilon>0$ there exists a subsequence ${p_{n_k}}$ that has $\varepsilon$-close terms. But this subsequence depends on the choice of $\varepsilon$; it is not necessarily a Cauchy sequence. You need a subsequence that, for any $\varepsilon>0$ given, eventually has $\varepsilon$-close terms: that's a Cauchy sequence. Here's a link to a nice correct proof: http://www.math.harvard.edu/~elkies/M55a.02/pdflatex/top6.pdf – Saúl Pilatowsky-Cameo Oct 16 '16 at 02:07