Consider a complete metric space E with the following property:

If $x_n$ is a bounded sequence, then $\forall \epsilon > 0$, $\exists i,j , i \neq j$ such that $d(x_i,x_j) < \epsilon$. Question:
Given any such bounded sequence in $E$, can we always find a convergent subsequence ? I am unable to construct a convergent subsequence for an arbitrary bounded sequence in E - the given property seems too weak! Am I missing something?

  • $\begingroup$ I think you need to add the assumption that such $i,j$ can be found for arbitrarily large indices $i$ and $j$. Else, just define $x_1=x_2$, and the sequence $\{x_n\}_{n=1}^{\infty}$ trivially satisfies your condition because $d(x_1,x_2)=0<\epsilon$ for all $\epsilon>0$. $\endgroup$ – Michael Apr 30 '16 at 23:33
  • $\begingroup$ I added the word "any" to the question to make it clearer. The problem asks that given any bounded sequence, can we find a convergent subsequence. Thanks. $\endgroup$ – sid Apr 30 '16 at 23:39
  • $\begingroup$ I suppose that if you make that property a property of hte space $E$, ratehr than of a particular sequence, then you can say it holds on any tail (since a tail of a bounded sequence is also a bounded sequence), so indeed for arbitrarily large $i,j$. Perhaps that is what you meant anyway, I read it differently. $\endgroup$ – Michael Apr 30 '16 at 23:40
  • $\begingroup$ So why not just choose $\epsilon_k$ sequentially (and decaying rapidly) so as to construct a Cauchy sequence? $\endgroup$ – Michael Apr 30 '16 at 23:42
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    $\begingroup$ Say we construct a subsequence like: For $\epsilon_1$, we can find $d(x_{n_1},x_{n_2}) < \epsilon_1$ and for $\epsilon_2 < \epsilon_1$, we can find $d(x_{n_3},x_{n_4}) < \epsilon_2$ , and so on..... This doesnt help in the construction of a Cauchy sequence, because there is no relation between $x_{n_2} $ and $x_{n_3}$. This is the principal problem I am facing. $\endgroup$ – sid Apr 30 '16 at 23:48

Let $(x_n)_{n=1}^\infty$ be a bounded sequence in $E$. We claim that $X := \{x_n\}_{n=1}^\infty$ is totally bounded. Indeed, suppose this were not the case. Then there is some $\epsilon >0$ with the following property:

For every finite set $F \subseteq X,$ one can find $x \in X$ such that $d(x,y) \geqq \epsilon$ whenever $y \in F$.

We construct a subsequence $(x_{n_k})_{k=1}^\infty$ as follows. Put $n_1 :=1$. Suppose $n_1<...< n_{k-1}$ were obtained so that $d(x_{n_i},x_{n_j})\geqq \epsilon$ whenever $1\leqq i,j <k$, $i\neq j$. Since we assumed $X$ is not totally bounded, the balls


must fail to cover $X$, and there must be an infinite amount of points of $X$ outside their union. Let $n_k>n_{k-1}$ be so that

$$x_{n_k} \not\in B(x_{n_1};\epsilon)\cup...\cup B(x_{n_{k-1}};\epsilon).$$

Thus, the bounded sequence $(x_{n_k})_{k=1}^\infty$ is such that $d(x_{n_i},x_{n_j})\geqq \epsilon$ whenever $i \neq j$, and the stated property of $E$ fails.

It follows that if $E$ has the property, $X$ must be totally bounded. Since $E$ is complete, $X$ must be relatively compact (its closure must be compact). Hence, the sequence $(x_n)_{n=1}^\infty$ in $X$ has a convergent subsequence.

  • $\begingroup$ Upvoted. I think this is a very elegant proof, except for some 'notational abuse'. Precompact is being used in place of 'totally bounded', and 'relatively compact' is the same as 'precompact'. Of course, for complete metric spaces they are equivalent, but I think it leades the reader down the wrong thought path. I'll take the liberty of editing your answer. $\endgroup$ – Fimpellizieri May 1 '16 at 1:14
  • $\begingroup$ I understood upto the point where $x_{n_k} \notin B(x_{n_1},\epsilon) \cup B(x_{n_2},\epsilon) \cup ... B(x_{n_{k-1}},\epsilon)$. But how does this imply the next statement ? Since $\{x_{n_k}\}_{k=1}^{\infty}$ is an infinite sequence, how do we pick $x_{n_{k+1}}, x_{n_{k+2}}$ etc? $\endgroup$ – sid May 1 '16 at 1:31
  • $\begingroup$ I accepted the edit. My main reference in what regards metric spaces is Dieudonné's "Foundations of Modern Analysis" (chapter 3). There, precompact stands for the same as totally bounded. But I will accept your criticism since this has been happening quite often with me. It seems like the terminology used in some classic books such as this one is getting outdated. Other thing that people usually get annoyed with is when I write $\max(\cdot)$ instead of $\max\{\cdot \}$. $\endgroup$ – user179514 May 1 '16 at 1:35
  • $\begingroup$ @sid I constructed the sequence inductively. I picked $x_{n_1}$, and supposing $x_{n_1},...,x_{n_{k-1}}$ had been chosen, I showed how to choose $x_{n_k}$. $\endgroup$ – user179514 May 1 '16 at 1:40
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    $\begingroup$ Ah I see it now! Awesome! $\endgroup$ – sid May 1 '16 at 1:50

Here is an approach to solving: Suppose $E$ has that property.

First prove: For any bounded sequence $\{x_k\}_{k=1}^{\infty}$ in $E$ and for any $\epsilon>0$, there exists a point $x_n$ that is within a distance $\epsilon$ of infinitely many other points in the sequence.

Next: Use this property to construct a Cauchy sequence.

  • $\begingroup$ How do we prove the existence of a cluster point in the sequence, when no details regarding the compactness of the metric space is given? $\endgroup$ – sid May 1 '16 at 0:26
  • $\begingroup$ Are you asking about my "next" suggestion? Or about my "first prove" suggestion? It seems to me that once we have Cauchy sequence, we are done, as complete metric spaces ensure Cauchy sequences converge. The above "next" and "first prove" things do not use compactness. $\endgroup$ – Michael May 1 '16 at 0:27
  • $\begingroup$ I meant on the "first prove" suggestion. I can't seem to be able to prove the existence of such a point based solely on the given property. $\endgroup$ – sid May 1 '16 at 0:29
  • $\begingroup$ What if you assume not, then proceed iteratively, throwing away an at-most-finite number of points when needed? $\endgroup$ – Michael May 1 '16 at 0:30
  • $\begingroup$ Note also that my "first prove" does not require you to find a cluster point. Given any $\epsilon>0$, there exists a point that works for that $\epsilon$. Different $\epsilon$ values are allowed to have different points that work. $\endgroup$ – Michael May 1 '16 at 0:35

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