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I have a certain sequence $x_n$ in a complete and bounded metric space and I would like to prove that it has a convergent subnet (not necessarily subsequence). The best that I was able to do, until now, is to prove that $x_n$ verifies the following property:

for all $K\in\mathbb N$ and for all $\epsilon>0$, there is $n_{\epsilon,K}$ such that $d(x_n,x_m)<\varepsilon$, for all $n,m\geq n_{\epsilon,K}$ such that $|m-n|\leq K$.

Question: Does this sequence have necessarily a convergent subnet.

I really would like this is true, even if I suspect it is not. Nevertheless, I am not able to find a counterexample. I have also tried to prove that this is indeed true: since $X$ is a complete and bounded metric space, it is paracompact and by a theorem of Howes (th. 6.20 of his book "Modern Analysis and Topology") it would suffice to prove that $x_n$ is almost Cauchy. But it seems to me that my hypotheses do not imply that $x_n$ is almost Cauchy, since none of my sets where $x_n$ is Cauchy is cofinal.

Thanks in advance for any help,

Valerio

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Well, to exclude existence of convergent subsequences, it should be enough to take $x_n$ such that $d(x_n,x_m)=\frac{m−n}{n}$, for all $m\geq n$. It verifies the hypotheses above but has no Cauchy subsequences... subnets? –  Valerio Capraro Jan 18 '12 at 15:49
    
Unless I am missing something, in a metric space (or any first countable space), shouldn't any sequence with a convergent subnet also have a convergent subsequence? So "subnet" here seems to be a red herring. –  Nate Eldredge Sep 5 '13 at 17:09
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2 Answers

Let the complete bounded metric space be the unit ball of the Hilbert space $l_2$, and let $\{e_i:i\in\mathbb N\}$ be an orthonormal basis. Consider the following sequence $(x_n)$, which goes from each $e_i$ to the next, but slower and slower. That is, $x_1=e_1$ and $x_2=e_2$ (so we went from $e_1$ to $e_2$ in one step), $x_3=(e_2+e_3)/2$ and $x_4=e_3$ (so we went from $e_2$ to $e_3$ in two equal steps), and so forth, going from $e_k$ to $e_{k+1}$ in $k$ equal steps. Because the steps get smaller and smaller, the sequence $(x_n)$ has the property specified in the question. But no subnet converges. Indeed, for each $i$, the inner products $\langle x_n,e_i\rangle$ are eventually zero, so the only possible limit for a subnet would be the zero vector. But each $x_n$ has norm at least $1/\sqrt2$, so convergence to zero is impossible.

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Yes, indeed! Thank you very much. –  Valerio Capraro Jan 18 '12 at 15:54
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The property you mention does not imply the existence of a convergent subnet: take e.g. the sequence $x_n:=\sqrt n$ on $\mathbb{R}$, as a metric space with the truncated standard distance, $d(x,y):=\min\{|x-y|, 1\}$, which is uniformly equivalent to the standard metric, and makes therefore $\mathbb{R}$ a bounded complete metric space. The property holds with $n_{\epsilon, K}:=\frac{K^2}{4\epsilon^2}\, ,$ although of course $(x_n)$ has no converging sub-net.

As a side remark, note that the property you wrote is equivalent to: $d(x_n,x_{n+1})\to0$ as $n\to\infty$.

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