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Suppose we have a sequence $\{x_n\}$. Consider the set $S$ of subsequential limits of $\{x_n\}$. Suppose, for any given $\epsilon$ I do the following experiment : For each subsequential limit $x^{(k)}$,$k=1,2,3,\dots,|S|$, I choose a point $x_{n_k}$ from the subsequence which converges to it such that $d(x_{n_k},x^{(k)}) < \frac{\epsilon}{k}$. Suppose $S$ is an infinite set. Then is it possible that the sequence $\{x_{n_k}\}$ converges to $x_{n_1}$ for some choice mentioned above ?

in other words, for each convergent subsequence, I am pulling out a point from it such that their distance with their limit points decreases. is it possible that all the points that I pull out are coming from the same subsequence and converge to some subsequential limit I used in previous steps.

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  • $\begingroup$ $x_{n_k}$ is a point of a subsequence then $\left\{x_{n_{k}}\right\}$ is a sequence. By the way, you didn't use the point $x_{n_k}$. I think you need to clarify the question and especially your notations. $\endgroup$ – user37238 Sep 18 '13 at 13:27
  • $\begingroup$ @user37238: Does it make sense now ? $\endgroup$ – aaaaaa Sep 18 '13 at 14:06
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Is it possible? If $x_n$ converges, I think it is certain, since all subsequences converge also, to the same limit.

An interesting question is whether the existence of such $n_1$ is a sufficient condition for convergence of $x_n$. I think the answer is no, since we could "shuffle together" sequences $a_n^{(k)}$, each converging to $1/k$, along with one that converges to zero, and then get a subsequence $a_{n_k}$ converging to zero, although the original sequence does not converge.

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  • $\begingroup$ Plz see the edit. Does it make sense now ? $\endgroup$ – aaaaaa Sep 18 '13 at 14:14
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There are a couple of problems here.

  1. If $y$ is subsequential limit of $\sigma=\langle x_n:n\in\Bbb Z^+\rangle$, you can’t talk about the subsequence of $\sigma$ that converges to $y$: $\sigma$ has infinitely many different subsequences converging to $y$. What you can do is assign to each $y\in S$ a specific subsequence $\sigma_y$ of $\sigma$ that converges to $y$, say $\sigma_y=\langle x_{n_k}^{(y)}:k\in\Bbb Z^+\rangle$

  2. If $S$ is finite, your construction produces only a finite sequence, which of course does not converge. If $S$ is uncountable, there is no way to enumerate $S$ as $\{y^{(k)}:k\in\Bbb Z^+\}$ so that your construction produces an ordinary infinite sequence: any set $\{y^{(k)}:k\in\Bbb Z^+\}$ is countable and therefore cannot be all of $S$. Both of these cases are possible.

If $S$ is countably infinite and you enumerate it as $S=\{y^{(k)}:k\in\Bbb Z^+\}$, and if for each $k\in\Bbb Z^+$ you choose a term $z_k$ of $\sigma_{y^{(k)}}$ whose distance from $y^{(k)}$ is less than $\frac{\epsilon}k$, then the sequence $\langle z_k:k\in\Bbb Z^+\rangle$ may converge to some point of $S$, or it may not converge at all.

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