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I have the following problem:

Let $\{x_k\}_{k \in \mathbb{N}}$ be a bounded and $\{x_{k_i}\}_{i \in \mathbb{N}}$ a convergent subsequence . Suppose each subsequence $\{x_{k_{i}}\}_{i \in \mathbb{N}}, \{x_{k_{i}+1}\}_{i \in \mathbb{N}}, \{x_{k_{i}+2}\}_{i \in \mathbb{N}}, \ldots, \{x_{k_{i}+p}\}_{i \in \mathbb{N}}, \ldots$, where $p \in \mathbb{N}$, are convergent to the same limit $L$. I want to know if with this hypothesis the sequence $\{x_k\}_{k \in \mathbb{N}}$ is convergent to $L$.

I though the main idea was prove that every subsequence is convergent but i get stuck at this part and I don't even get a good approach to prove that every subsequence is convergent. Also I tried to prove that sequence is convergent without use the previous ideia, however, the trouble that $p$ is not fixed for example: I know that following subsequences are convergent

$$ x_{u_i} \rightarrow x \text{ and } x_{v_i} \rightarrow x $$

but I can't conclude anything about $x_{u_i + v_i}$.

Any idea how to prove or a counter example

Thank you

Kutz
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  • "We can prove that each subsequence ${x_{k_{i}}}{i \in \mathbb{N}}, {x{k_{i}+1}}{i \in \mathbb{N}}, {x{k_{i}+2}}{i \in \mathbb{N}}, \ldots, {x{k_{i}+p}}_{i \in \mathbb{N}}, \ldots$, where $p \in \mathbb{N}$, are convergent to the same limit." That's definitely wrong. They might not even converge. – amsmath Aug 27 '19 at 17:43
  • I though that converge was guaranted by $x_{k+1} = H(x_{k}^{2})v$ using continuity. May be my explanation above it's not well to proof the fact ${x_{k_{i}}}{i \in \mathbb{N}}, {x{k_{i}+1}}{i \in \mathbb{N}}, {x{k_{i}+2}}{i \in \mathbb{N}}, \ldots, {x{k_{i}+p}}_{i \in \mathbb{N}}, \ldots$ are converging to the same limit. But it's proved in the article that I'm reading. So take that convergency as hypothesis of the problem. – Kutz Aug 27 '19 at 19:05
  • In your question you start with an arbitrary bounded sequence. So no. – amsmath Aug 27 '19 at 19:11
  • So take this convergency as hypothesis. EDIT: however I have $x_{k_i} \rightarrow x \Rightarrow H(x_{k_i}) \rightarrow H(x) $ after this i have another theorem which guarante that the limit is the same. I didn't write all thing cause it's not important for my question – Kutz Aug 27 '19 at 19:23
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    Say $f(x) = -x$ and $x_{k+1} = f(x_k)$. Start with $x_1=1$. Then your (bounded) sequence is $1,-1,1,-1,\ldots$, but not every subsequence converges, especially not the sequence itself. – amsmath Aug 27 '19 at 19:40
  • I got your point. I've edited my question. – Kutz Aug 27 '19 at 19:55
  • What I still don't understand: You wrote that $x_{k+1} = H(x_k^2)v$, so it follows that $x_{k_i+1} = H(x_{k_i}^2)v\to H(x^2)v$. But if $x_{k_i+1}$ also has the limit $x$, then $H(x^2)v = x$. Is this what you assume? – amsmath Aug 28 '19 at 04:17

2 Answers2

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Let $x_k=1$ if $k=2^i-1$ for some $i\in\mathbb{N}$, and $x_k=0$ otherwise. Also, let $k_i=2^i$. Then $x_{k_i+p}$ converges to $0$ for any $p\in\mathbb{N}$, because this subsequence has only finitely many nonzero terms. But $x_k$ clearly doesn't converge.

metamorphy
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I think that it's easier to prove that for every sub-sequence there is a sub-sub-sequence that converge, and conclued with this that the sequence converge.

tommaso faustini
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  • The problem it's the following: I have to take a subsequence of a generic subsequence and then proof it has the same limit of $x_{k_i+p}$, how do we construct the (sub)subsequence. – Kutz Aug 27 '19 at 19:13