# Proving that the mothersequence converges to $x$ if any subsequence contains a subsequence which converges to $x$ [duplicate]

Dear reader of this post,

I am currently working on some problems about sequences and their subsequences. I proved a claim and because this prove involves some elementary concepts, I would like to ask three related questions. The claim is as follows: For any real number $x$ and $x_m \in \mathbb{R}^{\infty}$, show that $x_m \rightarrow x$ if every subsequence of $(x_m)$ has itself a subsequence that converges to $x$.

My prove is as follows: Take any subsequence $x_{m_{k}}$. As stated, I know that $\exists$ a (sub)subsequence $x_{m_{k_{i}}} \rightarrow x$. Let $\bar{x}_{m_{k_{i}}}$ be the non-convergent part of the subsequence. Redefine $x_{m_{k}}$ as $\tilde{x}_{m_{k_{i}}} = \left\lbrace \bar{x}_{m_{k_{1}}},\bar{x}_{m_{k_{2}}},\ldots,x_{m_{k_{1}}},x_{m_{k_{2}}},\ldots \right\rbrace$. Redefine the mother sequence $x_m$ furthermore as $\tilde{x}_m = \left\lbrace \tilde{x}_{m_{1}},\tilde{x}_{m_{2 }},\ldots \right\rbrace$. Finally, take any $\epsilon >0$. I know that for any $\tilde{x}_{m_{k}}$ $\exists M \in \mathbb{N}$ for which $\forall i > M$ $| \tilde{x}_{m_{k_{i}}} - x | < \epsilon$. Thus, $x_m$ is convergent.

After having stated my prove, I'd like to ask my three questions:

• I think the prove requires each subsequence to contain infinitely many elements. Is this correct and does every subsequence contain indeed infinitely many elements in general?
• Do I really need to redefine the sequence $x_m$ and its subsequences $x_{m_{k}}$ and is this legitimate?
• If these two questions are answered positively, is my prove correct?

Thank you very much for your support. I am looking forward for your replies.

• What is the 'non-convergent part' of a (sub)sequence? – Michael Albanese Jul 25 '13 at 11:03
• Dear Michael, thank you for your comment. I thought that I could separate the subsequence into one part which is convergent and another which does not converge. Writing these lines, I realize that this separation does not match the claim [...] does every subsequence contain indeed infinitely many elements in general?''. Ok, it seems to me now that I either cannot separate a subsequence or the assertion is not correct. Could you maybe comment on this a bit? – fabian Jul 25 '13 at 11:10
• Such a decomposition will not be unique without further restrictions. – Michael Albanese Jul 25 '13 at 11:27
• The title does not correspond to the post (and the result it asserts does not hold). – Did Jul 25 '13 at 12:39
• Note that the post requires a limit $x$ which can be achieved within every subsequence. This is essential, but is not flagged in the title. For example if $s_n=(-1)^n$ every subsequence has either an infinite number of terms equal to $1$ or an infinite number equal to $-1$ (possibly both) - and therefore has a convergent subsequence which is constant. It does not qualify under the "limit $x$" definition because $1\neq -1$. – Mark Bennet Jul 25 '13 at 13:29

## 1 Answer

Here is an alternative method to prove the desired claim.

Theorem: If every subsequence of $(x_n)$ has a subsequence which converges to $x$, then $(x_n)$ converges to $x$.

Proof: Suppose $x_n$ does not converge to $x$. Then there is $\varepsilon > 0$ such that $|x_n - x| \geq \varepsilon$ for infinitely many $n$. Therefore, there is a subsequence $(x_{n_k})$ with $|x_{n_k} - x| \geq \varepsilon$ for all $k \in \mathbb{N}$. This is a contradiction as $(x_{n_k})$ is a subsequence of $(x_n)$ which does not have a subsequence which converges to $x$.

• Could you explain what this is contradicting exactly? Subsequences very much confuse me. – Remy Oct 12 '16 at 4:09
• Why does your third sentence imply that the subsequence converges to x? Doesn't that say it does NOT converge to x? – Remy Oct 12 '16 at 4:15
• The third sentence of the proof implies that the subsequence doesn't converge to $x$. The fourth sentence explains why this is a contradiction. – Michael Albanese Oct 16 '16 at 13:03
• How do we know $(x_{n_k})$ does not have a subsequence which converges? – user46372819 Oct 3 '17 at 17:23
• @OGC: It's a contradiction to the assumption that every subsequence of $(x_n)$ has a subsequence which converges to $x$. Namely, $(x_{n_k})$ is a subsequence of $(x_n)$ without the assumed property. – Michael Albanese Aug 24 '18 at 23:54