1

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.

fabian
  • 499
  • 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
  • 1
    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
  • Dear Did and Mark: thanks; I hope the title is now more adequate – fabian Jul 25 '13 at 13:38

1 Answers1

10

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 $x$.

Michael Albanese
  • 99,526
  • Could you explain what this is contradicting exactly? Subsequences very much confuse me. – Remy Oct 12 '16 at 04:09
  • 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 04:15
  • 1
    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? – user5826 Oct 03 '17 at 17:23
  • @AlJebr: It does not have a subsequence which converges to $x$ because for any subsequence $(x_{n_{k_l}})$, we have $|x_{n_{k_l}} - x| \geq \varepsilon$ for all $l$. – Michael Albanese Oct 03 '17 at 18:17
  • @MichaelAlbanese in the third sentence it is written :"therefore,there is a subsequence ..." can we write "for every subsequence ..." instead ? And if not, then why? – Maths Survivor Nov 11 '17 at 13:29
  • @Linda: No we can't. If we choose an arbitrary subsequence $(x_{n_k})$, it may not be the case that $|x_{n_k} - x| \geq \varepsilon$ for all $k$. For example, suppose $|x_n - x| \geq \varepsilon$ if and only if $n$ is odd (in particular, it is true for infinitely many $n$). Then the subsequence $(x_{n_k})$ where $n_k = 2k$, does not satisfy $|x_{n_k} - x| \geq \varepsilon$ for any $k$. – Michael Albanese Nov 11 '17 at 21:22
  • I'm finding the wording of your last sentence as to why this is a contradiction to be quite confusing. You are contradicting the assumption that every subsequence is converging to $x$? – OGC Aug 24 '18 at 22:47
  • 1
    @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