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Let $(x_n)$ be a sequence of real numbers.

Prove that if there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the original sequence $(x_n)$ itself converges to $x$ .

Thanks for any help.

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marked as duplicate by Guy Fsone, Giuseppe Negro, Error 404, Marcus M, kccu Dec 5 '17 at 17:40

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    $\begingroup$ As you wrote it it can be a little misleading, imo. The claim is actually: a sequence converges to some limit $\,l\,$ iff every infinite subsequence converges to the very same limit $\,l\,$ . Going from here to subsequences of subsequences is easy, though a little messy with the sub-sub-indexes. $\endgroup$ – DonAntonio Jul 21 '12 at 12:27
  • $\begingroup$ @Davide I was trying to proceed by contradiction by making 2 cases- a(n) is bounded and unbounded . $\endgroup$ – Ester Jul 21 '12 at 12:29
  • $\begingroup$ Notice that you can have sequences where every subsequence has a convergent subsequence, but said subsequences have different limits, and (hence) the overall sequence does not converge... $\endgroup$ – Ben Millwood Jul 21 '12 at 15:26
  • $\begingroup$ @DonAntonio no there is nothing wrong with the proposition. You should read the question more carefully. he is trying to say, if every subsequence has a sub-subsequence which converges to $l$, then the sequence converges to $l$ $\endgroup$ – Lost1 Apr 9 '13 at 12:43
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    $\begingroup$ @DonAntonio Done, I have balanced the downvote. $\endgroup$ – Julien Apr 9 '13 at 14:07
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First notice that your condition implies that your sequence is bounded.

Indeed, if $(x_n)$ is unbounded, we can find a subsequence $(x_{n_k})$ such that $|x_{n_k}|\ge k$. This subsequence does not have a convergent subsequence.


So we know that $(x_n)$ is bounded and it is not convergent. This means that $$M=\limsup x_n > \liminf x_n =m.$$ (Both $M$ and $m$ are real numbers, since $(x_n)$ is bounded.)

We know (from the properties of limit superior and limit inferior) that there is a subsequence $(x_{n_k})$ which converges to $M$ and there is a subsequence $x_{n_l}$ which converges to $m$. (And every subsequence of any of these two subsequences has, of course, the same limit $M$ resp. $m$.)

We have found two subsequences with different limits, which contradicts your assumptions about the sequence $(x_n)$.

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Suppose $x_n$ does not converge to $x$, but every subsequence of $x_n$ has a sub-subsequence which converge to $x$.

Since $x_n$ does not converge to $x$ we must be able to find a subsequence such that every term is more than $\epsilon$ away from $x$ for some $\epsilon>0$, but clearly this does not have a sub-subsequence which converges to $x$, by definition.

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    $\begingroup$ The second paragraph appears to be extraneous. $\endgroup$ – user147263 Nov 12 '15 at 1:09
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Let every subsequence of $ x_n$ has a convergent subsequence to $ x$ and suppose by way of contradiction that $x_n$ does not converges to $x$ . Then there exists $ ε>0$ such that for every $ n_0$ , $ |x-x_n|\geq ε$ for some $n\geq n_0$. Thus $ |x-x_n|\geq ε$ for an infinite number of $n$ . This implies that there exists a subsequence $ y_n$ of $ x_n $ , such that for each $n$, $ |x-y_n|\geq ε $ . However the latter contradicts the fact that $ y_n$ has a subsequence that converges to $ x$ .

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  • $\begingroup$ Please use MathJax in your answers. $\endgroup$ – Kamil Jarosz Jan 3 '16 at 11:26

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