# Sufficient condition for convergence of a real sequence

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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What have you tried? Hint: try to show it by contradiction. –  Davide Giraudo Jul 21 '12 at 12:23
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. –  DonAntonio Jul 21 '12 at 12:27
@Davide I was trying to proceed by contradiction by making 2 cases- a(n) is bounded and unbounded . –  Ester Jul 21 '12 at 12:29
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... –  Ben Millwood Jul 21 '12 at 15:26
@DonAntonio Done, I have balanced the downvote. –  1015 Apr 9 '13 at 14:07

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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Thanks for your help. –  Ester Jul 21 '12 at 12:33

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

Then, first we can see that $x_n$ must be bounded, because otherwise, we can easily construct a subsequence $x_{n_k}$ such that $|x_{n_k}|\rightarrow\infty$.

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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