# Sufficient condition for convergence of a real sequence [duplicate]

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.

• 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 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 $l$ – Lost1 Apr 9 '13 at 12:43
• @DonAntonio Done, I have balanced the downvote. – Julien 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 $(x_n)$.

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.

• The second paragraph appears to be extraneous. – user147263 Nov 12 '15 at 1:09

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 $$x$$ .