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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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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 $(x_n)$.

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.

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|>=ε for some n>=n_0. Thus|x-x_n|>=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|>=ε. However the latter contradicts the fact that y_n has a subsequence that converges to x .

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