Real Analysis – Proving Convergence of a Sequence with Convergent Subsequences

convergence-divergencemetric-spacesreal-analysissequences-and-series

Is the following true?
Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a
further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.

I guess that it is true but I am not sure how to prove this.

Best Answer

True. If not, there exists an $\epsilon > 0$, such that for all $k$, there exists an $n_k > k$ satisfying $|x_{n_k}−x| \ge \epsilon$ since if there is some $k$ which doesn't have such $n_k$, then we can take it as $N$, so $x_n$ converges to $x$. The subsequence $x_{n_{k}}$ does not have any subsequence converging to $x$.

Related Question