# Sequence and subsequence convergence

Let {$x_n$} be a sequence and $x \in \mathbb{R}$. Prove {$x_n$} converges to $x$ if and only if for every subsequence {$x_{n_k}$} of {$x_n$} there exists a subsequence {$x_{n_{k_j}}$} of {$x_{n_k}$} that converges to $x$.

So left to right is pretty simple. I need some help with right to left. I believe I want to suppose that {$x_n$} does not converge to $x$. Then I use the negation of convergence which is there exist $\epsilon > 0$ for every $N$ there exist an $n \ge N$ such that $|x_n - x| \ge \epsilon$. I don't know what epsilon to choose or if there is a better way.

## 1 Answer

As you said, begin by assuming there exists $\epsilon >0$ such that for every $K$ there exists $n_K \geq K$ with $|x_{n_K}-x| \geq \epsilon$. But don't choose an $\epsilon$. In fact, you can't - the $\epsilon$ depends on the specific sequence, which you don't know anything about.

Use this to create a subsequence $\{x_{n_K}\}_{K=1}^\infty$ such that $|x_{n_K} - x| \geq \epsilon$ for all $K$. To make sure it's a subsequence, you additionally restrict $n_K$ so that $n_K > \min\{n_{K-1}, K\}$.

Now assume the second part of the statement holds, i.e. for every subsequence $\{x_{n_k}\}$ there is a subsequence $\{x_{n_{k_j}}\}$ which converges to $x$.

Apply this to the subsequence you just made to get a contradiction.