# Proving that any subsequence of a convergent sequence in a metric space is itself convergent

I am doing practice exercises in preparation for a midterm and I am stuck on the following questions (there are in fact two questions, but I thought it would be better to split them apart). I will first list the question followed by the attempts at a solution.

Q. Prove that any subsequence of a convergent sequence in a metric space is itself convergent, and has the same limit as the sequence.

Attempt. Let $\{x_n\}$ be a convergent sequence in a metric space $(X,d)$. Then for all $\epsilon >0$ there exists N such that $n \geq N$ implies $d(x_n,x) \leq \epsilon$

Consider a subsequence {$x_{n_{k}}$} of $\{x_n\}$. Then...

And this is where I am no longer certain what to do. I thought that maybe if I could prove that the subsequence converges, I could simply use the theorem that states that any convergent subsequence of a Cauchy sequence converge with the same limit (given that in a metric space, any convergent sequence is a Cauchy sequence). Would anyone be able to guide me through the final steps? Please note that we are not permitted to use the definition of a neighbourhood for this question.

• think about subsequences they only skip points... right? – Matthew Levy Oct 19 '14 at 23:42

## 1 Answer

HINT: If something is valid for $n \geq N$ is valid for $n_k \geq N$