2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ think about subsequences they only skip points... right? $\endgroup$ – Matthew Levy Oct 19 '14 at 23:42
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.