# Thread: subsequence converging to p...

1. ## subsequence converging to p...

Let (an) be a sequence of real numbers, and let E = {an : n ∈ N} be the range of (an). Prove that (an) has a subsequence converging to p iff either p appears infinitely many times in (an) or p is an accumulation point of E.

For this if and only if is it showing that either it appears infinitely many times or p is an accumulation point of E?

To be perfectly honest I'm not sure how to even think about this one.

2. Originally Posted by dfanforlife
Let (an) be a sequence of real numbers, and let E = {an : n ∈ N} be the range of (an). Prove that (an) has a subsequence converging to p iff either p appears infinitely many times in (an) or p is an accumulation point of E.

For this if and only if is it showing that either it appears infinitely many times or p is an accumulation point of E?

To be perfectly honest I'm not sure how to even think about this one.
If $\displaystyle p$ appears infinitely often you can just take the subsequence to be that point repeated. If not, then for every $\displaystyle \varepsilon>0$ there exists some point of the sequence different from $\displaystyle p$ in $\displaystyle B_\varepsilon(p)$ and so in particular $\displaystyle B_{\varepsilon}\cap P\supsetneq\{p\}$. Since $\displaystyle \varepsilon$ was arbitrary the conclusion follows.