# Convergence and accumulation points proof.

Suppose $\{a_n\}_{n=1}^\infty$, converges to $A$ and $\{a_n: n \in J\}$ is an infinite set. Show that $A$ is an accumulation point of ${a_n: n \in J}$.

So far I just have the basics down for convergence but the accumulation point stuff is confusing to me.

So far I have:

If $\{a_n\}_{n=1}^\infty$ converges to $A$, then $\exists \epsilon>0$, $N \in \mathbb N^+$ such that $\forall n>=N$ we have $|a_n-A| < \epsilon$.

Let S=$\{a_n: n \in J\}$...?

The definition I'm trying to use is:

Let $S$ be a set of all real numbers. A real number $A$ is an accumulation point of $S$ iff every neighborhood of $A$ contains infinitely many points of $S$.

I feel like this should be easy for me to do but its not clicking yet.

This problem is #24 from the Intro to Analysis textbook by Edward D. Gaughan

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An accumulation point of a sequence is a more general concept that a limit. For example, the sequence $+1,-1,+1,-1,...$ has no limit, but two accumulation points, $\pm 1$.

(If you prefer to have distinct points, take the sequence $+1+\frac{1}{1},-1+\frac{1}{2},+1+\frac{1}{3},-1+\frac{1}{4},...$)

Think of accumulation points as limits of subsequences. A point is an accumulation point of a sequence iff you can find a subsequence converging to that point.

It should be clear that if $a_n \to A$, then all subsequences must also converge to $A$.

Suppose $J$ is an infinite set, and $U$ an open set containing $A$. Since $a_n \to A$, we have some $N$ such that $a_n \in U$ for all $n \ge N$. Then we see that the set $J'=J \cap \{N,N+1,...\}$ is also infinite (otherwise a quick contradiction), and for all $n \in J'$, $a_n \in U$. Hence $A$ is an accumulation point of the subsequence $a_n$, $n \in J$.

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Thank you very much for your help. I'm actually starting to understand this better –  Milton Green Sep 20 '13 at 1:35
Good luck! ${}{}{}$ –  copper.hat Sep 20 '13 at 1:37
I upvoted this answer, but I don't agree with your first example: $1$ and $-1$ are cluster points of the sequence, but not accumulation points of the set consisting of the terms of the sequence, at least according to the OP's definition –  egreg Oct 24 '13 at 20:49
@egreg: I added a slightly modified sequence that should address your concern. I have seen varying definitions for limit, cluster, accumulation points of sets and sequences, and sometimes forget to read the germane fine print... –  copper.hat Oct 24 '13 at 22:26
@copper.hat Yes, math terminology is not standardized. In this case using the term “accumulation” instead of “cluster” might confuse the OP. –  egreg Oct 24 '13 at 22:29