Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

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 $a_n$, $n \in J$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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