# Subsequence and Accumulation Point Proof

Suppose x is an accumulation point of {$a_n : n \in J$}. Show that there is a subsequence of
{${a_n : n \in J}$}$_{n=1}^\infty$ that converges to x.

I understand the general idea behind the problem that if you take a neighborhood (x - $\epsilon$, x + $\epsilon$) and examine an area of that neighborhood from (x - $\epsilon$, x) with x being the supremum, as well as the accumulation point of the original set, there will be infinitely points in that interval converging to x. I'm just not sure how to define the necessary subsequence.

Any help would be greatly appreciated.

• choose a sequence $\epsilon=1/k$. Since you know that there is at least one point inside $a_{n_k}\in (x-1/k,x)$. You may choose the smallest possible $k$ for that Commented Sep 11, 2014 at 8:06
• I remember another student mentioning going about this proof in a manner such as you have stated. However, I'm not sure how to choose k. Commented Sep 11, 2014 at 8:19

Instead of picking a single neighborhood of $x$ and looking for infinitely many points in that single neighborhood, try to build a sequence of neighborhoods with shrinking radii, choosing a single point from the set in each of the neighborhoods. Since $x$ is an accumulation point, you can do that. And since the radii are shrinking these points will converge to $x$.
For each $n$ you take a neighborhood centered at $x$ with radius $1/n$. That's your sequence of neighborhoods with shrinking radii.
• Pick the sequence $1/k$ for the radii :) Commented Sep 11, 2014 at 8:21
• Alright, to make sure that I've got this. You $\epsilon$ = 1/n and create a set (x - $\epsilon$, x) which since is an accumulation point will have infinitely many values of $\epsilon$ eventually converging to x? Commented Sep 11, 2014 at 8:31