# Accumulation point proof

1. Nov 10, 2011

### autre

I need to show that if every neighborhood of some $x\in A$ for some $A\subseteq\mathbb{R}$ contains infinitely many points of $A$, then $x$ is an accumulation point of $A$.

So far, I have:

Let $A\subseteq\mathbb{R}$. I want to show that if every neighborhood of $x\in A$ has infintely many points of A, there exists a $y\in\mathbb{R}$ such that $y\in((x-\epsilon,x+\epsilon)\bigcap A$\{x}).

Am I on the right track?

2. Nov 10, 2011

### mathman

Looks good.

3. Nov 10, 2011

### autre

I'm having some trouble finding that $y$.

Let A⊆R. Define the neighborhood of x∈A as (x−ϵ,x+ϵ) $\forall\epsilon>0$. Since x∈A, (x−ϵ,x+ϵ)⋂A is not empty. Let y∈(x−ϵ,x+ϵ)⋂A...how do I show that $y\not=x$?

Thanks!

4. Nov 10, 2011

### micromass

You can't in general. It could very well be that you chose y=x. Your choice doesn't disallow this. However, you know that (x−ϵ,x+ϵ)⋂A is infinite. So it doesn't only contain x, does it??

5. Nov 10, 2011

### autre

Right, but don't I have to prove that this intersection is infinite? I'm trying to show that x is an accumulation point, but (x−ϵ,x+ϵ)⋂A being infinite presupposes x being an accumulation point, right?

6. Nov 10, 2011

### gb7nash

I might be missing something, but you're trying to prove the definition of an accumulation point (which you can't do). The assumption is that every neighborhood of some x in A contains infinitely many points of A. Therefore, it is an accumulation point by definition.

7. Nov 10, 2011

### micromass

Isn't that given?? The first line of the post is

So you are given that the neigbourhood (x−ϵ,x+ϵ) contains infinitely many points of A.

Or am I totally misunderstanding your question?

8. Nov 10, 2011

### autre

No, you're right. the proof seemed too simple so I thought I was missing something. Makes sense now.