How do you prove that some point is the *only* accumulation point in a set?

[killian]

1. The problem statement, all variables and given/known data

I'm using the following definition of an accumulation point:

A point a \in \textbb{R} is an accumulation point of a set A\subset \textbb{R} if every \epsilon-neighborhood of a contains at least one element of A distinct from a.

Now, given the set A=\{\frac{1}{n}:n\in \textbb{Z}^+\}, I'm trying to prove that the only accumulation point of A is 0.

2. The attempt at a solution

I was able to prove that 0 is an accumulation point, but my question is about proving there can't be any others.

Intuitively, it makes sense to me because any element between 0 and 1 is either an element of A or between two elements of A. In either case, you can take \epsilon small enough to not include any elements of A.

The problem I have is that I'm not sure how to formulate this in a way that would be considered rigorous.

[Dick]

Having the right picture of the problem is the first step. And you've got that. Given n, what's an epsilon the would be less than both (1/n-1/(n+1)) and (1/(n-1)-1/n)? If you can find one, that would fulfill your intuitive sense in a rigorous way. Yes?

[HallsofIvy]

Let \alpha be any number other than 0. Show that \alpha is not an accumulation point.

You probably should do two cases: (1) alpha> 0 and (2) \alpha< 0. Sjhow that there are only finite number of n such that |1/n|> \alpha.

[arildno]

Well, how about letting z be an accumulation point, and then prove that it must equal zero?
That is, prove the uniqueness of zero as an accumulation point!