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

[killian]

Homework Statement

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!