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

## 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
Homework Helper
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
Homework Helper
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