# Finding accumulation points of $S=\{\frac{1}{n}+\frac{1}{m}\}$

1. Nov 26, 2013

### mahler1

The problem statement, all variables and given/known data.
Let $S==\{\frac{1}{n}+\frac{1}{m}, n,m \in \mathbb N\}$, find all the accumulation points of $S$ and decide whether $S$ is compact or not.

The attempt at a solution.

I've found that the points of the form $\frac{1}{n}$ and $0$ are accumulation points. For the first, given $ε>0$, take the point $\frac{1}{n}+\frac{1}{m}$ such that $m>\frac{1}{ε}$.
Then, $|\frac{1}{n}-(\frac{1}{n}+\frac{1}{m})|=|\frac{1}{m})|<ε$. To prove $0$ is an accumulation point, given $ε>0$, take $n: n>\frac{1}{ε} \implies \frac{1}{n}<ε$.

How can I prove that these are the only accumulation points of $S$?

For compactness, by the Heine Borel theorem, $S$ is compact $\iff$ $S$ is closed and bounded. But $0 \in \overline S$ and clearly $0$ doesn't exist in $S$, this means $S$ is not closed so $S$ is not compact.

2. Nov 26, 2013

### haruspex

Without loss of generality, assume m <= n. If a > 0 is an accumulation point, can you find an upper bound on m?