Accumulation points of the set $S=\{(\frac {1} {n}, \frac {1} {m}) \space m, n \in \mathbb N\}$ The exercise is to find the accumulation points of the set  $S=\{(\frac {1} {n}, \frac {1} {m}) \space m, n \in \mathbb N\}$
I'm trying to prove that if $A$={accumulation points of the set $S$}, then $A=\{(0,0), (\frac {1} {n},0),(0,\frac {1} {m}) \space n,m \in \mathbb N\}$. I could prove that the set of the right side of the equality is included in $A$. I don't know how to prove the other inclusion, which means that if $x$ is an accumulation point of $S$, then $x$ has to be $(0,0)$, or of the form $(\frac {1} {n},0)$ or $(0,\frac {1} {m})$. 

Find the distance between points $\left(\frac{1}{n}, \frac{1}{m}\right)$ and $\left(\frac{1}{n'}, \frac{1}{m'}\right)$.  Use an argument based on Cauchy sequences to conclude that any Cauchy sequence of $S$ (that isn't on the diagonal) will eventually only vary in one of its components.  Find the limit.
It's been a loooong time since I did this problem (10 years?) but that should at least be a big hint.
