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 $(0,\frac {1} {m})$. 

Here's another solution: let $$B=\{(\frac1n,0)\mid n\in\mathbb N\}\cup\{(0,\frac1n)\mid n\in\mathbb N\}\cup\{(0,0)\}.$$ Now, note that horizontal and vertical strips of the form $(a,b)\times\mathbb R$ and $\mathbb R\times(c,d)$ are open sets. The complement of $B\cup S$ is a union of such strips. Therefore, $B\cup S$ is closed, so it must contain every accumulation point of $S$. To complete the proof, it remains to verify that each point in $S$ is isolated (since we already know that every point in $B$ is an accumulation point).
