Find all limit points for the set $A=\lbrace (\frac{1}{n},\frac{1}{m}):n,m\geq 1\rbrace$. This was a question on an exam I took recently. When taking the exam I thought the only limit point was $(0,0)$ but that evidently was not the answer considering the amount of credit I lost. Thinking about the problem after the fact, I have found other limit points, such as any $(\frac{1}{2},\frac{1}{n})$, but I can't seem to find a concise notation for ALL the limit points. Can anybody help?
 A: As you've worked out, if you fix either $n$ or $m$ in $\{(1/n, 1/m):n,m≥1\}$, you get a limit point where the other variable is zero. As you already know the answer, I'll just give you a concise way of writing it:
\begin{align*}
\{(1/n, 0):n≥1\}\cup\{(0, 1/m):m≥1\}\cup\{(0,0)\}
\end{align*}
EDIT A bit more detail:
For a point $a$ to be a limit point of the set, given any ball of radius $\epsilon$ around the point, you need to be able to find another point $b$ so that it's not equal to $a$ but it's also inside the ball.
Consider any point of the form $(\frac{1}{p}, \frac{1}{q})$ where $p, q \in \mathbb{N}$. Then the "nearest points" are $(\frac{1}{p}, \frac{1}{q-1}), (\frac{1}{p}, \frac{1}{q+1}), (\frac{1}{p-1}, \frac{1}{q}),$ and $(\frac{1}{p+1}, \frac{1}{q})$. However, you now know that the distance between them is some constant in terms of $p$ and $q$. For example, the distance from $(\frac{1}{p}, \frac{1}{q})$ to $(\frac{1}{p}, \frac{1}{q-1})$ is $\frac{1}{q(q-1)}$. By choosing an $\epsilon$ less than the minimum of these four "closest values" it's clear that $(\frac{1}{p}, \frac{1}{q})$ can't be a limit point.
Now consider points of the form $(0, \frac{1}{q})$. Then given any $\epsilon > 0$, if you set $p = \lceil\frac{1}{\epsilon}\rceil+1$, you'll see that $(\frac{1}{p}, \frac{1}{q})$ is within epsilon of $(0, \frac{1}{q})$, and it's also not equal to $(0, \frac{1}{q})$. The same proof works for $(\frac{1}{p}, 0)$ and $(0, 0)$
