Limit points of the set $\{\frac{1}{2m} - \frac{1}{2n}\mid n, m \in \mathbb{N}\}$ Let S := $\{\frac{1}{2m} - \frac{1}{2n}\mid n, m \in \mathbb{N}\}$
Then, the limit points will be:
$\lim\limits_{n \to \infty} S = \frac{1}{2m}$
$\lim\limits_{m\to \infty} S = \frac{-1}{2n}$.
$\lim\limits_{m, n \to \infty} S = 0$. 
Thus, the limit points of S will be, at least, the following:
$ C := \{\frac{1}{2m}, \frac{-1}{2n}, 0\} \subset S'$
Yet, I am not sure that I can to prove that there are no other limit points, that is, that $S' \subset C$. Here's what I've got:
Let $x \notin C$ be a limit point of S $\Rightarrow x = \frac{1}{2m} - \frac{1}{2n} \text{ for some } n, m \in \mathbb{N}$. Let $\epsilon = \frac{\sqrt{2}}{2} * (\frac{1}{2m} - \frac{1}{2(m + 1)}) \Rightarrow \nexists s \in S : s \in B(x, \epsilon) \Rightarrow s \notin S'. $ 
Though I am not sure that the procedure was correct. Any help will be appreciated.
Thank you beforehand.
 A: Actually, your description of $C$ is not correct. It should be$$C=\{0\}\cup\left\{\frac1{2n}\,\middle|\,n\in\mathbb N\right\}\cup\left\{\frac{-1}{2n}\,\middle|\,n\in\mathbb N\right\}.$$
Now, let $(x_n)_{n\in\mathbb N}$ be an injective convergent sequence of elements of $S$; I will prove that $\lim_{n\to\infty}x_n\in C$. (I will deal with injective sequences only because the elements of $S'$ are those real numbers which are the limit of an injective sequence of elements of $S$.) Each $x_n$ can be expressed as $\frac1{2a_n}-\frac1{2b_n}$ with $a_n,b_n\in\mathbb N$. Since the sequence is injective, at least one of the sequences $(a_n)_{n\in\mathbb N}$ or $(b_n)_{n\in\mathbb N}$ is unbounded. Let us suppose that it is the first one. Since it is unbounded, it as a subsequence that tends to $+\infty$. We can assume, without loss of generality, that $\lim_{n\to\infty}a_n=+\infty$. But then, since the sequence $\left(\frac1{2a_n}-\frac1{2b_n}\right)_{n\in\mathbb N}$ converges and $\lim_{n\to\infty}\frac1{2a_n}=0$, it can only converge either to $0$ or to a number of the form $\frac{-1}{2k}$ ($k\in\mathbb N$). A similar argument applies if $(b_n)_{n\in\mathbb N}$ is unbounded; in this case $\lim_{n\to\infty}x_n$ will either be $0$ or a number of the type $\frac1{2k}$ ($k\in\mathbb N$).
