Find the limit points of the following set Find the limit points of the following set :
$$S=\left\lbrace\left(m+\dfrac{1}{4^{|p|}},n+\dfrac{1}{4^{|q|}}\right):m,n,p,q\in \Bbb Z\right\rbrace$$
My try:
For any $(m,n)\in \Bbb Z \times \Bbb Z$ we can find a sequence
$$\left(m+\dfrac{1}{4^{|p|}},n+\dfrac{1}{4^{|q|}}\right)\to (m,n)$$
Hence the set of limit points of $S$ is $\Bbb Z \times \Bbb Z$.
Is the answer correct?
 A: You need to show that if $x$ is a limit point of $S$, we must have $x \in \mathbb{Z}^2$. Your current result only shows that $\mathbb{Z}^2$ contains some limit points of $S$.
A: Your attempted proof would show that $\mathbb{Z}^2 \subseteq S'$, where $S'$ is the set of limit points of $S$. You do have to be more precise in how you mean that the sequence converges to $(m,n)$, however. Which variables are fixed? Which variable is the index for your sequence? I'm staring in particular at the $p$ and $q$ in your formula there, which are looking mighty ambiguous.
Also, we do not have containment in the other direction. Consider the sequence $$\left(m + \frac1{4^{|p|}}, n + \frac1{4^{|q|}}\right)$$ as $m,n,p$ remain fixed and $q \to \infty$. This converges to $\left(m + \frac1{4^{|p|}}, n\right) \not\in \mathbb{Z}^2$. Similarly, anything of the form $\left(m, n + \frac1{4^{|p|}}\right)$ is also a limit point. These however, together with $\mathbb{Z}^2$, do in fact comprise the entirety of $S'$. To show this, you need to show that any limit point of $S$ is actually of the form described above.
