Discrete set in $\mathbb{R^2}$ Just I was solving some multiple choice question and stuck in this question.
Question: Let $S⊂\mathbb{R^2}$ be defined by, 
$S=\{(m+\frac{1}{4^{|p|}} ,n+\frac{1}{4^{|q|}}): m,n,p,q∈\mathbb{Z}\}$
then,
(a) $S$ is discrete subset of $\mathbb{R^2}$
(b) the set of limit points of $S$ is the set $\{(m,n): m,n∈\mathbb{Z}\}$
(c)$S^c$ is connected but not path connected
(d) $S^c$ is path connected
My attempt: By the first sight, it is clear that derived set of $S$ is the set $S' =\{(m,n): m,n∈\mathbb{Z}\}$. Hence (b) is true! 
Now, I get some problem in other options, I know that,a set $S$ is discrete in larger topological space if, every point $x∈S$ has neighborhood $U$ such that $S∩U=\{x\}$. But here I am unable to find such a neighborhood and further I see that, when $p,q→∞, $ we get the limit point $(m,n)$ but $(m,n)∉S$, so that it forces me to think is $S$ is discrete! and further for connectedness, I know that, $\mathbb{Z}$ discrete subset of $\mathbb{R}$ and so I think It's complement will not be connected! But I am not sure about $S$ and 
 am not much sure about other options as well, Please help me :-(
 A: Let us keep your original definition of
$$
S=\bigl\{ (m+\tfrac{1}{4^{|p|}} ,n+\tfrac{1}{4^{|q|}}): m,n,p,q∈\mathbb{Z} \bigr\} \subset \mathbb{R}^2
$$
Note that $\mathbb{Z}^2=\{(m,n)\in\mathbb{R}^2 : m,n\in\mathbb{Z}\}\subset S$ as I pointed out in the comment above.
(a) $S$ is not discrete because all points of $\mathbb{Z}^2$ are not isolated.
(b) It is true that $\mathbb{Z}^2$ is the set of limit points of $S$.
(c) $S^c=\mathbb{R}^2\setminus S$ is path-connected. (It's easy to check if you draw the points of $S$.)
[Update: Proof of (c)]
Choose an irrational real number, for example, $\pi=3.141592\dotsc$
If $(x_0,y_0)\notin S$, then either $x_0$ or $y_0$ is not represented as $m+1/4^{|p|}$ for any $m,p\in\mathbb{Z}$. Assume that $x_0$ is not represented as $m+1/4^{|p|}$. Then $(x_0,y)\notin S$ for all $y\in\mathbb{R}$ so that we can find a continuous path in $\mathbb{R}^2\setminus S$ from $(x_0,y_0)$ to $(x_0,\pi)$. Similarly, if $y_0$ is not represented as $n+1/4^{|q|}$, then there is a continuous path in $\mathbb{R}^2\setminus S$ from $(x_0,y_0)$ to $(\pi,y_0)$.
Now we can assume that two points (that we want to find a continuous path in $\mathbb{R}^2\setminus S$) are located in the union of two lines $x=\pi$ and $y=\pi$. These lines are contained in $\mathbb{R}^2\setminus S$ so that it is trivial to find a continuous path connecting the two points inside the union of two lines.
