Let $S\subset \mathbb{R^2}$ be defined by $S=\{(m+\frac{1}{4^{|p|}}, n+\frac{1}{4^{|q|}}):m,n,p,q \in Z\}$ Let $S\subset \mathbb{R^2}$ be defined by $S=\{(m+\frac{1}{4^{|p|}}, n+\frac{1}{4^{|q|}}):m,n,p,q \in \mathbb{Z}\}$. Then


*

*S is discrete in $\mathbb{R^2}$

*The set of limit points of S is the set $\{(m,n): m,n \in\mathbb{Z}\}$

*$S^c$ is connected but not path connected 

*$S^c$ is path connected
It clear limit points of S is the set $\{(m,n): m,n \in\mathbb{Z}\}$ but about the other options 
 A: *

*Think about the case $p=0=q$.

 The set $S$ contains a lot of not isolated points. I.e. $(0,0)$ is not isolated. Define $(x_n,y_n)=\left(0+\frac1{4^{n}},0+\frac1{4^{n}}\right)\in S$. Then $(x_n,y_n)\to (0,0)$.


*If $A$ is the set of limit points of $S$, then $S\subseteq A$.

 Consider $(x_1,x_2)\in\mathbb{R}^2\setminus S$. W.l.o.g. consider $0\leq x_1<1$ and $0\leq x_2<1$. Consider, that either $x_1>0$ or $x_2>0$. Case 1: $x_1>0$. Then there exists $k\in \mathbb{N}$ such $\frac1{4^{k+1}}<x_1<\frac1{4^{k}}$. Define $\delta<\min\left\{x_1-\frac1{4^{k+1}},\frac1{4^{k}}-x_1\right\}$. Then $B_x(\delta):=\{(y_1,y_2)\in\mathbb{R}^2~:~\max\{|x_1-y_1|,|x_2-y_2|\}<\delta\}$ is an open neighbourhood of $(x_1,x_2)$. For $(y_1,y_2)\in B_x(\delta)$, you have $|x_1-y_1|<\delta$ and therefore $\frac1{4^{k+1}}<y_1<\frac1{4^{k}}$ and $(y_1,y_2)\notin S$. All in all $\mathbb{R}^2\setminus S$ is open and $S$ is closed and contains all of its limit points.


*Consider two points $(x_1,x_2),(y_1,y_2)\in\mathbb{R}^2\setminus S$. If $x_1,x_2,y_1,y_2\in\mathbb{R}\setminus\mathbb{Z}$, then you can build a path from $(x_1,x_2)$ to $(x_1,y_2)$ to $(y_1,y_2)$. Otherwise you have to disturb your points a little bit to get es path the same way. Therefore 3. is wrong, but 4. is true. (Especially as $S^c$ is path connected, it is connected).
