# 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 :-(

• OP said "... but $(m,n)\notin S$, ...". However, $(m,n)\in S$ for all $m,n\in\mathbb{Z}$ since $(m,n)=((m-1)+1/4^0,(n-1)+1/4^0)\in S$. Do you mean $S=\{ (m+1/4^p,n+1/4^q) : m,n\in\mathbb{Z}, p,q\in\mathbb{Z}_{>0} \}$?
– ChoF
Commented Mar 6, 2018 at 4:59
• First, of all thank sir, for your reply, by your comment I can see $(m,n)∈S$. Sir is then $S$ is not discrete? What about other options? Commented Mar 6, 2018 at 5:04
• Further sir, by $S$ I mean, $S=\{(m+\frac{1}{4^{|p|}} ,n+\frac{1}{4^{|q|}}): m,n,p,q∈\mathbb{Z}\}$ Commented Mar 6, 2018 at 5:05
• If $(m,n)\in S$ for all $m,n\in\mathbb{Z}$, these points are not isolated. It means that $S$ is no longer discrete.
– ChoF
Commented Mar 6, 2018 at 5:08
• Actually $S=\{(m+\frac{1}{4^{|p|}},n+\frac{1}{4^{|q|}}) : m,n,p,q\in\mathbb{Z}\}=\{(m+\frac{1}{4^p},n+\frac{1}{4^q}) : m,n\in\mathbb{Z},p,q\in\mathbb{Z}_{\geq0}\}$. The latter is simpler.
– ChoF
Commented Mar 6, 2018 at 5:11

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.

• Sir, first of all thanks for your answer and update :-). Sir, after spending some more time on option (c), I develop another way to "check option (c)" here it is: as $S$ contains all its limits points, so that $S$ is closed subset of $\mathbb{R^2}$. Hence $S^c$ must be an open subset of $\mathbb{R^2}$ and we know that, open subsets of $\mathbb{R^n}$ or $\mathbb{C^n}$ are connected if and only if they are path-connected! Which forces us to conclude that,option(c) in given question is false! :-) Commented Mar 6, 2018 at 13:13
• @AkashPatalwanshi You may consider "Theorem. If $X$ is locally path-connected and connected, then $X$ is path-connected." Note that any open set of $\mathbb{R}^n$ and $\mathbb{C}^n$ is locally path-connected. So you are right. If you prove that the open set $S^c$ is connected, then it is also path-connected. But how do you prove $S^c$ is connected? It is easier to prove that $S^c$ is path-connected.
– ChoF
Commented Mar 6, 2018 at 13:23
• Sir, my previous comment is just to "discard option (c)" if we suppose $S^c$ is connected set, then (by my privious comment) it "must be " path-connected!!. Hence option (c) in "given question" is false! For (d) we must prove it. Commented Mar 6, 2018 at 13:26
• @AkashPatalwanshi You are right. (c) does not make sense since $S^c$ is open in a locally path-connected space.
– ChoF
Commented Mar 6, 2018 at 13:28
• Apart from $\mathbb Z^2$, why $(m+\frac{1}{4},n)$ for some $m,n \in \mathbb Z$, is not a limit point of S?
– Riaz
Commented Jan 31, 2022 at 14:16

a set 'S' is Discrete iff S-S'=S where S' is set of all limit point. In this question S'=(m,n) so we conclude that (m+1÷4|p|, n+1÷4|q|) - (m, n) not equals to S. Hence s is not discrete