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

(A) $S$ is a discrete set.

(B) $\mathbb R^2\setminus S$ is path connected.

I think $S$ is a discrete set. If we fix any three of $m,n,p,q$ then the set which we get is countable. Thus we get $S$ as the union of four countable sets. So $S$ is countable & so $S$ is discrete. But I am not sure about it..If I am wrong please detect my fallacy and give what happen?

If $S$ is a discrete set then $\mathbb R^2\setminus S$ is path connected. But if NOT then what about the set $\mathbb R^2\setminus S$ ?

Edit : I know that a set $S$ is said to be discrete if it is closed and all points of it are isolated.

Am I correct ? Please explain.

  • $\begingroup$ If by "discrete" you mean that the subspace inherits the discrete topology, then countable does not imply discrete. $\endgroup$ Commented Mar 17, 2015 at 7:13
  • $\begingroup$ But discrete set in an Euclidean space is the set which is atmost countable $\endgroup$
    – Empty
    Commented Mar 17, 2015 at 7:18
  • $\begingroup$ What I'm asking is whether you want to know if the set is countable or has the discrete topology. For example, $\mathbb Q \subset \mathbb R$ is countable but does not inherit the discrete topology. $\endgroup$ Commented Mar 17, 2015 at 7:21

2 Answers 2


No you are wrong...

$S$ is not discrete... $(m,n) \in S$ ( when $p=q=0$) where $m,n \in \mathbb{Z}$, and it is a limit point of a sequence in $S$.

But $S$ is countbale, and so $\mathbb{R^2}-S$ is path connected... you can find a general proof here If $A\subset\mathbb{R^2}$ is countable, is $\mathbb{R^2}\setminus A$ path connected?

  • $\begingroup$ $(m,n)\notin S$ for $m,n \in \mathbb{Z}$ $\endgroup$
    – s.harp
    Commented Jan 1, 2016 at 15:48
  • $\begingroup$ @s.harp consider $p=q=0$ $\endgroup$ Commented Jan 1, 2016 at 15:54
  • $\begingroup$ Funny, sometimes you miss the most obvious things. $\endgroup$
    – s.harp
    Commented Jan 1, 2016 at 16:03
  • $\begingroup$ @ s.harp ) What is the most obvious thing ? $\endgroup$
    – Empty
    Commented Jan 1, 2016 at 16:06
  • $\begingroup$ @S.Panja-1729 He mean to say, he missed out 0 $\endgroup$ Commented Jan 1, 2016 at 16:10

Hint: Try to think of the graph of S near any of its limit points (m,n). It will somewhat look like kitchen sink filter which has more and more holes as you approach towards its centre (a limit point).


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