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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.

Empty
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  • If by "discrete" you mean that the subspace inherits the discrete topology, then countable does not imply discrete. – manthanomen Mar 17 '15 at 07:13
  • But discrete set in an Euclidean space is the set which is atmost countable – Empty Mar 17 '15 at 07:18
  • 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. – manthanomen Mar 17 '15 at 07:21

2 Answers2

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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?

Community
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Anubhav Mukherjee
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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).

Nitin Uniyal
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