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Let $S \subset R^2$ be defined by

$S$ = {$(m+ \frac{1}{4^{|p|}} , n+ \frac{1}{4^{|q|}}): m,n,p,q \in Z$}

Then,

  1. $S$ is discrete in $R^2$.

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

  3. $S^c$ is connected but not path connected.

  4. $S^c$ is path connected.

This question appeared in CSIR Dec 2015.

I don't have any idea about how to do this question. Please help!

Thanks in advance!

Shivani Goel
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1 Answers1

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Hint/Solution:

$(1).$ Note that $(m,n)$ is in $S$ and is also limit point of $S$ hence $S$ is not discrete.

$(2).$ Note that elements of the form $(m+ \frac{1}{4^p},n)$ are also limit points of $S$

$(3,4.)$ If A is countable then $\mathbb R^2$-$A$ is path connected

Community
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Arpit Kansal
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  • Can you please explain hint for 2 a little more? I am not good at finding limit points. – Shivani Goel May 31 '16 at 19:07
  • Dear @ShivaniGoel Recall that a limit point of a subset $S$ of a topological space $X$ is a point that can be "approximated" by points of $S$.Since we are working in metric space therefore a point $x \in X$ is a limit point iff there is a sequence of elements of $S$ that goes to $x$.So can you find a sequence in $S$ that converges to $(m+ \frac{1}{2^p},n)$?(Hint;For fixed $m,n,p$ consider the sequence $(m+ \frac{1}{2^p},n+\frac {1}{2^q})$ where does this sequence when $q \to \infty) $ – Arpit Kansal May 31 '16 at 20:18
  • I was thinking in the same way but I got confused when i saw that if we replace S by the set $S$ = {$(m+ \frac{1}{2^{|p|}} , n+ \frac{1}{2^{|q|}}): m,n,p,q \in Z$} , then the set given in 2) is the set of limit points of new set S. Can you explain this also? – Shivani Goel May 31 '16 at 20:29
  • Yes I am asking for all the limit points of the set S. – Shivani Goel May 31 '16 at 20:47
  • @ShivaniGoel $S$ given in problem or $S$ in your comment? – Arpit Kansal May 31 '16 at 20:48
  • S in my comment. – Shivani Goel May 31 '16 at 20:49
  • Note that instead of $4$ in your question you can use any non negative integer greater than $1$.I think all the limit points are $(m,n)$,$(m+ \frac{1}{2^p},n)$ and $(m,n+\frac{1}{2^q})$ – Arpit Kansal May 31 '16 at 20:52
  • Let us continue this discussion in chat. – Shivani Goel May 31 '16 at 20:54