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- Mar 10, 2012

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**A subset $D$ of $\mathbb R$ is said to be**

*Definition:***discrete**if for every $x\in D$ there exists $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\cap D=\{x\}$.

__Does there exist a discrete subset $D$ of $\mathbb R$ such that the set of limit points of $D$ is an uncountable set.__

**Question:**___

**Progress:**We claim that the set of limit points, $D'$, of a discrete set $D$ has to be nowhere dense.

To prove the above claim we just need to show that:

1. D' is a closed set (This doesn't even require $D$ to be discrete).

2. $\text{Int}(D')$ is empty.

(1) is obvious.

To show (2) assume the contradictory. Let $(a,b)\subseteq D'$. Now let $x\in (a,b)\cap D$ (such an $x$ has to exist). Now clearly since $D$ is discrete, $x\notin D'$. Also, using the property of discrete sets, we know that there is an open set $O$ which contains $x$ and satisfies $O\cap D=\emptyset$. Thus $O\cap D'=\emptyset$. This means $O\cap (a,b)=\emptyset$ and we get our contradiction.

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So if we are construct a counterexample we should be aiming towards constructing a nowhere dense set of limit points of a discrete set.