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Discrete sets and uncountability of limit points

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Definition: A subset $D$ of $\mathbb R$ is said to be discrete if for every $x\in D$ there exists $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\cap D=\{x\}$.

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


 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
As an off-the-cuff suggestion, can you take $D$ to consist of the midpoints of the intervals in the construction of the Cantor set?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
As an off-the-cuff suggestion, can you take $D$ to consist of the midpoints of the intervals in the construction of the Cantor set?
Thanks Opalg. I think this is a counterexample.

$D=\displaystyle \bigcup_{m=0}^{\infty} \left( \bigcup_{k=0}^{3^m-1} \left\{\frac{2k+1}{2\cdot 3^m}\right\}\right)$

Let me wrestle with the details though. The first thing is to verify that this actually is a discrete set.