# Constructing Infinite countable bounded subset of irrationals without accumulation point

I was solving some problems in countabilty,That's when this question arises.

$$\mathbb{R}-\mathbb{Q}$$ is uncountable, I think of a countable subset of irrationals, obviously finite set is countable, I think of a infinite subset, i.e $$\{n\sqrt2:n\in\mathbb{N}\}$$.I thought If we can add condition , that is that set should be bounded. The set that comes to my mind is $$\{x\in(0,1):x\in \mathbb{Q}^c\}$$ but that is not countable. After thinking longtime I found the set $$\{\sqrt2+\frac{1}{n}:n\in \mathbb{N}\}.$$ I thought to add more conditions. In this set I found that lot of elements are very close to $$\sqrt2$$ i.e $$\sqrt2$$ is a accumulation point. That's Why I want a set without accumulation point.

I am so curious to know whether to construct such set! Can someone help with this?

• Closure in WHERE????? – S.D. Nov 8 at 13:31
• Never. Every bounded sequence in $\Bbb R$ has a convergent subsquence. – S.D. Nov 8 at 13:32
• Every bounded infinite subset of $\Bbb R$ has a limit point in $\Bbb R$. – S.D. Nov 8 at 13:34
• The closure of a set includes the set itself, so any non-empty set has non-empty closure. Are you looking for a set with no accumulation points? – Robo300 Nov 8 at 13:35
• If you omit BOUNENDED, then it is possible. Otherwise not. – S.D. Nov 8 at 13:38

Define a relation $$R$$ on $$[0,1]$$ as $$xRy$$ if $$x-y\in \mathbb{Q}$$
Then the equivalance class of $$[1/\sqrt2]$$ is bounded, infinite$$(\{1/\sqrt2+\epsilon/2^n\}\subset [1/\sqrt2]$$ ), countable( since we can map every element to the distance from $$1/\sqrt2$$, i.e every element to maps to a rational number), no accumulation point(Since every class is dense in $$[0,1]$$
• Since every class is dense, every point in $[0,1]$ is an accumulation point. – Andrés E. Caicedo Nov 9 at 16:47