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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?

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  • $\begingroup$ Closure in WHERE????? $\endgroup$ – S.D. Nov 8 at 13:31
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    $\begingroup$ Never. Every bounded sequence in $\Bbb R$ has a convergent subsquence. $\endgroup$ – S.D. Nov 8 at 13:32
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    $\begingroup$ Every bounded infinite subset of $\Bbb R$ has a limit point in $\Bbb R$. $\endgroup$ – S.D. Nov 8 at 13:34
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    $\begingroup$ 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? $\endgroup$ – Robo300 Nov 8 at 13:35
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    $\begingroup$ If you omit BOUNENDED, then it is possible. Otherwise not. $\endgroup$ – S.D. Nov 8 at 13:38
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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 $[0,1]$

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    $\begingroup$ Since every class is dense, every point in $[0,1]$ is an accumulation point. $\endgroup$ – Andrés E. Caicedo Nov 9 at 16:47

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