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If $D\subset \mathbb{R}\setminus\mathbb{Q}$. How can we find a countable dense subset of $D$?

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To paraphrase @MattE’s comment, what you meant to say was: “Let $D$ be a set of irrational real numbers. How do I find a countable dense subset of $D$?” – Lubin May 13 '14 at 3:08

How about $\left\{q+\sqrt{2}\,\big|\,q\in\mathbb Q\right\}$?

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Not work. Because I need $\left\{q+\sqrt{2}:q\in \mathbb{Q}\right\}\subset D$... – jhon May 13 '14 at 2:53
Dear triple_sec, The OP wants a countable dense subset in an arbitrary subset $D$ of irrational numbers (not just the set of all irrational numbers). Regards, – Matt E May 13 '14 at 3:01
@MattE, after rereading the question, I realize that you have indeed correctly restated the request. And of course this problem is much more difficult! – Lubin May 13 '14 at 3:04
I see your point. In this case, there is no guarantee one can find a countable dense subset of $D$ without further information on the set $D$. For example, nothing specified in the problem excludes the case $D=\{\pi\}\subseteq\mathbb R\setminus\mathbb Q$, which clearly has no countable subset that is dense in $\mathbb R$. – triple_sec May 13 '14 at 3:28

Suppose $\emptyset\ne D\subseteq\mathbb R$. Choose $d\in D$. Let $\langle I_n:n\in\mathbb N\rangle$ be an enumeration of the open intervals with rational endpoints. Use the axiom of choice to get a sequence $\langle s_n:n\in\mathbb N\rangle$ such that $s_n\in D\cap I_n$ whenever $D\cap I_n\ne\emptyset$, otherwise $s_n=d$; and define $S=\{s_n:n\in\mathbb N\}$. Then $S$ is a countable dense subset of $D$.

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It is worth pointing out that this use of the axiom of choice is essential. For instance, it is consistent with the other axioms of set theory (without choice) that there is an infinite Dedekind finite set of reals. But no such set $D$ has a countable subset, much less a dense one. – Andrés E. Caicedo May 13 '14 at 3:20

Let’s see whether I can do this fairly efficiently, without getting egg all over me. Without loss of generality, we may assume that $D\subset[0,1]$. Now, for each $n\ge1$, consider the $2^n$ subintervals $I_{n,j}$ of length $1/2^n$, namely $[j/2^n,(j+1)/2^n]$, with $0\le j<2^n$. For $j,n$ such that $D\cap I_{n,j}\ne\emptyset$, let $x_{n,j}$ be a point of $D$ in this intersection. The set of all such $x_{n,j}$ is certainly countable, and seems to me to be dense in $D$.

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In line with @AndresCaicedo’s comment, notice that I did use choice here, but without acknowledging my debt to it. – Lubin May 13 '14 at 3:27

Fix an irrational number $\alpha$. Denoting by a bar the fractional part of a real number, take the set $\{ m + \overline{n\alpha} \mid m,n\in \mathbf{Z}, n\neq0 \}$.

EDIT: After clarification from Matt, I am revising my answer: it depends on the choice of $D$. Take $D$ to be all positive integral multiples of $\sqrt2$, This consists of irrationals , and is a discrete set.

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But I need a dense countable $S$ for example, with $S\subseteq D$, and $D$ is arbitrary set... – jhon May 13 '14 at 2:58
This set is indexed by a pair of integers, hence countable. – P Vanchinathan May 13 '14 at 3:00
Dear P Vanchinathan, The issue is not the countability, but that the OP needs a countable dense subset of an aribtrary set $D$ of irrational numbers (not just the set of all irrational numbers). Regards, – Matt E May 13 '14 at 3:02
@Matt: See my revised answer. – P Vanchinathan May 13 '14 at 3:08
Dear P Vanchinathan, I'm not sure I understand the point of your edit; the $D$ you describe is countable and dense in itself. Regards, – Matt E May 13 '14 at 3:14

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