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We know that $\Bbb R$ is separable i.e. it contains a dense subset which is countable. We have $\Bbb Q$ and ${\Bbb R} - {\Bbb Q}$ to be dense subsets respectively countable and uncountable. I was looking for a countable dense subset of $\Bbb R$ which is a proper subset of either (i) $\Bbb Q$ or (ii) ${\Bbb R} - {\Bbb Q}$ .

For (i), by considering the set of dyadic rationals i.e. $\{\frac{a}{2^b} | a \in \Bbb Z , b \in \Bbb N \}$ or more generally for any fixed prime $p \in \Bbb N$, consider, $\{\frac{a}{p^b} | a \in \Bbb Z , b \in \Bbb N \}$ . It is a countable proper subset of $\Bbb Q$ which is dense.

But I could not come up with any example for (ii) . Thanks in advance for help.

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    $\begingroup$ I think that the algebraic irrationals are countable and dense $\endgroup$ – Masacroso Dec 29 '17 at 10:26
  • $\begingroup$ For $(i)$ you can also delete a single point from $\mathbb Q$ or even remove the whole of $\mathbb Z$. For $(ii)$ you have a countable dense set already - how can you modify that to satisfy the other condition you need? $\endgroup$ – Mark Bennet Dec 29 '17 at 10:27
  • $\begingroup$ @Masacroso The examples in answers below and comments are true. But your choice looks extremely interesting. Would you kindly prove that the set of algebraic irrationals is dense in $\Bbb R$ $\endgroup$ – Utsav Dewan Dec 29 '17 at 10:37
  • $\begingroup$ @ThatIs: Paolo Leonetti's example is dense, and is a subset of the algebraic irrationals. So is my answer, provided $b$ is an algebraic irrational (e.g., $b=\sqrt{2}$). Thus, the set of algebraic irrationals has a dense subset, so it must be dense. $\endgroup$ – quasi Dec 29 '17 at 10:42
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Choose any rational number $a$. Then $\mathbb{Q}-\{a\}$ is a proper, dense subset of $\mathbb{Q}$.

Choose any irrational number $b$. Then $\{a+b \mid a \in \mathbb{Q}\}$ is a countable, dense subset of $\mathbb{R}-\mathbb{Q}$.

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  • $\begingroup$ Yes, absolutely! I studied a set which I missed by chance, if $t$ is a fixed irrational number then, $\{a+bt : a ,b \in \Bbb Z, b \neq 0 \}$ fits the bill right? $\endgroup$ – Utsav Dewan Dec 29 '17 at 10:31
  • $\begingroup$ No, $b$ has to be different from $0$.. $\endgroup$ – Paolo Leonetti Dec 29 '17 at 10:32
  • $\begingroup$ @PaoloLeonetti thanks. $\endgroup$ – Utsav Dewan Dec 29 '17 at 10:36
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$$ \left\{q\sqrt{2}: q\neq 0 \text{ rational }\right\}. $$

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You may also pick a countable dense set whose elements are all trascendental (and in such a way that this set is not the set of all rational multiples of some fixed transcendental, which would be the right boring answer). As algebraic numbers are countable, the set $T$ of all transcendentals is dense in $\mathbb{R}$. "Having a countable base" is a hereditary topological property, so the space $T$ of transcendentals has a countable base. Spaces with a countable base are separable, that is, have a countable dense subset (just pick one point in each basic open set). This means that $T$ contains a countable dense subset $D$. As $D$ is dense in $T$ and $T$ is dense in $\mathbb{R}$, $D$ is dense in $\mathbb{R}$. So, $D$ is a countable dense subset of the real line whose elements are all transcendental. From time to time I pose such question to my students...

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