Dense subset of $Q$

Question

Does there exist any dense subset of the set of rationals $Q$? In fact $Q$ is a (countable) dense subset of $R$ (hence $R$ is separable). If there exist any dense subset of $Q$ then $Q$ is also separable. But the problem is that if we assume $Q$ subset $Q$ then closure of $Q$ is not $Q$ (it is $R$)!

Answer 1

You are wrong. $\mathbb{Q}$ is dense in itself, as every topological space is equal to its own closure.

Answer 2

Pick $N \in \mathbb{N}$. Then $S = \{p/q \in \mathbb{Q} : \gcd(p,q) = 1, q \geq N\}$ is a dense subset of $\mathbb{Q}$. (If we pick $N = 1$, we get the other answers.)

The closure of $S$ in $\mathbb{Q}$ is $\mathbb{Q}$. (That the closure of $S$ in $\mathbb{R}$ is $\mathbb{R}$ does not change that $S$ is dense in $\mathbb{Q}$.)

Answer 3

Closure is something you do to a subset, not to a space (that would be (sequential) completion). So "the closure of $\Bbb Q$" is really an uncomplete phrase since it's missing a reeference to the ambient space.

"The closure of $\Bbb Q$ in $\Bbb R$" is $\Bbb R$, while "the closure of $\Bbb Q$ in $\Bbb Q$" is $\Bbb Q$.