If $A$ is dense in $\Bbb Q$, then it must be dense in $\Bbb R$.

Solution 1:

$A$ is dense in $\mathbb{Q}$ if for any two rationals $q_1 < q_2$ there is some $a\in A \cap \mathbb{Q}$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $\mathbb{Q}$ is dense is $\mathbb{R}$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $\mathbb{Q}$ to finish the job.

We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $a\in A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$

Solution 2:

Since $A$ is dense in $\Bbb Q$ so $\overline A \cap \Bbb Q = \Bbb Q \subseteq \overline A.$ So $\Bbb R = \overline {\Bbb Q} \subseteq \overline A \subseteq \Bbb R.$ Therefore $\overline A = \Bbb R.$ This shows that $A$ is dense in $\Bbb R.$

Solution 3:

There is a more general statement of this theorem. Let $X$ be a topological space and suppose $Z \subset Y \subset X$. If $Y$ is dense in $X$ and $Z$ is dense in $Y$ (w.r.t. the subset topology) then $Z$ is dense in $X$.

To prove this, suppose $Z$ is not dense in $X$. Then there exists $A$ open and non-empty in $X$ such that $Z \cap A = \emptyset$. Then we have two cases depending on if $A' = Y \cap A$ is non-empty. If $A'$ is empty then $A \cap Y = \emptyset$ therefore $Y$ is not dense in $X$. Else $A'$ is non-empty and open in $Y$ w.r.t. the subset topology, and $A' \cap Z = \emptyset$. Hence $Z$ is not dense in $Y$. Either way we have a contradiction.

Solution 4:

Does that imply that between any two rational numbers, there exists a real number?

Well, if $a$ and $b$ are distinct rationals, then $(a+b)/2$ is a real number that's between them. It's also rational. And, if you want an irrational number that's between them, take something like $a+(b-a)/\sqrt{2}$.