If $A$ is dense in $\Bbb Q$, then it must be dense in $\Bbb R$. I have $A$ is a subset of $\mathbb{R}$. If $A$ is dense in $\mathbb{Q}$, then it must be dense in $\mathbb{R}$. I am confused because $A$ is dense in $\mathbb{Q}$. Does that imply that between any two rational numbers, there exists a real number? I understand for anything to be dense in $\Bbb R$, there must exist something that lies between any two real numbers. However, how does knowing something is dense in $\mathbb{Q}$ prove that it must be dense in the reals? Any help is appreciated.
 A: $A$ 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: $$r_1<q_1<a<q_2<r_2$$
A: 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 $\Bbb R.$
A: 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.
A: 
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 $a+(b-a)/\sqrt{2}$.
A: Another proof. $A$ being dense in $\mathbb{Q}$ means that for any $q\in\mathbb{Q}$ there is a sequence in $A$ converging to $q$.
Let $r\in\mathbb{R}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there is a sequence $\{q_n\}_{n=1}^\infty$ converging to $r$. For each $n$, pick a sequence $\{a_{n,i}\}_{i=1}^\infty$ in $A$ converging to $q_n$. For each $n$, choose $k_n$ such that for all $i\ge k_n$ we have
$$|a_{n,i}-q_n|<2^{-n}.$$

Claim: The sequence $\{a_{n,k_n}\}_{n=1}^\infty$ converges to $r$.

Proof: Let $\epsilon>0$. Choose $N$ such that for each $n\ge N$ we have
$$|r-q_n|<\frac{\epsilon}{2}\qquad\text{and}\qquad2^{-N}<\frac{\epsilon}{2}\ .$$
Then for each $n\ge N$ we have
$$|r-a_{n,k_n}|\le|r-q_n|+|q_n-a_{n,k_n}|<\epsilon\ ,$$
consluding the proof.
A: Another definition of "dense" is that every open neighborhood of $\mathbb Q$ has a member of $A$. And using that definition, we want to prove that every open neighborhood of $\mathbb R$ has a member of $A$. So suppose we take a neighborhood $N_1$ of $r$ in $\mathbb R$. Since $\mathbb Q$ is dense in $\mathbb R$, there is rational $q$ in $N_1$. We can take a neighborhood $N_2$ of $q$ that is a subset of $N_1$, and there will be $a$ in that neighborhood, and thus $a$ will be in $N_1$ as well.
Basically, $A$ being dense in $\mathbb Q$ means that for every $q$, there is $a$ "close" to $q$, and $\mathbb Q$ being dense in $\mathbb R$ means that for every $r$, there is $q$ "close" to $r$. So given any $r$, we take $q$ "close" to $r$, then we take $a$ "close" to $q$, and $a$ is "close" to $r$. 
It's analogous to "If everyone lives close to a school, and every school is close to library, then everyone lives close to a library.", although there's some additional rigor regarding the term "close" that has to be introduced.
