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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.

$A$ is dense in $mathbb{Q}$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbb{Q}$ 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 $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$

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 $Bbb R.$

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 $a+(b-a)/sqrt{2}$.

Another proof. $A$ being dense in $mathbb{Q}$ means that for any $qinmathbb{Q}$ there is a sequence in $A$ converging to $q$.

Let $rinmathbb{R}$. 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 $ige k_n$ 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 $nge N$ we have
$$|r-q_n|<frac{epsilon}{2}qquadtext{and}qquad2^{-N}<frac{epsilon}{2} .$$
Then for each $nge N$ we have
$$|r-a_{n,k_n}|le|r-q_n|+|q_n-a_{n,k_n}|<epsilon ,$$
consluding the proof.

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.

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 $X$.