Ggtkuk

I have $A$ is a subset of $mathbbR$. If $A$ is dense in $mathbbQ$, then it must be dense in $mathbbR$. I am confused because $A$ is dense in $mathbbQ$. 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 $mathbbQ$ prove that it must be dense in the reals? Any help is appreciated.

$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ 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 $mathbbQ$ is dense is $mathbbR$. 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 $mathbbQ$ 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 $a+(b-a)/sqrt2$.

Another proof. $A$ being dense in $mathbbQ$ means that for any $qinmathbbQ$ there is a sequence in $A$ converging to $q$.

Let $rinmathbbR$. Since $mathbbQ$ is dense in $mathbbR$, 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
$$|r-q_n|<fracepsilon2qquadtextandqquad2^-N<fracepsilon2 .$$
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 $X$.

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.