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.

• Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many. – user334732 Mar 20 at 15:47
• And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them. – user334732 Mar 20 at 15:48

$$A$$ 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 $$q_1. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $$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

• Isn't it 'for any two $q_1, q_2 \in \mathbb Q$ there exists $a\in A\color{red}{\cap\mathbb Q}$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$) – CiaPan Mar 20 at 15:49
• @CiaPan. I think you are right. This was written hastily. – Mason Mar 20 at 15:51
• :) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part. – CiaPan Mar 20 at 15:59
• I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept. – Mason Mar 20 at 16:01

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 $$\Bbb R.$$

• Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1 – Mason Mar 20 at 15:45
• This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A \subseteq X.$ Then the closure of $A$ in $Y$ say $\overline {A}^Y = \overline A \cap Y,$ where $\overline A$ is the closure of $A$ in $X.$ – Dbchatto67 Mar 20 at 15:49

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

• @RingØ Or $a=b=\mathrm{anything}$. Edited to add the necessary "distinct". Thanks! – David Richerby Mar 20 at 23:18

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.

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.