# Disjoint sets of rationals both dense in $\mathbb{R}$

There exists a pair of disjoint subsets of $\mathbb{Q}$ such that both are dense in $\mathbb{R}$. True or false?

The statement is true but how can I find an example of such disjoint subsets of $\mathbb{Q}$?

• What have you thought of? What examples have you tried? – Guido A. Aug 8 '18 at 5:57
• Is$(\frac{m}{2^k})_{m, k}$ dense? – dEmigOd Aug 8 '18 at 5:59
• Are m and k integers? Then this is dense in R. – Mathsaddict Aug 8 '18 at 6:02
• First I took example - A = {m/2^k, m - Z , k - N union 0} and B = { m/3^k, m - z, k - N union 0} – Mathsaddict Aug 8 '18 at 6:08
• You're looking more generally for ways to partition $\mathbb{Q}$ into pieces not based on size (e.g. you don't want $(-\infty,\sqrt{2})\cap\mathbb{Q}$ versus $(\sqrt{2},\infty)\cap\mathbb{Q}$). The two obvious places to look for a distinction to draw are numerator and denominator. For example, we could let $EN$ be the set of rationals which (in lowest terms) have an even numerator. Or, we could let $DD$ be the set of rationals whose denominator (in lowest terms) is a power of $2$ ("dyadic" fractions). Or, so on. (cont'd) – Noah Schweber Aug 8 '18 at 6:08

Let $x \in \mathbb{R}$. Now,

$$0 < 2^nx - [2^nx] < 1$$

and thus

$$0 < x - \frac{[2^nx]}{2^n} < \frac{1}{2^n}$$

This says that $D = \{\frac{m}{2^n}\}_{m \in \mathbb{Z}, n \in \mathbb{N}}$ is dense in $\mathbb{R}$. Now, let's consider a subset of this set,

$$A = \{\frac{m}{2^n} : n \in \mathbb{N}, m \in \mathbb{Z} , 2 \not |n \}$$

If $A$ were dense in $D$, it would be dense in $\mathbb{R}$. For this, we only have to show that $D \setminus A$ can be approximated by elements of $A$. Let $\frac{m}{2^n} \in D \setminus A$ and write $m = 2^ks$ with $2 \not | s$. It can't be that $k \leq n$, because that would mean $\frac{m}{2^n} = \frac{s}{2^{n-k}} \in A$. Therefore, $k \geq n$ and thus $\frac{m}{2^n} = 2^{k-n}s \in \mathbb{Z}$. This means that it is sufficient to show that $A$ can approximate integers: if $a \in \mathbb{Z}$ and $\varepsilon > 0$, then taking $\frac{1}{2^n} < \varepsilon$ we have that

$$\left|a - \frac{2^na + 1}{2^n}\right| = \frac{1}{2^n} < \varepsilon$$

and $\frac{2^na + 1}{2^n} \in A$ because $2 \not | \ 2^na+1$. In conclusion, we have shown that $A$ is dense in $D$ and since $D$ is dense in the reals, so is $A$. With an identical construction (which I encourage you to write out) one can show that the same holds for

$$B = \{\frac{m}{3^n} : n \in \mathbb{N}, m \in \mathbb{Z} , 3 \not |n \}$$

To conclude, then, we just have to observe that $A \cap B = \emptyset$. In effect, if

$$\frac{m}{2^n} = \frac{j}{3^k}$$

for some $n,m,j,k$, then $3^km = 2^nj$ and thus $2 | 2^nj = 3^km$. But since $2$ and $3$ are coprime, this would imply $2 | m$ which is absurd.

HINT.-It is enough to show an example in $I=[0,1]$. All rational is periodic after a finite number of digits $$r=0.a_1a_2....a_m[b_1b_2....b_n]\in I$$ It is not hard to prove that $A$ and $B$ below are disjoint and dense in $I$

$$A=\{r\text{ such that } b_n=1\}\\B=\{r\text{ such that } b_n\ne1\}$$