If a set $E\subset [a,b]$ has possitive measure, then $x-y\in \mathbb{R\setminus Q}$

If a set $E\subset [a,b]$ has possitive measure, show that they exist $x,y\in E $ s.t $x-y\in \mathbb{R\setminus Q}$

Because $m(E)>0 $ then $\exists ε>0 $ s.t $B(x,ε)\subset E$ if not then $E=\{x_1,x_2,...\}$ and that can't be happening.

Now I can choose $ε'=ε/\sqrt{5}$ such that $d(x,y)=ε'$ and without loss of gennerality i can choose $x,y$ accordingly s.t $x-y>0$. Thus I get $d(x,y)=ε/\sqrt{5} \Rightarrow x-y \in \mathbb{R\setminus Q}$

Is the solution ok ? How can I justify better that $B(x,ε)\in E$?

Solution 1:

That is wrong. Take a fat Cantor set. Its measure is greater than $0$, but it contains no interval $(x-\varepsilon,x+\varepsilon)$.

Take $x\in E$. If $(\forall y\in E):y-x\in\Bbb Q$, then $E\subset x+\Bbb Q$, which is countable. Therefore, $m(E)=0$.