If a set E ⊂ [ a , b ] has possitive measure, show that they exist x , y ∈ E s.t x − y ∈ R ∖ Q Because m ( E ) &gt; 0 then ∃ ε &gt; 0 s.t B ( x , ε ) ⊂ E if not then E = { x 1 , x 2 , . . . } and that can't be happening.Now I can choose ε ′ = ε / 5 such that d ( x , y ) = ε ′ and without loss of gennerality i can choose x , y accordingly s.t x − y &gt; 0. Thus I get d ( x , y ) = ε / 5 ⇒ x − y ∈ R ∖ Q Is the solution ok ? How can I justify better that B ( x , ε ) ∈ E?

Question

If a set   E  ⊂  [  a  ,  b  ] has possitive measure, show that they exist   x  ,  y  ∈  E s.t   x  −  y  ∈      R    ∖    Q  Because   m  (  E  )  &amp;gt;  0 then   ∃      ε    &amp;gt;  0 s.t   B  (  x  ,      ε    )  ⊂  E if not then   E  =  {      x    1    ,      x    2    ,  .  .  .  } and that can&#039;t be happening.Now I can choose             ε        ′    =      ε        /        5   such that   d  (  x  ,  y  )  =            ε        ′   and without loss of gennerality i can choose   x  ,  y accordingly s.t   x  −  y  &amp;gt;  0. Thus I get   d  (  x  ,  y  )  =      ε        /        5    ⇒  x  −  y  ∈      R    ∖    Q  Is the solution ok ? How can I justify better that   B  (  x  ,      ε    )  ∈  E?

Kamora Greer · Accepted Answer

That is wrong. Take a fat Cantor set. Its measure is greater than 0, but it contains no interval   (  x  −  ε  ,  x  +  ε  ).Take   x  ∈  E. If   (  ∀  y  ∈  E  )  :  y  −  x  ∈      Q  , then   E  ⊂  x  +      Q  , which is countable. Therefore,   m  (  E  )  =  0.

If a set E ⊂ [ a , b ] has possitive measure, show that the

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