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In $\mathbb{R}^2$ consider the 2 dimensional positive Lebesgue measure set, $$D=\{re^{it}:t\in[0,2\pi),r\in[0,1]\setminus\mathbb{Q}\} . $$ Does there exist sets $A,B\subset\mathbb{R}$ of 1 dimensional positive Lebesgue measure such that $$A\times B\subset D?$$

Note: The above is not true if $$D=\{(x,y)\in [0,1]\times [0,1]:x-y\notin\mathbb{Q}\}.$$

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No, it there were two such sets $A$, $B$, then with the map $(x,y) \mapsto (x^2, y^2)$ ($t \mapsto t^2$ bi-Lipschitz on compacts inside $(0, \infty)$) , it would transform into $A_1$, $B_1$ of measure non-zero inside $\{(x,y)\ | \ x+y \not \in \mathbb{Q}^2\}$, which is not possible.

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  • $\begingroup$ $r\notin \mathbb{Q}$ doesn't imply $r^2\notin \mathbb{Q}$. Then how we conclude that $D$ maps inside $\{(x,y):x+y\notin \mathbb{Q}\}?$ $\endgroup$ Dec 9 '20 at 14:25
  • $\begingroup$ I guess of I defined $D$ to be $\{re^{it}:r^2\notin \mathbb{Q}\}$ then your argument works perfectly. Am I right? $\endgroup$ Dec 9 '20 at 14:30
  • $\begingroup$ @Oh, I see, it transforms into $x + y \not \in \mathbb{Q}^2$, and again get a contradiction ( only use $\mathbb{Q}^2$ is dense, and nothing else) $\endgroup$
    – orangeskid
    Dec 9 '20 at 21:40
  • $\begingroup$ I am not getting why you need to take $A,B$ away from 0? $\endgroup$ Dec 10 '20 at 2:09
  • $\begingroup$ @Prof.Hijibiji: You want to the image not to be of measure $0$. You guarantee that if the inverse $t\mapsto\sqrt{t}$ is Lipschitz. $\endgroup$
    – orangeskid
    Dec 10 '20 at 2:14
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Let $A$ and $B$ be defined by \begin{align*} A & =\big\{a\in\mathbb{R}: a^2=\sum_{i=1}^\infty a_i10^{-2i}, a_i\in\{1,2\,\dots,9\}, a^2\not\in \mathbb{Q}\big\} \\ B & =\big\{b\in\mathbb{R}: b^2=\sum_{i=1}^\infty b_i10^{-2i+1}, b_i\in\{1,2,\dots,9\}, b^2<\frac12, b^2\not\in \mathbb{Q}\big\}. \end{align*} Note that if $r^2=a^2+b^2\in \mathbb{Q}$ then the decimal expansion of $r^2$ will eventually have a finite length repeating sequence. Any such sequence would have to come from a unique repeating sequence of even place decimals from $a^2$ and repeating odd place decimals from $b^2$. Since $a^2,b^2\not\in \mathbb{Q}$, such repeating sequences do not exist, thus $r^2\not\in\mathbb{Q}$. Not entirely sure about the measure of $A$ and $B$, but for their squares, I think the measures are positive.

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    $\begingroup$ Actually you need to show that $A\times B\cap D$ is positive Lebesgue measure in $\mathbb{R}^2$ which could not be true as we can see from the other answer. Thank you for your participation. Appreciation $\endgroup$ Dec 11 '20 at 9:16

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