Consider two sets on the plane $A=\mathbb{Q}\times \mathbb{R}$ and $B=\mathbb{R}\times \mathbb{Q}$. We know that $A\cap B=\mathbb{Q}\times \mathbb{Q}\neq\emptyset$. What about the general cases?
That is, will $A\cap B\neq\emptyset$ if $A,B\subset\mathbb{R}^2$ satisfy that
- each vertical fiber $A_y$ of $A$ is dense in $\mathbb{R}\times y$,
- each horizontal fiber $B_x$ of $B$ is dense in $x\times\mathbb{R}$?
What if we replace the assumption by
a. each vertical fiber $A_y$ of $A$ is of positive volume,
b. each horizontal fiber $B_x$ of $B$ is of positive volume?
The only case that I know is if the positive volume assumption is replaced by full vulmue (Fubini theorem).
Thanks!