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

  1. each vertical fiber $A_y$ of $A$ is dense in $\mathbb{R}\times y$,
  2. 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).



A simple way to construct sets with the required properties is to first pick $U,V\subseteq\mathbb{R}$ and set $$ \begin{align} &A = \left\{ (x,y)\in\mathbb{R}^2\colon x+y\in U\right\},\\ &B = \left\{ (x,y)\in\mathbb{R}^2\colon x+y\in V\right\}. \end{align} $$ Then the vertical fibres $A_y$ are all translates of $U$ and the horizontal fibres $B_x$ are translates of $V$ (and, similarly, for the horizontal fibres of $A$ and vertical fibres of $B$). Choosing $U,V$ disjoint, then $A,B$ are also disjoint. You can take $U=\mathbb{Q}$ and $V=\mathbb{R}\setminus\mathbb{Q}$ for the first example and $U=(0,1)$, $V=(1,2)$ for the second. Or, if you prefer, take $U=\left((0,1)\setminus\mathbb{Q}\right)\cup\left(\mathbb{Q}\setminus(0,1)\right)$ and $V=\mathbb{R}\setminus U$ to give a simultaneous counterexample to both.

If you want to venture into non-measurable sets, which requires the axiom of choice, then taking $U$ to be any subset of $\mathbb{R}$ with full outer measure and zero inner measure (e.g., a Vitali set) and $V=\mathbb{R}\setminus U$ then $A$ and $B$ (as a subset of $\mathbb{R}^2$) together with their horizontal and vertical fibres (as subsets of $\mathbb{R}$) will have full outer measure, but have zero inner measure. In fact, again using the axiom of choice, its possible to find uncountably many pairwise disjoint sets whose horizontal and vertical fibres all have full outer measure.


For your second question, the answer is easily "no." Let $A = \{(x,y)\ |\ x < y\}$ and let $B$ be its complement. Every $A_x,\ A^y,\ B_x,\ B^y$ has infinite length, but $A \cap B = \emptyset$.


Corrected in light of Andres' and Jacob's comments: Let us define $\mathbb{Q}_{odd}$ as the subset of $\mathbb{Q}$ that has odd denominator when expressed in lowest terms and $\mathbb{Q}_{even}$ similarly with even denominator. If you let $A=[\mathbb{Q}_{odd}\times (\mathbb{R} \setminus \mathbb{Q}_{odd})] \cup [\mathbb{Q}_{even}\times (\mathbb{R} \setminus \mathbb{Q}_{even})]$ and $B=[(\mathbb{R} \setminus \mathbb{Q}_{even})\times \mathbb{Q}_{odd}] \cup [\mathbb{R} \setminus \mathbb{Q}_{odd})\times \mathbb{Q}_{even}]$ they have no intersection.

Added: For the second part, if we don't still need dense, we can use a checkerboard: $A=[2m,2m+1)\times[2n,2n+1)\cup[2m+1,2m+2)\times[2n+1,2n+2), m,n \in \mathbb{Z}$

$ B=[2m,2m+1)\times[2n+1,2n+2)\cup[2m+1,2m+2)\times[2n,2n+1), m,n \in \mathbb{Z}$

Added yet again for the second part: Let me work in $(0,1)\times(0,1)$ as we can then biject to $\mathbb{R}^2$. Following Andres' comment, how about $C=[\mathbb{Q}\cap(0,\frac{1}{2})\cup[\mathbb{R}\setminus \mathbb{Q} \cap (\frac{1}{2},1)], D=(0,1)\setminus C$, then $A=(C\times C) \cup (D \times D), B=(C\times D) \cup (D \times C)$. Both are dense and of positive measure.

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    $\begingroup$ Ross, many sections $A_y$ and $B_x$ are empty, so your example does not satisfy the requirements. $\endgroup$ – Andrés E. Caicedo Apr 3 '11 at 15:33
  • $\begingroup$ @Andres: see if the new version works. Thanks. $\endgroup$ – Ross Millikan Apr 3 '11 at 15:55
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    $\begingroup$ Wouldn't $A_1$ and $B_2$ have non-empty intersection? Where $A_1$ refers to the firm term in the union. More specifically I think $(1/3,1/2)$ is in both sets. $\endgroup$ – JSchlather Apr 3 '11 at 16:18
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    $\begingroup$ After a little more thought $\mathbb Q_{odd} \times \mathbb Q_{even} \subset \mathbb Q_{odd} \times (\mathbb R \setminus \mathbb Q_{odd})$ but we also have that $\mathbb Q_{odd} \times \mathbb Q_{even} \subset (\mathbb R \setminus \mathbb Q_{even}) \times \mathbb Q_{even}$. A similar argument works for the other two so $A \cap B= \mathbb Q_{even} \times \mathbb Q_{odd} \cup \mathbb Q_{odd} \times \mathbb Q_{even}$. Which I suppose is slightly better than $\mathbb Q \times \mathbb Q$. $\endgroup$ – JSchlather Apr 3 '11 at 16:37
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    $\begingroup$ Yes, that's fine. You could just take $$A=({\mathbb Q}\times{\mathbb Q}^c)\cup ({\mathbb Q}^c\times {\mathbb Q})$$ and $$B=({\mathbb Q}\times{\mathbb Q})\cup({\mathbb Q}^c\times{\mathbb Q}^c).$$ $\endgroup$ – Andrés E. Caicedo Apr 3 '11 at 16:58

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