Does there exist a set $M \subset \mathbb{R}^2$ which has the following two properties:

  • Forall $x \in \mathbb{R}$ the set $\{y \in \mathbb{R} \mid (x,y) \in M\}$ is countable.
  • Forall $y \in \mathbb{R}$ the set $\{x \in \mathbb{R} \mid (x,y) \notin M\}$ is countable.
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    $\begingroup$ If the continuum hypothesis holds, then yes: let $M=< ^{-1}$, where $< $ is a well ordering of $\mathbb R$ of type $\omega_1$. $\endgroup$ – Emil Jeřábek Dec 7 '11 at 14:21

It is a theorem of Sierpinski (Sur un theoreme equivalent a l'hypothese du continu) that the existence of such a set is equivalent to the continuum hypothesis.


Here's an easy proof of the equivalence of the statement to CH (I think). One direction is just what Emil said. For the other direction:

Suppose continuum is $\geq \aleph_2$. Restrict attention to an $\aleph_2$-sized subset of $\mathbb{R}$, and suppose we had a relation with the properties you want on that subset. (Note: if there is a relation with those properties on $\mathbb{R}^2$ then it will retain those properties when we restrict to a smaller set.) Take $\aleph_1$ many $x$-coordinates from this set; each of them only has countably many $y$'s that it gets paired with in the relation, so in total there are at most $\aleph_1$ many $y$'s that get paired with any of these $x$'s. So, take some $y$ which doesn't get paired with any of these $x$'s. (There will be such a $y$ because we have $\aleph_2$ many $y$'s in total.) This $y$ has at least $\aleph_1$ many $x$'s that it does not get paired with; and this contradicts the properties of the relation.

Ramiro, I'm guessing this is the same argument Sierpinski gave?


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