11
$\begingroup$

The background theory here is at least ZFC and probably ZFC+MA+$\neg$CH, if it matters. Here's the question:

Suppose that $C\subseteq\omega_2\times\omega_2$. Must there exist uncountable sets $A,B\subseteq\omega_2$ such that either $A\times B\subseteq C$ or $(A\times B)\cap C=\emptyset$?

Note that the usual issue with coloring ordered pairs, namely coloring them based on whether they agree with underlying ordering, doesn't present itself here since I'm not asking for $A=B$ or $|A|=|B|=\aleph_2$.

My other thought for showing this to be false would be to do a kind of "Bernstein-like" construction, enumerating all pairs $A,B\subseteq\omega_2$ of size $\aleph_1$ and diagonalizing against the products $A\times B$. Assuming MA+$2^{\aleph_0}=\aleph_2$, we'd have that $|\mathcal{P}_{\aleph_1}(\omega_2)|=\aleph_2^{\aleph_1}=2^{\aleph_1}=2^{\aleph_0}=\aleph_2$, but the construction breaks down once $\geq\aleph_1$ many points have been chosen.

If this statement turns out to be true, I'd also be interested in higher dimensional forms (sets of n-tuples, etc).

Edit: After looking around some more, I think what I am really asking is whether the polarized partition relation $\binom{\aleph_2}{\aleph_2}\to\binom{\aleph_1}{\aleph_1}^{1,1}$ (just a restatement of my question above) is consistent with MA+$\neg$CH. Notably, Hajnal proved this relation under GCH.

$\endgroup$
2
  • 3
    $\begingroup$ This sort of question has been studied by various people (including Stevo, Saharon, Justin). For the case you ask, take a look for instance at Stevo's work on colorings, in particular, chapter 9 of his "Walks" book. $\endgroup$ Jul 6 at 3:06
  • 1
    $\begingroup$ The Walks book suggests these things are close to Chang's Conjecture. In fact, in a different paper (Can. J. Math., 1991), Stevo showed that under MA($\aleph_1$), CC is equivalent to $\binom{\aleph_2}{\aleph_2}\to\binom{\aleph_0}{\aleph_0}^{1,1}_\omega$. But maybe this is a red herring, as under GCH, $\binom{\aleph_2}{\aleph_2}\to\binom{\aleph_1}{\aleph_1}^{1,1}$ (partitions with 2 colors) holds, but $\binom{\aleph_2}{\aleph_2}\to\binom{\aleph_1}{\aleph_1}^{1,1}_4$ (partitions with 4 colors) may fail (this is due Prikry). Perhaps I should email one of the people you mentioned. $\endgroup$ Jul 6 at 14:02
2
$\begingroup$

I think Harvey Friedman in "A consistent Fubini-Tonelly theorem for nonmeasurable functions", Illinois Journ. Math., 24(1980), 390--395. proves that if $\aleph_2$ random reals are added to a model of CH, then each $C\subseteq [0,1]\times [0,1]$ has $A,B\subseteq [0,1]$ of outer Lebesgue measure 1 such that either $A\times B\subseteq C$ or else $(A\times B)\cap C=\emptyset$. This implies what you asked.

$\endgroup$
1
  • 1
    $\begingroup$ I looked in the paper but rectangles do not appear there and how would this work with $C=\{(x,y):x\le y\}$? $\endgroup$
    – KP Hart
    Aug 5 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.