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.

Improve this question
$\endgroup$
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. $\endgroup$
    – Andrés E. Caicedo
    Jul 6, 2021 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. $\endgroup$
    – Iian Smythe
    Jul 6, 2021 at 14:02

1 Answer 1

Reset to default
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.

Improve this answer
$\endgroup$
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, 2021 at 15:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .