1
$\begingroup$

$\DeclareMathOperator{\Ord}{Ord}$Let $c: \Ord^{[2]} \rightarrow 2$ be a $2$-coloring of the class of pairs of ordinals.

Is there a definable class-sized subclass $H$ of $\Ord$ which is $c$-homogeneous, that is, such that $c(H^{[2]}) = \{0\}$ or $c(H^{[2]}) = \{1\}$?

$\endgroup$
9
  • $\begingroup$ Define "class" in this context? $\endgroup$
    – Asaf Karagila ♦
    Jul 12, 2017 at 20:03
  • $\begingroup$ I assume the coloring is also definable, right? $\endgroup$
    – Asaf Karagila ♦
    Jul 12, 2017 at 20:07
  • $\begingroup$ Not necessarly. But I'm also interested in the case when $c$ is definable. $\endgroup$
    – nombre
    Jul 12, 2017 at 20:09
  • $\begingroup$ I'm fairly sure that you can arrange a coloring which diagonalizes over all the definable classes to get a counterexample in that case. But I'm not sure, maybe I'm talking out of my fanny. $\endgroup$
    – Asaf Karagila ♦
    Jul 12, 2017 at 20:10
  • 1
    $\begingroup$ A paper by Enayat and Hamkins would be relevant. $\endgroup$
    – Hanul Jeon
    Nov 4, 2020 at 17:16

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .