4
$\begingroup$

For any set $A\neq\emptyset$, denote by $[A]^A$ the collection of sets $B\subseteq A$ such that there is a bijection $\varphi:B\to A$. If ${\cal S}\subseteq [A]^A$, we say that $B\in[A]^A$ is monochromatic with respect to ${\cal S}$ if either

  1. ${\cal P}(B)\cap {\cal S} = \emptyset$, or
  2. $\big({\cal P}(B)\cap[A]^A\big) \subseteq {\cal S}$.

Consider the following statement in ${\sf (ZF)}$:

(Non-Mono) If $A$ is an infinite set, then there is ${\cal S}\subseteq [A]^A$ such that no $B\in[A]^A$ is monochromatic with respect to ${\cal S}$.

(Non-Mono) is false for finite sets $A$ because $[A]^A = \{A\}$. On the other hand, (Non-Mono) is a theorem of ${\sf (ZF) + (AC)}$.

Question. In ${\sf (ZF)}$, does (Non-Mono) imply the Axiom of Choice, the Partition Priniple or the weaker Dual Cantor-Bernstein theorem?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ The second question, while understandable, is phrased in a fairly annoying way. You've been told before, that PP and its related principles are all very difficult to tackle and we know very little about them. Knowing that a principle does not imply the axiom of choice, but does imply PP would solve an open problem, similarly with the dual CB theorem. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 2, 2023 at 15:15
  • 1
    $\begingroup$ Apologies - I have rephrased the question. $\endgroup$
    – Dominic van der Zypen
    Commented Aug 2, 2023 at 17:46

0

Reset to default

You must log in to answer this question.