0
$\begingroup$

Let $ * $ range over cardinal relations $ \{<,<>\}$; if we add the following axiom to $\sf ZF$, would that prove a known form of choice?

Parallelism: $ |x| * |y| \leftrightarrow |\mathcal P(x)| * |\mathcal P(y)| $

Where, $<>$ stands for incomparability; that is, absence of an injection in both directions.

Improve this question
$\endgroup$
3
  • $\begingroup$ I don't understand why you have so many relations with the bidirectional implication and universal quantifier. $=$ and $<$ are enough. $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 1, 2023 at 9:39
  • $\begingroup$ @AsafKaragila, OK, I've edited it. I kept the bidirectional because it's more in spirit with this assertion. $\endgroup$
    – Zuhair Al-Johar
    Commented Feb 1, 2023 at 10:04
  • $\begingroup$ I didn't say to remove the bidirectional implication, I just said that with it are $=$ and $\neq$ are equivalent (and with the universal quantifier, $<$ and $>$ are equivalent). $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 1, 2023 at 10:08

0

Reset to default