1

Working in bi-sorted $L_{\omega_1,\omega} (=, \in)$, if we write $\sf ZFC + Classes$ as it is; i.e., in bi-sorted $L_{\omega, \omega} (=,\in)$, and add the following definability rule written in bi-sorted $L_{\omega_1,\omega} (=, \in)$ :

Definability rule: if $\phi_1,\phi_2,\phi_3,...$ are all formulas in bi-sorted $L_{\omega,\omega}(=,\in)$, in which only symbol "$y$" occurs free, and it never occurs bound, then: $$\forall X: \bigvee_{i \in \mathbb N} X=\{y \mid \phi_i\}$$ This would ensure that all classes are pointwise definable in bi-sorted $L_{\omega,\omega}(=,\in)$

Would "$\sf ZFC + Classes + Definability \ rule$", manage to prove that all classes are countable?

Zuhair Al-Johar
  • 9,152

1 Answers1

2

Since I don't have a good understanding of your Classes axiom/theory, let me answer instead for Gödel-Bernays set theory GBC, for which the answer is negative.

We know that there are pointwise definable models of GBC, and every such model will satisfy your definability axiom, since every class there is definable. But none of these models think that every set or class is countable.

The main explanation is that merely knowing that every class is definable is not sufficient to build the definability map $$\text{class }X\quad\to\quad\text{defining formula }\phi.$$ It is this map and not the pointwise definability itself that leads to the conclusion that there are only countably many classes.

We discuss this issue at length in our paper:

But I am unsure how much of this analysis applies to your theory Classes.

Martin Sleziak
  • 4,608
Joel David Hamkins
  • 224,022
  • 1
    Both MK and GBC are not compatible with all classes being countable because $V$ is provably uncountable in them! Actually $V$ is provably inaccessible. This is a consequence of the axiom of limitation of size. And even its weaker version stating that the range of every class partial function from a set is a set is still powerful enough to prove that $V$ is uncountable. With ZFC + Classes we don't have that. – Zuhair Al-Johar Dec 27 '22 at 00:50
  • 1
    Yes, I know all that. My point was that the definability property doesn't in general imply countability. – Joel David Hamkins Dec 27 '22 at 01:06
  • 1
    Ah! I see, Nice! – Zuhair Al-Johar Dec 27 '22 at 01:08
  • 1
    I thought the stronger logic background used here may make such a proof possible? – Zuhair Al-Johar Dec 27 '22 at 01:12