1

At the very abstract level, the idea here is to try find a kind of choice that is of intermediate strength between set choice and global choice?

What I'm trying to capture is the idea of having a choice function over any set sized family of proper classes, which would of course imply choice over any set sized family of sets, but I hope that it won't imply global choice!?

The base theory is MK - Foundation - Limitation of size + Replacement for sets.

To capture the above notion we'll define the notion of rows of relations as in this posting.

Define (row): $$ z \text{ is a row of }R \iff \\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if every row of it is a proper class.

Axiom of set sized choice over proper classes :$$\forall \ relation \ R \ \text{ with proper class rows } \\ [set (dom(R)) \to \exists f \subset R \ (f: dom(R) \to rng(R))] $$

is that weaker than Global choice?

Zuhair
  • 4,555
  • 2
    What do you mean by "Replacement for sets"? The whole point of MK is that it has Replacement for classes. – Asaf Karagila Feb 04 '20 at 13:04
  • @AsafKaragila, if we replace all elements of a set by sets after a definable function (parameters allowed) then the resulting class is a set. Yes! of course class comprehension of MK ensures that there are such definable replacement classes for every set, but it doesn't guarantee them being sets. Replacement for sets states that those are sets. – Zuhair Feb 04 '20 at 13:45

1 Answers1

2

We do something similar to How can one prove the axiom of collection in ZFC without using the axiom of foundation?, starting with a model of $\sf MK$ without Foundation and with a class of Quine atoms. We may assume that Global Choice holds in this model.

Next build a permutation model where the class of atoms is "set-amorphous" in the sense that every class of atoms is a set or its complement is a set. Easily, Global Choice fails.

Now suppose that you have a relation $R$ whose domain is a set, by the definition of the permutation model, it is fixed when we fix a set of atoms. Choose a large enough set of atoms $a$ fixing $R$, and now for every $x$ in the domain of $R$, if $y$ satisfies $R(x,y)$, then either $y$ is fixed by all the permutations fixing $a$, or its orbits under these permutations also satisfy $R(x,\pi y)$.

But now for every $x$ we can find a slightly larger $a_x$ which fixes also some $y$ such that $R(x,y)$. As the domain of $R$ is a set, there is a set $a$, without loss of generality, might as well assume that it was the firstly chosen set of atoms, containing all of these $a_x$'s, and thus we have that in this $V(a)$ we have enough representatives for choosing from $R$'s classes.

Finally, we can use the fact that $V(a)$ does satisfy a weak form of Foundation (that is by starting the von Neumann hierarchy with $a$ rather than $\varnothing$ we can generate $V(a)$) to ensure there is a Collection of images, and thus choice for sets is enough to finish the proof.

Note that this is an argument to show that Collection is weaker than Global Choice, as one would expect.

Asaf Karagila
  • 393,674
  • should I infer from this that collection + set choice is equal to the here mentioned choice axiom, over the base theory mention in this posting? – Zuhair Feb 04 '20 at 19:59
  • Probably. I'll get back to you in a while. My brain is on a roll with some other stuff right now. – Asaf Karagila Feb 04 '20 at 20:10
  • Yes, this equivalence is trivial. Leave me a comment if you don't see it. – Asaf Karagila Feb 05 '20 at 08:32
  • @Yes, I can see it! that choice here implies collection and set choice is clear. For the other direction suppose we have a set sized family of proper classes, then that can be readily captured by a relation with those proper classes being its rows and whose domain is a set, now we pair each element in its domain with the chosen element from the intersection of the collection set by the row indexed by that element in the domain, thus proving the form of choice here! – Zuhair Feb 05 '20 at 18:54