11

Say I want to prove the axiom(s) of collection from the axiom(s) of replacement. If you have the axiom of foundation, then you can use Scott's trick to do this.

But suppose I'm working in a context without the axiom of foundation. How can I prove it then? It certainly seems like it ought to be possible using the axiom of choice instead. In particular, if you allow the axiom of global choice, it's quite easy. And if it's possible with global choice, it certainly ought to be possible with ordinary local choice! And yet so far I have not been able to make it work (again, without using the axiom of foundation).

How can you prove collection from replacement, without using foundation, and just using ordinary, local choice?

Thanks all!

Harry Altman
  • 4,652
  • Do you know at all that it's possible? There's a reason why collection is stronger than replacement over weaker variants of ZF. – Asaf Karagila Jan 20 '16 at 08:12

1 Answers1

13

The answer is that you can't.

Instead of omitting foundation, I'll add atoms. You can replace them by Quine atoms and have the same results without foundations to your liking.

We construct the same permutation model as in this answer: start with a proper class of atoms and global choice, and make the class of atoms have only finite subsets while preserving local choice.

Now consider the predicate $\varphi(x,y)$ stating that $x<\omega$ and $y$ is a set of atoms which is equinumerous with $x$. Since $\omega$ is still a set, and we have finite sets of atoms in every finite size, $\forall x\in\omega\exists y(\varphi(x,y))$.

However, if there is some $B$ such that $B$ is a set and for every $x\in\omega$ there is some $y\in B$ such that $\varphi(x,y)$ then $\bigcup B$ is an infinite set of atoms. Contradiction.

Community
  • 1
Asaf Karagila
  • 393,674
  • Well, that's certainly unexpected. Thanks! – Harry Altman Jan 20 '16 at 15:47
  • I agree! I made the comment on the question, and then this solution hit me on the head. Or maybe it's the mild fever from having a cold... :-P – Asaf Karagila Jan 20 '16 at 15:49
  • I have a question on your model: I and Shervin discussed how to prove Collection from ZFC without replacement on Twitter, and Shervin argues that we may use hereditary hierarchy $H_\kappa$ to prove Collection. As far as I checked, his proof has no problem, and it collides with your proof. Thus I think your model could invalidate either choice or replacement. I wonder how do you think about it. – Hanul Jeon Feb 05 '20 at 18:16
  • 2
    @Hanul: In this model the sets which are hereditarily $<\omega$ form a proper class, unless you require them to be atom-free, in which case it means nothing about the whole universe. – Asaf Karagila Feb 05 '20 at 22:41
  • 2
    @Hanul: Also, just as a remark to Shervin, you can define the hereditary hierarchy without choice in a proper way: $H(\kappa)$ is the set of all $x$ such that $\aleph^*(\operatorname{tcl}(x))<\kappa^+$. This is a set since any such $x$ must be in $V_{\kappa^+}$, and indeed every set lies in some $H(\kappa)$. It is no longer continuous, but you can modify this by defining $H(<\kappa)$, and then $H(\kappa)=H(<\kappa^+)$ and the hierarchy is continuous. In either case, this doesn't help with a proper class of atoms. – Asaf Karagila Feb 05 '20 at 22:51
  • @AsafKaragila Thank you for detailed answer. – Hanul Jeon Feb 06 '20 at 05:34
  • @AsafKaragila, thanks for pointing out the flaw. Also I am not familiar with this $\aleph^*$ operator. Is it the same as taking the cardinality of some set? I guess not, since we want to avoid choice, right? – Shervin Sorouri Feb 08 '20 at 09:53
  • 1
    @Shervin: The Lindenbaum number of $x$, denoted by $\aleph^*(x)$, is the least non-zero ordinal that $x$ does not map onto. – Asaf Karagila Feb 08 '20 at 09:54
  • @Asaf, Oh. It seems interesting. May I ask for a reference for studying more on this? I really need a break from this large cardinal and extender business! :) – Shervin Sorouri Feb 08 '20 at 09:59
  • 1
    @Shervin: Start with Jech "Axiom of Choice", I guess? – Asaf Karagila Feb 08 '20 at 11:00
  • @Asaf, I was looking more for an expository paper, type of situation. But that works I guess. Thanks! – Shervin Sorouri Feb 08 '20 at 11:25
  • @Shervin: Exposition to what exactly? A definition? You need context... Try my masters thesis maybe? – Asaf Karagila Feb 08 '20 at 11:27
  • @Asaf, oh, sorry. Now I realized I didn't mention the permutation models part. I was refering to your answer here. – Shervin Sorouri Feb 08 '20 at 11:29
  • @Shervin: Definitely Jech's book. – Asaf Karagila Feb 08 '20 at 11:31
  • @Asaf, Thanks for helping! – Shervin Sorouri Feb 08 '20 at 11:32