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Let ${\sf ZFCU}$ be the axioms of ${\sf ZFC}$ modified to allow for urelements in the usual way. We do not assume that the urelements form a set.

I want to know whether ${\sf ZFCU}$ plus the axiom schema of collection---i.e.:

(Collection) $\forall x\exists y \phi(x, y) \to \forall z\exists w\forall x\in z\exists y\in w \phi(x, y)$

implies schematic dependent choice---i.e.:

(SDC) $\forall x\exists y\phi(x, y) \to \forall z\exists f(f(0) = z \wedge \forall n\phi(f(n), f(n+1)))$

  • Maybe I'm missing something, but isn't collection already part of $\mathsf{ZFCU}$, and for that matter SDC already a consequence of choice in $\mathsf{ZFU}$? – Noah Schweber Mar 05 '21 at 16:10
  • Didn't you ask this? https://math.stackexchange.com/questions/1337583/does-the-principle-of-schematic-dependent-choice-follow-from-zfcu?rq=1 – Asaf Karagila Mar 05 '21 at 16:16
  • @NoahSchweber See the question Asaf linked! –  Mar 05 '21 at 16:19
  • @AsafKaragila Yes. Why? Unless I'm missing something (totally possible), the models you gave there don't model collection. –  Mar 05 '21 at 16:20
  • https://math.stackexchange.com/questions/3533971/is-set-sized-choice-over-proper-classes-weaker-than-global-choice works? – Asaf Karagila Mar 05 '21 at 16:26
  • @AsafKaragila I don't see it yet (I had a somewhat similar idea but couldn't make it work). In the model you given global choice fails, but why will SDC fail? –  Mar 05 '21 at 16:35
  • (Ah, my previous comment is correct, so I re-state it) $\mathsf{SDC}$ is equivalent to the following statement: for every formula $\phi(x,y)$, if $\forall x\exists y \phi(x,y)$ then there is a set $A$ such that $\forall x\in A\exists y\in A \phi(x,y)$. The proof is almost identical to that of Theorem 2.4 of Aczel's The Relation Reflection Scheme. (But we need some modification for proving the above statement from $\mathsf{SDC}$.) – Hanul Jeon Mar 05 '21 at 16:51
  • @HanulJeon Then I'll repost my reply! The equivalence is right, of course (given collection). But I don't see immediately how that helps. –  Mar 05 '21 at 17:01
  • @GME That's embarrassing - I posted an answer to that question too! :P – Noah Schweber Mar 05 '21 at 17:16
  • @NoahSchweber haha no worries! I've done worse; easily. –  Mar 05 '21 at 17:18
  • I think your $\mathsf{SDC}$ also implies the reflection principle (at least a partial form of it.) I think the results in Section 4 of Blechschmidt and Swan works. (But I have not checked that how can reduce some $\mathsf{RRS}_2$ kinds of stuff to $\mathsf{SDC}$.) – Hanul Jeon Mar 07 '21 at 12:27
  • (Or I may go too far: Lemma 11.1 of Gitman-Friedman-Kanovei would work.) – Hanul Jeon Mar 07 '21 at 12:29
  • Indeed, it does! In fact, I‘m interested in SDC /because/ it implies a particular reflection principle. What I want to know is whether collection could be used instead. (The reflection principle implies SDC). But again, I don‘t see how that helps. –  Mar 07 '21 at 20:08

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