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Timeline for Existence of finite powerset

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May 13 at 6:10 comment added Elliot Glazer You could probably get a counterexample to the original question (with the usual foundation axiom) by starting with a ZFA model with a strongly amorphous set of atoms A and taking all the sets hereditarily a quotient of some $A^n,$ and reintroducing Foundation with the tricks in mathoverflow.net/a/314490/109573 or arxiv.org/abs/2312.11902
May 13 at 2:46 history became hot network question
May 12 at 20:14 vote accept Paul Blain Levy
May 12 at 20:02 comment added Hanul Jeon Okay, now I see the point. I misread the question...
May 12 at 19:43 comment added Joel David Hamkins @HanulJeon What was the argument you had in mind with "the usual argument" by $\in$-induction. I don't see it. (And regarding your earlier suggestion about $m<n$, yes, one can prove that every finite set has a power set even without $\in$-induction, but that doesn't help with the question if there are infinite sets, since one seems to need $\omega$ in order to get the set of all finite subsets of the set.)
May 12 at 19:35 comment added Paul Blain Levy The Finite Powerset statement is indeed equivalent to "If there is an infinite set, then $\omega$ exists". That equivalence holds even if both parts of $\in$-induction (i.e. Transitive Containment and Regularity) are excluded. For ($\Rightarrow$), let $X$ be an infinite set, then take the set of cardinalities of all finite subsets of $X$, and this is $\omega$.
May 12 at 19:30 answer added Joel David Hamkins timeline score: 8
May 12 at 19:27 comment added Joel David Hamkins This makes a complete argument then.
May 12 at 19:26 comment added Paul Blain Levy I have edited the question to make clear that $\in$-induction is assumed, not just Regularity. Sorry for not noticing this issue.
May 12 at 19:26 history edited Paul Blain Levy CC BY-SA 4.0
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May 12 at 19:20 comment added Hanul Jeon I see the point. But can we apply the same argument for "every set is finite" even when there is an infinite set? We should do it by proving "For every $m<n$, the power set of $m$ exists" for each finite ordinal $n$.
May 12 at 19:15 comment added Joel David Hamkins I don't how to implement your "usual argument" by $\in$-induction without this. If every set is finite, then induction on the finite ordinals shows every set has a power set. If there is an infinite set, then inductively for every $n$ we get the collection of size $n$ subsets of $X$ as a set, and then if $\omega$ exists by collection we can put them all together. But don't we need $\omega$ to do this? The argument I've sketched does not seem to use $\in$-induction, but it does use existence of $\omega$ if there is an infinite set.
May 12 at 19:13 comment added Hanul Jeon @JoelDavidHamkins I am not sure if I understand your question correctly: Do you mean if I claim/assume $\mathsf{ZFC}$ without Powerset and Infinity proves the implication "If there is an infinite set, then $\omega$ exists?" This theory proves $\omega$ is definable as a proper class, and I have not thought this claim in my comment.
May 12 at 19:12 history edited YCor CC BY-SA 4.0
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May 12 at 19:07 comment added Joel David Hamkins @HanulJeon Are you claiming/using that this theory proves that if there is an infinite set, then $\omega$ exists?
May 12 at 18:59 comment added Hanul Jeon The answer is yes if we assume $\in$-Induction which is stronger than the usual Foundation without Infinity. (The combination of Infinity and Replacement implies the existence of a transitive closure of a set, which is equivalent to $\in$-induction over the remaining axioms.) The proof follows from the usual induction argument. I am not sure what happens without $\in$-Induction.
May 12 at 18:43 history asked Paul Blain Levy CC BY-SA 4.0