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Is it possible to prove, in first-order ZFC, that the axiom of replacement can't generate a set that includes members going arbitrarily high in the cumulative hierarchy, but without using the axiom of replacement itself for the proof? Or in other words, to prove without using replacement that the kind of mapping used by replacement won't include elements in its output image going arbitrarily high in the hierarchy?

If it's helpful, part of the motivation for this question is the issue raised here [1] arguing that the replacement axiom doesn't follow directly from the iterative conception of sets: "When Replacement has been justified according to the iterative conception, the reasoning has in fact been circular as it was in Zermelo [1930a], with some feature of the cumulative hierarchy picture newly adduced solely for this purpose." In particular, one concern you might have with the axiom of replacement, from the standpoint of the iterative conception of sets, is that the mapping in replacement might map to sets arbitrarily high in the hierarchy, so there might be no valid set formed by the image of the mapping (just a class). Now, if you have full ZFC including replacement, then you can show that the mappings used by replacement can never map arbitrarily high in the hierarchy, but what I'm unclear on is whether this can be shown without relying on the axiom of replacement for any part of the proof (since in the context of the linked paper the goal would be to intuitively justify replacement from the iterative conception, and avoid circularity in this justification).

Here is a related question, but for 2nd order ZFC [2]. Note, the answer to this previous question assumes that there is an inaccessible cardinal in the metatheory, but the current question should rely only on the ZFC axioms.

Edit: remove proof sketch for now, since not sure if it was on the right track.

[1] https://www.jstor.org/stable/41472440

[2] https://math.stackexchange.com/questions/1263848/are-categorical-second-order-axiomatizations-of-set-theory-inconsistent-due-to-t]]

[3] Proof that we can't get $\omega + \omega$ without Replacement

Andy
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    How are you going to prove the existence of a hierarchy without Replacement? – Asaf Karagila Feb 08 '23 at 06:36
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    @AsafKaragila Also, regarding proving the hierarchy's existence, isn't this issue similar with or without replacement, namely that either way we can only prove the existence of some initial fragment of the hierarchy, i.e. it's true that replacement allows you to get much higher, but you still can't use it to show that the hierarchy reaches the inaccessible cardinals, even though most set theorists seem to think the hierarchy goes far beyond that initial fragment. Or am I misunderstanding? – Andy Feb 08 '23 at 18:29
  • I think you are misunderstanding, because I can't make a lot of sense of your argument. Either there are inaccessible cardinals, in which case, they are a set, and then Replacement can "reach them" quite vacuously (although $V_\kappa$ is a model of second-order Replacement when $\kappa$ is inaccessible). Or there are none, in which case the whole thing is vacuous at best. It's also an issue of mixing what seem to be "true" and "provable" in a given universe, which makes it even more... confusing. – Asaf Karagila Feb 09 '23 at 00:48
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    @AsafKaragila I was just saying that ZFC w/ replacement can't be used to prove the existence of inaccessible cardinals (whether or not it's true that they exist as part of the hierarchy, as many set theorists seem to think), i.e. analogous to how much smaller cardinals can't be proven in ZFC w/o replacement (i.e. both axiom systems can prove the existence some initial fragment of the hierarchy). But more importantly, I guess I don't understand what your objection is to my original question. Are you saying the question itself is not well-defined, or just so obvious as to be vacuous? – Andy Feb 09 '23 at 01:13
  • @AsafKaragila Also, I just updated the question text with an attempt at the proof, but my set theory knowledge is quite rusty, so I'm not sure if this is correct. But hopefully it at least clarifies what I'm asking? – Andy Feb 09 '23 at 04:57
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    Your post makes me wonder about something more primitive: Is it possible to prove, in first-order ZFC, that there does not exist any set that includes members going arbitrarily high in the cumulative hierarchy? In other words, is the following statement true: For every set $S$ there exists an ordinal $\alpha$ such that for every $x \in S$, if there exists an ordinal $\beta$ such that $x \in V_\beta$ then $x \in V_\alpha$. – Lee Mosher Feb 09 '23 at 15:07
  • @LeeMosher I think if you had a set going arbitrarily high in the hierarchy (in ZFC), then you could create a mapping to a collection of arbitrarily large ordinals and then show the existence of that set with replacement. And if you have a set with arbitrarily large ordinals, you can form the transitive closure, which would be the set of all ordinals, which is a contradiction (by burali-forti). So I don't think you can have such a set, but I could be misunderstanding something. – Andy Feb 09 '23 at 15:33
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    @LeeMosher Yes, that is provable (and is a standard exercise). For instance, we can prove that every set occurs in some level of the cumulative hierarchy, and then use the fact that no set of rank $\alpha$ has any elements of rank $\ge\alpha$ for any ordinal $\alpha$. (This is a bit of a nuke; a simpler proof is just to consider the set of ranks of elements of a given set, and hit that with Union and apply Burali-Forti.) – Noah Schweber Feb 09 '23 at 17:01

2 Answers2

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No such proof can exist: a counterexample is given by one of the standard examples of a failure of replacement, $V_{\omega + \omega}$, with the function-class $\{(n,\omega+n)\ |\ n \in \omega\}$.

$V_{\omega+\omega}$ is a model of (ZFC–replacement); moreover, it has an exhaustive cumulative hierarchy, i.e. it believes “every set is in some $V_\alpha$”. The given function-class is provably a function-class in (ZF–replacement), and its domain is a set in $V_{\omega+\omega}$, but its image goes arbitrarily high within the cumulative hierarchy there.

Peter LeFanu Lumsdaine
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The replacement axiom is equivalent (over the other axioms of ZF) to the assertion that in every instance, the ranks of witnesses are bounded in the ordinals. Specifically, replacement is equivalent to the following principle:

Bounded witness rank axiom. If $A$ is a set and every element $a\in A$ has a unique witness $b$ for some property $\varphi(a,b,c)$, possibly with some fixed parameter $c$, then the ranks of the witnesses $b$ are bounded.

Certainly the replacement axiom implies the bounded witness rank axiom, since if the can be collected into a set, then the rank of that collecting set will be a bound on the ranks of the witnesses themselves.

Conversely, however, if you have the bounded witness rank axiom, then replacement holds, since if you can bound the ranks of the witnesses, then you have found a collecting set for them. Namely, all witnesses appear in some sufficiently large $V_\alpha$, and then you can pick them out exactly as a subset using the separation axiom.

JDH
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