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Randall Holmes has made a quite convincing argument against the fact that the full axiom schema of replacement should be considered as “intuitively obvious”—even though he does believe ZFC to be probably consistent. Instead, he claims that only $\Sigma_2$ replacement axioms should be considered as obvious.

As far as I understand it, the rationale behind believing in the full axiom schema of replacement is the belief that any collection of sets (whether it is first-order definable or not) the same size of a set should be a set. We will call this intuitive belief “limitation of size” (LOS)—but note that it is not an axiom strictly speaking, because we are thinking in Platonist, not formal, terms.

At this point, it is important to notice that, similarly, the rationale behind believing in the axiom schema of separation is that we believe that all the “subsets” (in the intuitive sense) of a set are sets too! I will call this belief the “Platonist subset” (PS) “axiom”.

Now, if we believe in a Platonist world of sets in which all the axioms of ZFC are satisfied, with separation replaced by PS and replacement replaced (!) by LOS, the collection of all the ordinals of this world has the structure of a inaccessible (in the Platonist sense) cardinal. For that reason, I think that anyone who believes in the consistency of ZFC because they believe in the Platonist truth of LOS should also believe in the consistency of (ZFC + inaccessible cardinal).

But one might also believe in the existence of an inaccessible cardinal without believing in LOS! My question is, would that belief (together with the Platonist belief in PS and in $\Sigma_2$ replacement) imply that you believe in the consistency of ZFC? Well, this question is actually ill-stated; but it is essentially equivalent to the perfectly precise following question, which is the point of my post:

Is (ZC + $\Sigma_2$ replacement + inaccessible cardinal) equiconsistent with (ZFC + inaccessible cardinal)?

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Let $\kappa$ be an inaccessible cardinal. By reflection principle, we can find $\lambda>\kappa$ such that $V_\lambda$ is a model of $\mathsf{ZC}$ with $\Sigma_2$-replacement: since the truth relation for $\Sigma_2$ formulas are definable on $\mathsf{ZFC}$, we can postulate $\Sigma_2$-replacement as a single formula.

Note that we only need to reflect $\Sigma_2$-replacement and the fact that $\mathrm{Ord}$ does not have a maximum. Then $\lambda$ will be an limit ordinal so $V_\lambda\models \mathsf{ZC}$. Furthermore, $\kappa\in V_\lambda$ and $\kappa$ is still an inaccessible in $V_\lambda$. Hence $\mathsf{ZFC}$ with an inaccessible cardinal proves there is a transitive model of $\mathsf{ZC}$ + $\Sigma_2$-replacement with an inaccessible.

By the same argument, we can see that $\mathsf{ZFC}$ with a set-many inaccessibles proves the consistency of $\mathsf{ZC}$ + $\Sigma_2$-replacement with a set-many inaccessibles.


The result is not different even if we assume there are proper class many inaccessibles. We can reflect $\Sigma_2$-replacement with the statement "the class of all inaccessibles is unbounded" into some $V_\lambda$. By the same argument, $V_\lambda$ is a model of ZC with $\Sigma_2$-replacement. Moreover, $V_\lambda$ thinks the class of all inaccessibles is unbounded, so is a proper class.

Note that $\lambda$ must be singular: the existence of such regular $\lambda$ implies the existence of an 1-inaccessible, which implies the consistency of $\mathsf{ZFC}$ + "there are proper class many inaccessibles".

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    $\begingroup$ @Noah: At a glance at your comment, you're wrong. Cut your universe at the first inaccessible limit of inaccessible cardinals. There are plenty of worldly cardinals which are limit of inaccessible cardinals. So in fact every set is an element of a transitive model with a proper class of inaccessible cardinals! So clearly you need more. $\endgroup$ – Asaf Karagila Nov 27 '19 at 3:11
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    $\begingroup$ @Noah: Sorry, bro. Worldly cardinals. :-) $\endgroup$ – Asaf Karagila Nov 27 '19 at 4:33
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    $\begingroup$ @Noah: Just to clarify my previous comment, your second comment is also mistaken. $\endgroup$ – Asaf Karagila Nov 27 '19 at 22:25
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    $\begingroup$ Wow, I was an idiot yesterday (not saying I'm not one today, mind you :P). @RémiPeyre please ignore everything I said. $\endgroup$ – Noah Schweber Nov 28 '19 at 16:37
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    $\begingroup$ @Noah: Does that mean that they should also ignore that you say to ignore everything you said? $\endgroup$ – Asaf Karagila Nov 29 '19 at 13:38

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