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Say that a long model is an $\mathfrak{A}\models\mathsf{I\Sigma_1}$ such that $\mathfrak{A}$ has strictly greater cardinality than each of its proper initial segments (in the case $\vert\mathfrak{A}\vert=\aleph_1$ this is "$\omega_1$-like"ness). If $\mathfrak{A}$ is a long model and $\vert\mathfrak{A}\vert$ is regular, then in fact $\mathfrak{A}\models\mathsf{PA}$ since the bounding scheme is trivially satisfied in $\mathfrak{A}$ (see Lemma 2.15 of Hajek/Pudlak). Moreover, by the MacDowell/Specker theorem (see Kossak, 63 years of the MacDowell-Specker theorem) the converse holds: if $\kappa$ is an uncountable cardinal and $\varphi$ is true in every long model of $\mathsf{PA}$ of size $\kappa$, then $\mathsf{PA}\models\varphi$ already. (Given $\mathsf{PA}\not\models\varphi$, let $\mathfrak{B}\models\mathsf{PA}+\neg\varphi$ be countable and iterate MD/S $\kappa$-many times.)

Interestingly, the first observation seems to need regularity:

Is there a singular cardinal $\kappa$ such that the common theory of long models of size $\kappa$ is strictly weaker than $\mathsf{PA}$?

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    $\begingroup$ I'm not sure about the details yet but I highly suspect that you can use the ideas in the proof of the second Paris-Mills theorem to build a long model of size $\beth_{\omega}$ in which tetration fails to be total. (In particular there will be a countable initial segment whose image under tetration is cofinal in the model.) I think this will still be a model of $\mathsf{I\Sigma_1}$, but if not you should be able to do it with some hyperoperator. $\endgroup$ Jun 5, 2023 at 3:18
  • $\begingroup$ @JamesHanson What is the second Paris-Mills theorem? I'm not familiar with it. $\endgroup$ Jun 5, 2023 at 3:33
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    $\begingroup$ Oh actually it looks like I don't need to write it up. There's a relevant result mentioned in the Kossak paper you linked. It says Kaye showed that for each $n \geq 1$ and each singular $\kappa$, there is a $\kappa$-like model of $\mathsf{B\Sigma^0_n} + \mathrm{exp}+\neg \mathsf{I\Sigma^0_n}$. That answers your question, right? $\endgroup$ Jun 5, 2023 at 4:22
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    $\begingroup$ @JamesHanson It's even worse/better than that - a couple sentences later Kossak mentions a beautiful result that's in Ian Haken's thesis (we overlapped at Berkeley), jointly with Slaman, which thoroughly nukes the question. That's what I get for not reading the paper before linking to it! Would you like to put this as an answer, or should I? $\endgroup$ Jun 5, 2023 at 4:35
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    $\begingroup$ @Stef Yes, $\mathsf{PA}$ is first-order Peano arithmetic (occasionally "PA" is used to refer to Presburger arithmetic, but that's a rarity in my experience). The standard source on subsystems of $\mathsf{PA}$ is the Hajek/Pudlak book linked in my question. $\endgroup$ Jun 6, 2023 at 19:04

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The answer to the question is strongly in the positive, since the following theorem of Kaye implies that for every singular cardinal $\kappa$, and for any fixed natural number $n$, the common theory of $\kappa$-like models of $\mathrm{I}\Sigma_{n}$ is strictly weaker than $\mathsf{PA}$.

Theorem. Let $\kappa$ be a singular cardinal, $n \geq 1$, and let $M\models \mathrm{B}\Sigma_n + \mathsf{exp} + \lnot\mathrm{I}\Sigma_n$. Then there is a $\kappa$-like model $K$ that is elementarily equivalent to $M$.

The above theorem appears as Theorem 2.4 of Kaye's paper Constructing κ-like models of arithmetic. J. London Math. Soc. (2) 55 (1997), no. 1, pp.1–10. According to Kaye, the paper arose out of a question of Slaman: Are there $\kappa$-like models of arithmetic for singular $\kappa$ that do not satisfy $\mathsf{PA}$?

A key ingredient of Kaye's proof is the technology of "doubly indexed indiscernibles" invented by Keisler, who showed that for any consistent extension $T$ of Zermelo set theory, and any given singular cardinal $\kappa$, $T$ has a $\kappa$-like model. This result of Keisler appears in his paper Models with orderings, Logic, methodology and philosophy of science III (eds. B. van Rootselaar and J. F. Staal; North-Holland, Amsterdam, 1968) 35–62.

It is worth noting that Kaye has another paper related to this topic: The theory of $\kappa$-like models of arithmetic. (English summary) Special Issue: Models of arithmetic. Notre Dame J. Formal Logic 36 (1995), no. 4, pp.547–559.

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    $\begingroup$ Looking at this from an outsider's perspective, so apologies if this is a stupid question: is there hope of a precise description of the theory of $\kappa$-like models for a specific singular cardinal $\kappa$ like, say, $\aleph_\omega$ or the first fixed point of $\gamma\mapsto\aleph_\gamma$ (maybe assuming GCH)? Is it finitely axiomatizable? Is this worth opening another question about? $\endgroup$
    – Gro-Tsen
    Jun 6, 2023 at 9:40
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    $\begingroup$ @Gro-Tsen As Kaye points out in the paper referenced in the last paragraph of my answer, the work of Keisler on singular-like models makes it clear that $T$ = the theory of $\kappa$-like models for strong limit singular cardinals is recursively axiomatizable (via the scheme dubbed INDISC in Kaye's paper). The latest work on this subject (that I know of) is that of Haken, referenced in Noah Schweber's answer; which includes a negative answer to a question of Kaye regarding the axiomatization of $T$. I think the finite axiomatizability of $T$ is an open question. $\endgroup$
    – Ali Enayat
    Jun 6, 2023 at 14:14
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To supplement Ali's answer, I'll also mention the following more recent result of Haken/Slaman: they built a $\kappa$-like model of $\mathsf{I\Sigma_1+\neg B\Sigma_2}$ for $\kappa$ an uncountable singular strong limit cardinal. This is Theorem 13 in Haken's thesis.

This leaves open, as far as I know, what can happen at singular cardinals which are not strong limits.

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    $\begingroup$ I changed $\neg B\Sigma_1$ to $\neg B\Sigma_2$, which is what Noah intended to write. $\endgroup$
    – Ali Enayat
    Jun 6, 2023 at 7:34
  • $\begingroup$ @AliEnayat Indeed, thank you! $\endgroup$ Jun 6, 2023 at 16:46

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