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Apologies in advance if this question is obvious/not research level.

Let $\preceq$ be the consistency strength relationship on theories. Working over $ZF$ or $ZFC$, is there some large cardinal notion $\psi(\alpha)$ with an ordinal parameter such that there exists some ordinal $\kappa$ such that $ZF(C)+\psi(\alpha)$ is consistent relative to some stronger theory for all $\alpha<\kappa$ and $$ZF(C)+\psi(\alpha)\prec ZF(C)+\psi(\alpha+1)$$ for all $\alpha<\bigcup\kappa$, but $ZF(C)+\psi(\kappa)$ is inconsistent? What about a $\psi'(\alpha)$ such that $ZF(C)+\psi'(\alpha)$ is consistent relative to some stronger theory and $$ZF(C)+\psi'(\alpha)\prec ZF(C)+\psi'(\alpha+1)$$ for all set-sized $\alpha$, but $ZF(C)+\psi'(O_n)$ is inconsistent?

The motivation is having a large cardinal notion that gets 'as close to inconsistency' as possible without becoming inconsistent, in a precise sense.

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  • $\begingroup$ The standard proof of the Kunen inconsistency shows that there can only be a non-trivial elementary embedding $j:V_\alpha\rightarrow V_\alpha$ if $\alpha=\lambda$ or $\alpha=\lambda+1$ where $\lambda$ is a strong limit cardinal of countable cofinality. In this case, $\lambda=\lim_{n\in\omega}j^n(\text{crit}(j))$. $\endgroup$ Apr 7, 2023 at 22:02
  • $\begingroup$ @JosephVanName Thank you, I'll edit to remove the error. $\endgroup$
    – Alec Rhea
    Apr 7, 2023 at 22:17

1 Answer 1

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The answer is yes.

One can make easy artificial examples. For example, let $\psi(\gamma)$ assert that there are (only) finitely many inaccessible cardinals, but at least $\gamma$ many.

From the assumption of infinitely many inaccessible cardinals, we can prove the consistency of ZFC+$\psi(n)$ for any particular finite $n$, and furthermore the consistency strength of $\psi(n)$ is steadily increasing as $n$ increases, since in fact $\psi(n+1)$ implies $\text{Con}(\text{ZFC}+\psi(n))$. But $\psi(\omega)$ is clearly inconsistent, so this is an instance of your requested phenomenon with $\kappa=\omega$.

One can make many other similar such examples using larger $\kappa$, simply by taking $\psi(\gamma)$ to assert that there are fewer than $\kappa$ many LC of a certain type, but at least $\gamma$ many.

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  • $\begingroup$ Thank you; I suspected there were artificial examples like this, and if I can't think of some way to formally exclude them I'll accept this answer. $\endgroup$
    – Alec Rhea
    Apr 8, 2023 at 1:35
  • $\begingroup$ Out of curiosity, are there any 'non artificial' examples you're aware of (in the sense I hope we both intuitively understand)? $\endgroup$
    – Alec Rhea
    Apr 8, 2023 at 16:36
  • $\begingroup$ @AlecRhea Erdős cardinals would give such examples if we work with $V=L$. $\endgroup$
    – Hanul Jeon
    Apr 10, 2023 at 22:17
  • $\begingroup$ @HanulJeon Interesting, $V=L$ seems to act as a 'ceiling' for large cardinals in many ways. $\endgroup$
    – Alec Rhea
    Apr 11, 2023 at 5:48

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