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A cardinal $\kappa$ is $\alpha$-inaccessible if $\kappa$ is inaccessible and such that for any ordinal $\beta < \alpha$, the set of $\beta$-inaccessibles less than $\kappa$ has cardinality $\kappa$.

For example, a cardinal is 1-inaccessible if it is an inaccessible limit of (0)-inaccessibles. A domain $V_{\kappa_1}$ for ${\kappa_1}$ a 1-inaccessible would have a proper class of 0- inaccessibles. Likewise, a domain $V_{\kappa_2}$ for $\kappa_2$ a 2-inaccessible would contain a proper class of 1-inaccessibles.

However, I think there must be some ordinals $\gamma$ for which there cannot be a proper class of $\gamma$-inaccessibles, but I cannot make this intuition precise enough to prove it. If my intuition is correct, for which ordinals does it fail?

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    $\begingroup$ There is no such $\gamma$. That is, we expect significantly stronger large cardinal assumptions to be consistent. $\endgroup$ Sep 29, 2018 at 17:24

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If $\kappa$ is Mahlo, then $\kappa$ is $\kappa$-inaccessible. So if there is a Mahlo cardinal $\kappa$, we get that $V_\kappa\models$ "There is a proper class of $\alpha$-inaccessibles for every ordinal $\alpha$." Since the existence of Mahlo cardinals is not currently known to be inconsistent with ZFC - to put it mildly (Mahlo cardinals are incredibly low down in the large cardinal hierarchy) - the answer to your question is (currently) no.

In fact, even the existence of a proper class of Mahlos isn't much to write home about: it holds in $V_\lambda$ whenever $\lambda$ is weakly compact.

And all of this is way, way, way weaker and smaller than even a single measurable cardinals.

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    $\begingroup$ If $\kappa$ is Mahlo, then it is a limit of $\lambda$-inaccessibles for all $\lambda<\kappa$. $\endgroup$
    – Asaf Karagila
    Sep 29, 2018 at 17:41
  • $\begingroup$ @AsafKaragila Derp, that was silly. Fixed! $\endgroup$ Sep 29, 2018 at 17:45

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