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This answer seems to imply that: for an ordinal $\alpha$, to be recursively inaccessible (i.e. $\alpha$ is admissible and limit of admissible) implies to be not locally countable (i.e. $L_\alpha \models \exists \beta \ ``\beta \text{ is uncountable"}$). Here is the relevant excerpt:

If there is some $r\in L_\alpha$ with $\omega_1^r=\alpha$, then $L_\alpha$ will be locally countable (= $L_\alpha\models$ "every set is countable"). But plenty of countable admissible $\alpha$s don't give rise to locally countable levels of $L$! In particular, if $\alpha$ is an admissible limit of admissibles (= "recursively inaccessible") then every real in $L_\alpha$ is contained in some admissible $L_\beta$ with $\beta<\alpha$.

However I believe that if by "admissible" we read here "$\Sigma_1$-admissible", this statement does not hold as in this sense the first recursively inaccessible ordinal appears way before the first ordinal that is not locally countable (see resp. 2.3 and 2.21 in DA Madore's zoo of ordinals). Whence my question:

Does the statement hold if by "admissible" we read "$\Sigma_n$-admissible for all $n$"? If no, can we describe the first ordinal which is not locally countable in term of higher recursive inaccessibility?

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    $\begingroup$ That sentence was misleadingly written, my apologies; I've edited it so the point is clearer. $\endgroup$ Jan 18, 2023 at 5:20

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Suppose $L_\alpha\models$"There is a largest cardinal". Then ($L_\alpha$ is $\Sigma_n$-admissible for all $n<\omega$) $\Leftrightarrow$ ($L_\alpha\models$ ZF$^-$) $\Leftrightarrow$ ($L_\alpha$ does not project ${<\alpha}$) $\Leftrightarrow$ ($\alpha$ is a cardinal in $L_{\alpha+1}$). So the least $\beta$ such that $L_\beta\models$"$\omega_1$ exists" is $\beta=\alpha+1$ where $\alpha=\omega_1^{L_\beta}$ is the least ordinal such that $L_\alpha$ is $\Sigma_n$-admissible for all $n<\omega$, hence the least such that $L_\alpha\models$ ZF$^-$.

However, if $\alpha$ is least such that ($L_\alpha\models$ ZF$^-$ and $\alpha$ is a limit of $\beta$'s such that $L_\beta\models$ ZF$^-$), then $L_\alpha\models$"Every set is countable". (For each of those $\beta$'s, $L_\beta$ is pointwise definable, so $\beta$ is countable in $L_{\beta+2}$).

If we let $\alpha$ be least such that $L_\alpha\models$" KP + $\omega_1$ exists", then $\omega_1^{L_\alpha}$ is a limit of limits of ZF$^-$ levels, a limit of limits of limits of ZF$^-$ levels, etc.

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  • $\begingroup$ About "$L_\alpha$ does not project $<\alpha$", is this a different form of the term "project" than the $\Sigma_1$-projectum? A $\xi$ where $\rho_1^{L_\xi}=\xi$ sounds smaller than the least $\alpha$ where $L_\alpha\vDash ZF^-$. OTOH in the appendix of "Gaps in the constructible universe" it's said that for such $\alpha$ "each $\beta<\alpha$ is $\alpha$-projectible to $\omega$", although maybe this is to trim down such $\alpha$ in order to satisfy the characterization theorem for gap ordinals. $\endgroup$
    – C7X
    Jan 18, 2023 at 5:45
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    $\begingroup$ It just means that $\mathcal{P}(\beta)\cap L_{\alpha+1}=\mathcal{P}(\beta)\cap L_\alpha$ for every $\beta<\alpha$. $\endgroup$
    – Farmer S
    Jan 18, 2023 at 9:40
  • $\begingroup$ And does this last sentence yields a characterization of this least $\alpha$? That is for the least $\beta$ such that $\beta$ if is a limit of limits of ZF$^-$ levels, a limit of limits of limits of ZF$^-$ levels, etc. then the next admissible $\alpha>\beta$ is such that $L_\alpha \models ``\omega_1 \text{ exists"}$? $\endgroup$
    – Johan
    Jan 18, 2023 at 15:32
  • $\begingroup$ And by "etc." I mean, probably as you meant, that for all $\delta < \beta$, $\beta$ is a $\delta$-limit of ZF$^-$ levels, where $\delta$-limits are inductively defined as limits of limits iterated $\delta$ times. $\endgroup$
    – Johan
    Jan 18, 2023 at 17:17
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    $\begingroup$ @Johan No, the least $\alpha$ is quite larger. In Arai97 there is a characterization of admissible $\alpha$ where $L_\alpha\vDash"\omega_1\textrm{ exists}"$, specifically a characterization of the adm. $\sigma$ where $L_\kappa\vDash"\sigma\textrm{ is an uncountable cardinal}"$ for adm. $\kappa>\sigma$, these are the $\sigma$ that start gaps of length $\kappa$ (we take $\kappa=\sigma^+$.) From Marek-Srebrny's "Gaps" we know for each $\delta$ the least such $\delta$-limit would be upper-bounded by the least ordinal starting a gap of length $\delta$. $\endgroup$
    – C7X
    Jan 19, 2023 at 1:56

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