Assume $\alpha$ is admissible, $R\in L_\alpha$ is a linear order that don't have any infinite descending chain in $L_\alpha$. Is there always an end extension $M$ of $L_\alpha$, such that $M\vDash KP$ and $R$ is $M$-isomorphic to an $M$-ordinal?
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asked Jun 13, 2023 at 4:03
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$\begingroup$ Is this situation even possible? $\endgroup$– Asaf Karagila ♦Commented Jun 13, 2023 at 4:40
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1$\begingroup$ @AsafKaragila yes. For any non Σ¹₁-reflecting countable admissible α, there is a Σ₁(L_α) linear order which is not well ordered but has no infinite descending chain in L_α⁺, this is the main theorem of Gostanian's paper "the next admissible ordinal". $\endgroup$– Reflecting_OrdinalCommented Jun 13, 2023 at 4:55
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$\begingroup$ Is it clear that this even works for a Harrison order with $\omega_1^{\mathrm{CK}}$? $\endgroup$– James E HansonCommented Jun 13, 2023 at 5:15
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$\begingroup$ If $\alpha$ is countable, then there are at least some illfounded initial segments $R'$ of $R$ for which there is such an $M$ (that is, $R'$ is isomorphic to an $M$-ordinal). $\endgroup$– Farmer SCommented Jun 13, 2023 at 13:43
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$\begingroup$ @FarmerS Can you prove the full proposition when α is countable? $\endgroup$– Reflecting_OrdinalCommented Jun 13, 2023 at 13:50
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1 Answer
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In Chapter 3 of his PhD thesis, Harvey Friedman showed that there is a recursive linear ordering $\prec$ which has no hyperarithmetic infinite descending sequence and which does not support a jump hierarchy. Then, $\prec$ is an element of $L_{\omega_1^{CK}}$ and has no infinite descending sequence in $L_{\omega_1^{CK}}$. If $M$ is an $\Sigma_1$-admissible end-extension of $L_{\omega_1^{CK}}$, then for every ordinal $a$ in $M$ which is countable in $M$ there is a jump hierarchy in $M$ supported by $a$. It follows that $\prec$ cannot be isomorphic to an ordinal in $M$.
