Timeline for Do all limit $\alpha \in \omega_1^L$ satisfy $L_\alpha \models V=HC$?
Current License: CC BY-SA 4.0
14 events
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| May 12, 2022 at 21:08 | comment | added | Martín S | @FarmerS Thank you so much for such a detailed answer! It's exactly what I was looking for and I understand the proof. In fact now, thanks to this insight, I've noticed an essentially equivalent proof can be confectioned from other results scattered across the paper. More concretely, its Lemma 8.1 states "If $\alpha$ is not a gap ordinal $L_\alpha$ is pointwise definable", and its Lemma 4.1.a "If $L_\alpha$ is pointwise definable, there's an arithmetical copy of it in $L_{\alpha+2}$". By checking this last proof, it is seen the isomorphism building the copy also belongs to $L_{\alpha+2}$. | |
| May 9, 2022 at 16:32 | comment | added | Farmer S | Now suppose that $\alpha$ ends a gap. Then $\alpha$ is a limit of $\beta$'s as above, and hence for each such $\beta<\alpha$, there is a surjection $\pi:\omega\to L_\beta$ with $\pi\in L_\alpha$, and so $L_\alpha$ models "$V=\mathrm{HC}$". | |
| May 9, 2022 at 16:32 | comment | added | Farmer S | Then $\pi$ is definable from the parameter $p$ over $L_{\beta+1}$, and therefore $\pi\in L_{\beta+2}$ (coding pairs appropriately, so that issues with ranks of pairs are avoided)... | |
| May 9, 2022 at 16:32 | comment | added | Farmer S | Then $H\preccurlyeq L_\beta$, and by condensation, the transitive collapse $M$ of $H$ (exists and) equals $L_{\beta'}$ for some $\beta'\leq\beta$. But by elementarity, $x$ is also definable over $M$ from the transitive collapse $\bar{p}$ of $p$, and so $x\in L_{\beta'+1}$. By choice of $x$, therefore $\beta'=\beta$. And so by the minimality of $p$, therefore $\bar{p}=p$. It follows that $L_\beta$ is pointwise-definable from the parameter $p$. Let $\pi:\omega\to L_\beta$ be the resulting surjection... | |
| May 9, 2022 at 16:31 | comment | added | Farmer S | For this, fix the lexicographically least $p\in[\beta]^{<\omega}$ such that there is a real $x\in L_{\beta+1}\backslash L_\beta$ which is definable over $L_\beta$ from parameter $p$, and fix a witnessing real $x$. Let $H\preccurlyeq L_\beta$ be the definable hull of $\{p\}$ over $L_\beta$ (that is, $H$ is the set of all elements of $L_\beta$ which are definable over $L_\beta$ from $p$)... | |
| May 9, 2022 at 16:31 | comment | added | Farmer S | @MartinS Fair enough. How about this? If $\beta$ ends a gap, then there is a surjection from $\omega$ to $L_\beta$ which is definable over $L_\beta$, without parameters. This follows from standard fine structure theory, and gives a surjection $\omega\to L_\beta$ which is in $L_{\beta+1}$. But for our purposes, it's enough to prove a slightly weaker thing: there is a surjection from $\omega$ to $L_{\beta}$ which is in $L_{\beta+2}$... | |
| May 9, 2022 at 10:36 | comment | added | Martín S | @FarmerS Thank you for your answer! The issue is, I'm not sure my proof of $L_\alpha \models ZF^-$ can go through without already knowing $L_\alpha \models V=HC$ (since I've changed the proof given by Marek to use more modern methods). And indeed, they present the proof of $L_\alpha \models ZF^-$ later, as if it were not necessary to show the truth of the Corollary. | |
| May 4, 2022 at 14:19 | answer | added | C7X | timeline score: 5 | |
| May 3, 2022 at 12:34 | comment | added | Farmer S | Regarding the "last fact" mentioned in the question, i.e. whether $L_\alpha$ has the isomorphism between $L_\beta$ and $E_\beta$: The Mostowski collapse can take longer than $\omega$ steps (consider e.g. recursive wellorders of $\omega$, which have Mostowski collapses cofinal in $\omega_1^{\mathrm{ck}}$). But if $\alpha$ starts a gap then $L_\alpha$ models ZF$^-$, and so can compute Mostowski collapses. | |
| May 3, 2022 at 1:21 | comment | added | François G. Dorais | The existence of the Mostowski collapse is often called Beta. Beta is not provable in KP but it is provable in KP + $\Sigma_2$-collection. So it's somewhere between $\Sigma_1$ and $\Sigma_2$ in the admissibility spectrum. | |
| May 2, 2022 at 21:49 | comment | added | Asaf Karagila♦ | @Noah: It's a $\Delta_1$ fortune. | |
| May 2, 2022 at 21:48 | comment | added | Noah Schweber | @AsafKaragila Pun intended, or just good fortune? | |
| May 2, 2022 at 21:11 | comment | added | Asaf Karagila♦ | It is absolutely false that every $L_\alpha$ satisfies $V=\rm HC$ for limit $\alpha<\omega_1^L$. Just take an elementary submodel of $L_{\omega_2}$ and collapse to some $L_\alpha$, for example. | |
| May 2, 2022 at 20:56 | history | asked | Martín S | CC BY-SA 4.0 |