Timeline for answer to Can there be an embedding j:V → L, from the set-theoretic universe V to the constructible universe L, when V ≠ L? by Joel David Hamkins
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15 events
| when toggle format | what | by | license | comment | |
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| Jun 26, 2021 at 11:21 | comment | added | Joel David Hamkins | Ah, I see. Thanks! I have now edited. | |
| Jun 26, 2021 at 11:20 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Jun 26, 2021 at 10:55 | comment | added | Farmer S | @JoelDavidHamkins Oh, I see it probably became confusing: the "computer", as I mentioned, was my typo; the typo that I originally tried to refer to was the whole phrase. I.e. why shouldn't $L$ compute successors correctly? And that also doesn't seem relevant to what comes after it... | |
| Jun 26, 2021 at 10:51 | comment | added | Farmer S | @JoelDavidHamkins Sorry, there was an errant "r"; i.e. should have been "Since $L$ does not compute successors correctly", in the proof of the "(Update) Improved Corollary". | |
| Jun 26, 2021 at 10:30 | comment | added | Joel David Hamkins | @FarmerS I am not seeing that phrase, and the word "computer" does not appear on this page except in your comment (and presumably now also in mine). | |
| Jun 26, 2021 at 10:15 | comment | added | Farmer S | @JoelDavidHamkins I think there's some typo in the proof of GCH $>\aleph_0$, where it says "Since $L$ does not computer successors correctly". | |
| Nov 10, 2013 at 13:03 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 10, 2013 at 12:38 | comment | added | Joel David Hamkins | Great! I like it very much. | |
| Nov 10, 2013 at 6:41 | comment | added | Yair Hayut | As a corollary to your corollary - if there is such $j:V\rightarrow L$ then $GCH$ hold everywhere (except maybe at $\aleph_0$). Since $0^\#$ doesn't exist, we can use the covering lemma: let $\lambda$ be uncountable cardinal in $V$. $j^{\prime\prime} \lambda \subset B \in L$ where $|B|^V=\lambda$. Now $|B|^L = \mu < (\lambda^+)^V$ but by the proof of the first theorem, $|P^V (\lambda)| \leq |P^L(B)| = (\mu^+)^L$, so $(2^\lambda)^V = (\mu^+)^L \leq (\lambda^+)^V$. | |
| Nov 9, 2013 at 13:31 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 9, 2013 at 13:23 | comment | added | Asaf Karagila♦ | One approach which might help to solve the case for $L[c]$ where $c$ is Cohen, is to see if $M\subseteq V$ is a model of $\sf ZF$ then $j\colon V\to L$ can be restricted to $M$ as well. In that case, I feel, we may be able to conclude that if $M\subseteq V$ is a model of $\lnot\sf AC$ then there is no such $j$. This will exclude Cohen extensions of $L$, but not Sacks or Miller extensions which are minimal. | |
| Nov 9, 2013 at 13:09 | comment | added | Asaf Karagila♦ | Bon appetit! :-) | |
| Nov 9, 2013 at 13:09 | comment | added | Joel David Hamkins | Yes, that's right. I guess I'll edit, once I'm done with breakfast... | |
| Nov 9, 2013 at 13:07 | comment | added | Asaf Karagila♦ | Not to be petty about terminology, but aren't the theorems about $0^\#$ and the stationary class where $\sf GCH$ fails are more of corollaries to the first theorem? | |
| Nov 9, 2013 at 12:28 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |