Timeline for Kunen (2011) exercise IV.8.17: $M[G]\vDash HOD^\mathbb{R}\subseteq L[\mathbb{R}]$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| May 30, 2017 at 13:51 | comment | added | Sam Roberts | Just came back to this, and realised I wasn't as clear as I thought I was :S How do we construct $H$, and how do we know that there is such a $\pi$ in $L$? Thanks again for all your help, Joel! | |
| May 22, 2017 at 10:27 | history | bounty ended | Sam Roberts | ||
| May 22, 2017 at 10:03 | comment | added | Sam Roberts | Thanks so much, Joel! Sorry it took me a while to work through your answer! | |
| May 22, 2017 at 10:02 | vote | accept | Sam Roberts | ||
| May 16, 2017 at 11:04 | comment | added | Joel David Hamkins | I agree with that. That is why I said it is "easy to see". The question is about the converse. | |
| May 16, 2017 at 10:57 | comment | added | Asaf Karagila♦ | The first inclusion is trivial. $L(\Bbb R)$ is the smallest model of ZF which contains all the reals, certainly any $N$ which contains all the reals satisfies $L(\Bbb R)\subseteq N$. In particular $\rm HOD_\Bbb R$. | |
| May 16, 2017 at 1:35 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| May 15, 2017 at 16:05 | comment | added | Joel David Hamkins | One can avoid the fussing with $\dot a$ if you just move to $L[a]$ as a new ground model. | |
| May 15, 2017 at 14:26 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 20 characters in body
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| May 15, 2017 at 14:15 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| May 15, 2017 at 14:10 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |