Timeline for Transitive models and CH
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 14, 2015 at 0:27 | vote | accept | Ashutosh | ||
| Nov 5, 2015 at 12:01 | comment | added | Joel David Hamkins | I suppose one can view that as a reduction, but I'm inclined to say that the symmetric model argument can be reduced to a direct forcing argument. | |
| Nov 5, 2015 at 11:54 | comment | added | Asaf Karagila♦ | Well, sure. Your argument, however, can be reduced to passing through $L(\Bbb R)$, which can be clearer and easier to see why these things work nicely (after forcing over $L(\Bbb R)$ to have say, $2^{\aleph_0}=\aleph_2$, you didn't add any new reals, so every real is in $L$ or Cohen over $L$, so this is again a Cohen extension). | |
| Nov 5, 2015 at 11:48 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 5, 2015 at 11:44 | comment | added | Joel David Hamkins | But haven't I also given a proof here that it happens? I find my argument to be more direct, since it doesn't use symmetric models. There is no need to consider $L(\mathbb{R})$ explicitly, although of course the models will have the same $L(\mathbb{R})$ since they have the same reals. | |
| Nov 5, 2015 at 7:25 | comment | added | Asaf Karagila♦ | The situation is possible. See my answer on math.SE | |
| Nov 5, 2015 at 3:26 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 5, 2015 at 3:10 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| S Nov 5, 2015 at 2:59 | history | answered | Joel David Hamkins | CC BY-SA 3.0 | |
| S Nov 5, 2015 at 2:59 | history | made wiki | Post Made Community Wiki by Joel David Hamkins |