Timeline for Can measures be added by forcing?
Current License: CC BY-SA 3.0
21 events
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| Sep 19, 2021 at 9:25 | history | wiki removed | Stefan Kohl♦ | ||
| Apr 19, 2021 at 11:58 | comment | added | Joel David Hamkins | Not necessarily, since for example perhaps $0^\dagger$ exists. | |
| Apr 19, 2021 at 11:34 | comment | added | Ur Ya'ar | @JoelDavidHamkins Thanks! In the first example, will $V$ be a class forcing extension of $L_\mu$? | |
| Apr 19, 2021 at 8:16 | comment | added | Joel David Hamkins | @UrYa'ar Your question is interesting. On the one hand, if $\kappa$ is measurable in $V[G]$, then one can always build $L_\mu$, where it is measurable, and this will be an inner model of $V$. So it was measurable before. But this might not be a ground. On the other hand, here is a negative answer. Take the example $V[G][g]$ of my answer, and do class forcing to $V[G][H]$, forcing the ground axiom, but the new forcing is all up high. So $V[G][H]$ thinks $\kappa$ is not measurable in any ground, but becomes measurable in $V[G][H][g]$. | |
| Apr 19, 2021 at 8:10 | comment | added | Ur Ya'ar | @JoelDavidHamkins is it the case that whenever a cardinal becomes measurable after forcing, it must have already been a measurable before? I.e. is it true that if $\kappa$ becomes measurable in a forcing extension then there is a ground model where $\kappa$ is measurable? | |
| Oct 28, 2012 at 17:51 | vote | accept | Trevor Wilson | ||
| Oct 28, 2012 at 15:17 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 28, 2012 at 14:06 | comment | added | Trevor Wilson | Hi Joel, thank you for your answer. I would also be interested in hearing about the issue François raises, either in seeing even one example of a new measure that does not lift an old measure, or the stronger property you mention. By the way, I think there may be a typo in the answer above immediately after "We may factor the forcing as". | |
| Oct 28, 2012 at 13:17 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 28, 2012 at 12:51 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 28, 2012 at 12:42 | history | edited | Andreas Blass | CC BY-SA 3.0 |
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| Oct 28, 2012 at 12:35 | comment | added | Joel David Hamkins | By the way, François's variations of the question are also very interesting, and I can explain further examples of that, e.g. where a measurable cardinal is preserved, but no new measure lifts an old measure, etc. | |
| Oct 28, 2012 at 12:33 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 28, 2012 at 12:28 | comment | added | Asaf Karagila♦ | Of course. I blame my hasty reading on the faulty LaTeX :-) | |
| Oct 28, 2012 at 12:28 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 28, 2012 at 12:22 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 28, 2012 at 12:21 | comment | added | Joel David Hamkins | Asaf, that would be for downwards absoluteness generally, rather than downwards absoluteness to ground models. | |
| Oct 28, 2012 at 12:16 | comment | added | Asaf Karagila♦ | It is obvious that any notion incompatible with $L$ is not downwards absolute. No? | |
| Oct 28, 2012 at 12:15 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 28, 2012 at 12:08 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 28, 2012 at 12:03 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |