Timeline for Which forcings preserve (some) determinacy?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 19, 2019 at 16:44 | answer | added | Noah Schweber | timeline score: 7 | |
| Oct 15, 2012 at 1:19 | history | edited | Noah Schweber | CC BY-SA 3.0 |
added 90 characters in body
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| Oct 15, 2012 at 1:18 | comment | added | Noah Schweber | @Andreas, good point. Edited. | |
| Oct 15, 2012 at 1:05 | comment | added | Andreas Blass | What you wrote about countably closed forcing applies more generally to any forcing that doesn't add reals (since that's all you used from the "countably closed" hypothesis) --- and there are forcings that don't add reals but aren't countably closed, for example the standard forcing to add a club subset in a statinoary, co-stationary subset of $\omega_1$. | |
| Oct 15, 2012 at 0:51 | answer | added | Joel David Hamkins | timeline score: 8 | |
| Oct 14, 2012 at 22:00 | history | edited | Noah Schweber | CC BY-SA 3.0 |
Incorporated comment
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| Oct 14, 2012 at 21:58 | comment | added | Noah Schweber | (Sorry, I screwed up my quotation marks: 'has property $P$."' should be 'has property $P$.""') | |
| Oct 14, 2012 at 21:57 | comment | added | Noah Schweber | Good points. One thing that might be doable is to ask when a theorem of the following form is provable in ZFC (+ large cardinals?): "$Det(\Gamma)\implies$ "if $\mathbb{P}$ is any poset such that $\Vdash_\mathbb{P} $"$Det(\Gamma)$", then $\mathbb{P}$ has property $P$." This is expressible in the language of ZFC, since forcing is definable, as long as $\Gamma$ is a sufficiently nice pointclass. Would this work? | |
| Oct 14, 2012 at 20:54 | comment | added | François G. Dorais | Statements like $(\ast)$ are always tricky to formulate. I think you need some more qualifiers on the transitive model $W$. As is, we could take a large $W$ where $\mathbb{P}$ is countable and then everything falls apart. But that's not really an answer since things probably fell apart in $W$ before forcing with $\mathbb{P}$. Working inside a fixed universe $V$ would help but then the assumptions of $(\ast)$ are likely to need more large cardinal power than what you intended. | |
| Oct 14, 2012 at 20:17 | history | edited | Noah Schweber | CC BY-SA 3.0 |
Fixed grammar
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| Oct 14, 2012 at 19:56 | history | asked | Noah Schweber | CC BY-SA 3.0 |