Timeline for On intermediate transitive models for ZFC between M an M[G]
Current License: CC BY-SA 3.0
10 events
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| Oct 27, 2016 at 10:01 | comment | added | Vladimir Kanovei | Regarding Zoorado Nov 20 '12 at 14:25, it is true nevertheless that an intermediate model is a Q-generic extension for some Q being an strengthening of P in the sense that the domain of Q is still equal to the domain of P but $\le_Q$ strengthens $\le_P$. | |
| Nov 21, 2012 at 3:33 | comment | added | Zoorado | Ah thanks again for the example. It does seem that we really cannot say much about the relation between intermediate extensions and forcing notions as partial orders. | |
| Nov 20, 2012 at 20:39 | comment | added | Joel David Hamkins | It's not quite right. I should say that in this tree, the only (necessarily nontrivial) complete suborders are forcing equivalent to the whole order... | |
| Nov 20, 2012 at 17:27 | comment | added | Joel David Hamkins | If you want $Q$ to be a complete suborder, so that maximal antichains in Q are also maximal in P, then you can make a counterexample by using the forcing to collapse $\omega_1$ to $\omega$ via $P=\omega_1^{\lt\omega}$, which is a tree. In a tree order, the only complete suborders are the whole order. But forcing with P also adds a Cohen real (and lots of other stuff, which does not collapse $\omega_1$), and so the corresponding intermediate extensions do not arise from complete suborders of $P$. | |
| Nov 20, 2012 at 16:44 | comment | added | Joel David Hamkins | I see. I think in this case, one can still make a counterexample, but I'd have to think about it. | |
| Nov 20, 2012 at 16:15 | comment | added | Zoorado | Ok I get why it is that way in the example. I think I messed up my second question. What I wanted to say is, if M[G] is a P-generic extension, then must an intermediate M[G'] be a Q-generic extension for some Q isomorphic to a suborder of P? My bad for the confusion. | |
| Nov 20, 2012 at 14:58 | comment | added | Joel David Hamkins | Yes, that is false. You have to go the Boolean algebra, and get your subalgebra there. The same example works, since every subalgebra of my forcing $P$, where the conditions have the same length, which determines all the bits of the first Cohen real $c$, also determines all the bits of the second one, just because conditions in $P$ have the same length. So $P$ has no suborder adding just the first Cohen real. But meanwhile, $\mathbb{B}=RO(P)$ does have a subalgebra adding just the first real. | |
| Nov 20, 2012 at 14:25 | vote | accept | Zoorado | ||
| Nov 20, 2012 at 14:25 | comment | added | Zoorado | Wow the second example is very illuminating. Thanks a lot. Would it even be false to say in general that if M[G] is a P-generic extension, then an intermediate M[G'] is a Q-generic extension for some Q being a suborder of P? | |
| Nov 20, 2012 at 3:00 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |