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Jan 24, 2020 at 2:56 history became hot network question
Jan 23, 2020 at 19:55 comment added Andrés E. Caicedo @Toby I edited $M[x]$ to $M(x)$. In modern notation, the former is by design a model of choice, so it would be confusing to use it in this setting.
Jan 23, 2020 at 19:54 history edited Andrés E. Caicedo CC BY-SA 4.0
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Jan 23, 2020 at 19:25 vote accept Toby Meadows
Jan 23, 2020 at 19:19 answer added Asaf Karagila timeline score: 10
Jan 23, 2020 at 19:16 history edited Toby Meadows CC BY-SA 4.0
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Jan 23, 2020 at 19:14 comment added Toby Meadows Hmmm .. I'll edit the question and see if you still think it's solved. Thanks for looking at it.
Jan 23, 2020 at 19:12 comment added Asaf Karagila Toby, a Cohen real is the most basic of generic extensions...
Jan 23, 2020 at 19:12 comment added Toby Meadows Thanks. I was aware of this result and had it in mind, but I'm not sure it does answer the question. This is why I added the condition that M[G] is a generic extension of M. Grigorieff's Theorem B shows that in such a case N=M[x] for some x \in N. If I understand the remarks on the Bristol model correctly, then it cannot be of the form M[x] (assuming that L=M). I probably should have mentioned the Theorem B think in the OP.
Jan 23, 2020 at 18:57 comment added Andrés E. Caicedo The answer is no! For a rather dramatic example, see for instance MR3878470 Karagila, Asaf The Bristol model: an abyss called a Cohen real. J. Math. Log. 18 (2018), no. 2, 1850008, 37 pp. (Take a look at section 7.2 for a brief comment on this matter.)
Jan 23, 2020 at 18:42 history asked Toby Meadows CC BY-SA 4.0