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Let $M$ be a countable model of $ZFC$ and $M[G]$ be a (set) generic extension of $M$. Suppose $N$ is a countable model of $ZF$ with

$$M\subseteq N \subseteq M[G]$$

and that $N=M(x)$ for some $x\in N$; i.e., it is the smallest inner model of $M[G]$ which contains $x$ and $M$.

Is $N$ a symmetric extension of $M$?

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    $\begingroup$ 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.) $\endgroup$ Jan 23 '20 at 18:57
  • $\begingroup$ 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. $\endgroup$ Jan 23 '20 at 19:12
  • $\begingroup$ Toby, a Cohen real is the most basic of generic extensions... $\endgroup$
    – Asaf Karagila
    Jan 23 '20 at 19:12
  • $\begingroup$ Hmmm .. I'll edit the question and see if you still think it's solved. Thanks for looking at it. $\endgroup$ Jan 23 '20 at 19:14
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    $\begingroup$ @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. $\endgroup$ Jan 23 '20 at 19:55
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Yes, if $N=M(x)$ (taking the modern notation over Grigorieff's $M[x]$), then it is a symmetric extension. This is a very recent result of Toshimichi Usuba (see this and that).

However, by the Bristol model construction, this symmetric extension need not be obtained by the forcing that was used to get $M[G]$.

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  • $\begingroup$ The Bristol model is pretty intimidating. Can you elaborate a little on your second comment? Happy to move to email if that is preferable. $\endgroup$ Jan 23 '20 at 19:22
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    $\begingroup$ The point is that $L(V_\alpha^M)$ where $M$ is the Bristol model are all symmetric models. $\endgroup$
    – Asaf Karagila
    Jan 23 '20 at 19:32
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    $\begingroup$ But regardless, feel free to drop me an email if you have any followups or any other choiceless questions... $\endgroup$
    – Asaf Karagila
    Jan 23 '20 at 19:44

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