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The following is known:

Theorem. Suppose $V[G]$ is a generic extension of $V$ by a set forcing, and let $N$ be a model of $ZFC$ with $V\subseteq N\subseteq V[G].$ Then $N$ is a generic extension of $V$ by a set forcing, in particular $N=V[A],$ for some set of ordinals.

It seems that the above theorem is not true if $N$ does not satisfy $AC$. In fact the following abstract is given in a talk by James Cummings (see

If $c$ is Cohen-generic over $L$, then there is a transitive class model $M$ of $ZF$ intermediate between $L$ and $L[c]$ which is not of the form $L(A)$ for any $A.$

Does anyone know a proof of this fact?

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The Bristol model, I see. Have you contacted James or Menachem? Asaf Karagila was working on an alternative approach for a while. – Andrés E. Caicedo Feb 13 '14 at 6:31
@AsafKaragila: Dear Asaf, have you completed your notes on Bristol model? – Mohammad Golshani Nov 13 '14 at 5:24

To my knowledge there is no written proof of this fact. I have all the available notes, which include a very very scattered description of $V_{\omega+1}$ and $V_{\omega+2}$ of this model $M$, and a single lemma which is used to proceed through successor of singular cardinals.

I am working on rebuilding this model in a cleaner method, and I am relatively close to finishing (some technical constructions are needed to finish the outline, but the idea itself is completely finished).

With luck I might actually finish this soon, and I could write a reasonable outline announcement (and then a full detailed accounts of the construction).

I should probably add that I asked all the people involved in the construction of this model, The Bristol model, and what I was told by everyone of them is that it started as some general idea to play with, and by the time they realized someone should be writing things down they already did a lot of the work, so it didn't survive into the notes.

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Thanks Asaf, I will wait to hear more about the answer. – Mohammad Golshani Feb 13 '14 at 8:38
Great! I'll be waiting, then. (I have hopes of actually finishing today, but as every other day showed, it might not be the case.) – Asaf Karagila Feb 13 '14 at 8:43

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