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 http://settheory.mathtalks.org/cmu-math-logic-seminar-tues-11-september/):

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?

  • $\begingroup$ The Bristol model, I see. Have you contacted James or Menachem? Asaf Karagila was working on an alternative approach for a while. $\endgroup$ Commented Feb 13, 2014 at 6:31
  • $\begingroup$ @AsafKaragila: Dear Asaf, have you completed your notes on Bristol model? $\endgroup$ Commented Nov 13, 2014 at 5:24

1 Answer 1


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.

Update (April 25, 2017):

It's on arXiv now.

  • $\begingroup$ 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.) $\endgroup$
    – Asaf Karagila
    Commented Feb 13, 2014 at 8:43
  • 1
    $\begingroup$ Wow. It only took me three years to finish this. Never say "finish this soon" on something complicated. The general proof is done, but I still need to write some additional consequences of the construction, and a nice blackboxed introduction to iterations of symmetric extensions. This, I hope, will take much much less than three years. But a few minutes ago I finished writing the actual proof. PHEW! I'll update my answer once I have posted the preprint on arXiv. $\endgroup$
    – Asaf Karagila
    Commented Mar 7, 2017 at 13:34
  • $\begingroup$ Mohammad, I posted the paper on arXiv. $\endgroup$
    – Asaf Karagila
    Commented Apr 25, 2017 at 4:59
  • $\begingroup$ For finishing my dissertation? Sure... :) $\endgroup$
    – Asaf Karagila
    Commented Apr 25, 2017 at 16:41
  • $\begingroup$ I read the introduction of the paper till now. You have said your strategy of the proof is different from the one suggested by the Bristol group. The idea presented by Bristol group also seems very interesting to me and somehow very strange. Do you possibly can give the basic idea of their proof in some details (or do you have any notes)? (maybe reading your paper gives the basic idea) $\endgroup$ Commented Apr 26, 2017 at 3:41

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