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In the vast majority of papers forcing is always developed over ZFC.

Not surprisingly too, since infintary combinatorial principles are often used to prove results based on properties such as chain conditions, closure, and so on.

I am looking for a good start on forcing over models of ZF. I have before me two papers which I have yet to read thoroughly, however may not be as useful for this purpose as I am hoping.

  • Grigorieff, S. Intermediate Submodels and Generic Extensions in Set Theory. The Annals of Mathematics, Second Series, Vol. 101, No. 3 (May, 1975), pp. 447-490
  • Monro, G. P. On Generic Extensions Without the Axiom of Choice. The Journal of Symbolic Logic, Vol. 48, No. 1 (Mar., 1983), pp. 39-52

While I do intend to read them either way, it seems that neither develops the theory of forcing in the absolute absence of choice. I am currently looking for references which deal with such situation, or with the relation between forcing theorems proved in ZFC and the amount of choice needed for them to hold.

Edit: I probably should have mentioned that I am quite familiar with permutation models of ZFA+embedding theorems and transfer theorems (Jech-Sochor, Pincus' theorem) as well with symmetric extensions.

I am not looking for ways to develop forcing extensions of ZF without the axiom of choice; rather I am looking for theorems such as c.c.c forcing does not collapse cardinals and similar theorems extended to the choiceless contexts if possible, or the strength of choice needed for these theorems to hold.

Consider two examples:

  1. Suppose a model of ZF in which the axiom of choice does not hold. Can we, by set forcing add the axiom of choice? If not, can it be done using a machinery similar to a symmetric extension? If we can in fact find such extension, does that mean the model without choice is a symmetric extension between two larger models?

  2. Suppose A is an infinite Dedekind-finite set, what can we say on a forcing poset based on A (either domain of functions are partial to A or the range is in A)? Can we "collapse" amorphous sets onto ordinals? Can we collapse one amorphous set onto another? And so on.

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  • $\begingroup$ Well, the canonical reference is books.google.com/…. $\endgroup$
    – user5810
    Commented Sep 26, 2011 at 15:42
  • $\begingroup$ (retracted, was looking at the wrong paper.) $\endgroup$
    – Not Mike
    Commented Sep 26, 2011 at 16:10
  • $\begingroup$ @Ricky: Are you sure? This is about consequences of AC, while I am looking for defining forcing in the lack thereof. $\endgroup$
    – Asaf Karagila
    Commented Sep 26, 2011 at 16:32
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    $\begingroup$ "or with the relation between theorems proved in ZFC and the amount of choice needed for them to hold." $\endgroup$
    – user5810
    Commented Sep 26, 2011 at 18:03
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    $\begingroup$ Hi Asaf, Only now I see this nice question. Busy life as usual, but I'll add something of substance if time permits. For now, the answer to 1 is no. Gitik's model where all cardinals are singular is a counterexample. Woodin has shown that in the choiceless setting, from very strong large cardinal assumptions (beyond embeddings from V to V) it follows that we can recover choice, but class forcing is needed in general. It would be fabulous if one could provide structure in Woodin's setting by identifying enough of it as a symmetric model. But V is in general a class symmetric extension of HOD. $\endgroup$ Commented Sep 28, 2011 at 6:50

5 Answers 5

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Friedman's book "Fine structure and class forcing" develops forcing over ZF, rather than ZFC, in chapter 2. Although chapter 1 is about fine structure, it is not used in chapter 2. Although the rest of his book is well above my level, I find Friedman's exposition of forcing quite helpful.

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  • $\begingroup$ The online catalog of the local library shows the copy is on the shelf somewhere in the library. I'll go look at it tomorrow and let you know how 'spot on' this suggest was. Thanks! $\endgroup$
    – Asaf Karagila
    Commented Sep 26, 2011 at 19:39
  • $\begingroup$ Actually, now that I look at it a bit more carefully (and with coffee), I'm not sure this is what you're looking for, although I think you probably will find it interesting. I don't think Friedman spends much time really getting into what happens when AC fails; I think he just works in a little more generality than, say, Kunen. Still, it might be helpful. $\endgroup$ Commented Sep 26, 2011 at 23:15
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The book "Theory of Semisets" by Vopenka and Hajek contains forcing constructions over models that violate AC. For one example: Start with the basic Fraenkel model (of ZF with atoms, ZFA --- I'm using here the terminology of Jech's "Axiom of Choice" book); it has an infinite Dedekind-finite set $A$ of atoms. Adjoin an $A$-indexed family of Cohen reals, by forcing with finite partial functions from $A\times\omega$ to 2. The pure part of the resulting model is the basic Cohen model. (In other words, instead of the usual procedure of passing to a symmetric submodel of a forcing extension, you can equivalently start with symmetry in the ground model and then just force.) This is how Vopenka and Hajek introduce the basic Cohen model.

Unfortunately, I think the only way to read the Vopenka-Hajek book is straight through from the beginning, because there's a lot of notation and terminology that will make no sense if you just open the book to the chapter you're interested in.

Another nice example of forcing over choiceless models of ZFA is that Mostowski's linearly ordered model of ZFA can be obtained from the basic Fraenkel model by adding a generic linear ordering of $A$ with finite conditions.

I second Francois's suggestion to look into Eric Hall's work, which builds on ideas like these and takes them a good deal farther.

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  • $\begingroup$ I looked over Eric Hall's work, and it indeed seemed impressive. I wanted in addition some more foundational book about this. As you say, his work builds on ideas like these; so I probably should start with this book. I'll grab it from the library and see what's in there later today. Thanks! $\endgroup$
    – Asaf Karagila
    Commented Oct 2, 2011 at 6:16
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Arnie Miller's "Long Borel hierarchies" specifically pp 8-12 may be of interest for you. See here

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  • $\begingroup$ Actually this entire paper is of interest to me. I also was wondering about facts true in the Feferman-Levy model. Many many many thanks! $\endgroup$
    – Asaf Karagila
    Commented Sep 26, 2011 at 18:29
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You may want to look at Eric Hall's papers.

Regarding your question 1, I think that if $M$ is a model of SVC with $S$ (see Blass, Injectivity, projectivity, and the axiom of choice, TAMS 255), then you can force AC by wellordering the set $S$ using finite functions from $\omega$ into $S$. On the other hand, if you can force AC with a poset $P$ then the original model should satisfy SVC with $P$. So it looks like SVC is the key to force AC (unless you allow class forcing).

Regarding your question 2, forcing with finite injections from a $\omega$ to any set $A$ which is of greater cardinality than every finite ordinal will force a bijection between $\omega$ and $A$. I suppose you could do the same to force a bijection between any two given sets $A$ and $B$ by using finite partial injections from $A$ into $B$, provided this poset satisfies the obvious density requirements. However, this might accidentally force both sets to lose certain properties.

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  • $\begingroup$ Thanks for the references. I will be sure to dig through them. As for satisfaction. The theorem is that $M\models AC\implies M[G]\models AC$. Your first paragraph seems to imply the inverse as well, which I would believe is not completely true. $\endgroup$
    – Asaf Karagila
    Commented Sep 27, 2011 at 17:57
  • $\begingroup$ I think you're misreading the first paragraph. The correct statement is that if $M[G] \vDash AC$ (for some generic $G$ over some forcing poset $P \in M$) then $M \vDash SVC$ (with parameter $P$). $\endgroup$ Commented Sep 27, 2011 at 18:04
  • $\begingroup$ I think I should start sleeping full nights, or at least a good schlafstunde! :-) $\endgroup$
    – Asaf Karagila
    Commented Sep 27, 2011 at 18:04
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Here is an equivalence of a forcing principle to the Axiom of Choice, courtesy of Arnold Miller, found in his preprint, "The maximum principle in forcing and the axiom of choice":

(Abstract) In this paper we prove that the maximum principle in forcing [ $p$ $\Vdash$ $\exists$$x$ $\theta$($x$) iff there exists a a name $\tau$ such that $p$ $\Vdash$ $\theta$($\tau$)--my quote from Miller's preprint, found on his homepage at www.math.wisc.edu/~miller/] is equivalent to the axiom of choice. We also look at some specific partial orders in the basic Cohen model [his model in which the axiom of choice fails].

This paper also discusses partial orders in Cohen's original model for which the maximum principle fails and one (at the end of this paper) for which the maximum principle succeeds.

It would be interesting to find out what difficulties (if any) the absence of the maximum principle causes for forcing in $ZF$ (and if these difficulties could be overcome).

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    $\begingroup$ Yes. This is a known theorem of Bell from the 1970s. One of the reasons I have worked so hard to develop a framework for iterating symmetric extensions is that it would allow to mitigation in restoring the maximum principle. At least you didn't cite my thesis. $\endgroup$
    – Asaf Karagila
    Commented Feb 10, 2018 at 16:37
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    $\begingroup$ (You did notice this is a question from six and a half years ago, right?) $\endgroup$
    – Asaf Karagila
    Commented Feb 10, 2018 at 16:38
  • $\begingroup$ @AsafKaragila: I did notice that this was an old question. However, I was not aware that Bell first proved the theorem (Miller notes that showing the necessity of the axiom of choice in proving the maximum principle was problem 1.30 in Bell's book on boolean-valued models--however, some authors have been known (Hartley Rogers Jr., for example) to stick open problems in their problem sets). Did Bell in fact prove this equivalence? When and in what paper did he prove this? Also, did you in fact use this equivalence in your thesis? At any rate, if you wish, I can delete this answer. $\endgroup$ Commented Feb 10, 2018 at 16:58

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