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I am still studying the topics in forcing and did not yet study much about forcing with a class of conditions.

I know from Jech's Set Theory that you can force that the class of ordinals in the world will be countable in the generic extension, which means that you can take a proper class and turn it into a set.

I was wondering if there is a nice characterization of classes that can be forced into sets.

Clearly it is sufficient to assume that the class is well-orderable, and if you assume AC then if you collapsed a class to a set, in the generic extension it will be well-orderable, but does that mean that you can construct from the function you've added one which shows that you actually collapsed $Ord^M$ to some $\dot\alpha$ in the generic extension? (i.e. is the sufficient condition is also necessary?)

Edit:

So I went to see my advisor today, and we talked about this question a little bit. He wasn't able to give me a full answer but he gave me a nice direction. We discussed generic extensions with global choice, namely you only add a class function which is a well-ordering of the world, without introducing new sets into the world. Jech wrote about it right between Easton Forcing and Levy Collapse in chapter 15 of his book.
I searched further and found some mailing list in which Solovay said (quite recently too!) that he and Jensen did something like that already way back.

In light of everything, I came up with some new questions which I will ponder on my own for a while; one of the question is worth mentioning here as it related very much to my original question - especially in light of Joel's answer:

Is it possible to collapse the entire universe to some inaccessible cardinal in a larger universe? (If so, doesn't that imply that $Con(ZFC)\rightarrow Con(ZFC+In)$? (where $In$ is the existence of an inaccessible cardinal)) If not in $ZFC$ how about $NBG$?

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Asaf, you know that $Con(ZFC)$ does not imply $Con(ZFC+In)$, so the answer to your previous to last question is no. Since $NBG$ is equiconsistent with $ZFC$ then the answer to your last question is also no. –  Andres Caicedo Feb 16 '11 at 3:23

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For any set $X$, we may consider the partial order $\mathbb{P}$ consisting of all finite partial functions from $\omega$ to $X$, ordered by extension. If $G\subset\mathbb{P}$ is $V$-generic for this forcing notion, then $f=\cup G$ is a function from $\omega$ onto $X$. It is a function, since conditions in $G$ are compatible; it is total, since it is dense to add any given point to the domain; and it is surjective, since for any $x\in X$ the set of conditions $p$ with $x\in\mathop{ran}(p)$ is dense. Thus, in the forcing extension $V[G]$, the set $X$ becomes countable. This is how any set can be made countable by forcing. Since the forcing notion $\mathbb{P}$ is a set, it also follows that $V[G]\models ZFC$.

In the case that $X$ is an arbitrary proper class, one can still form the partial order $\mathbb{P}$ consisting of all finite partial functions from $\omega$ to $X$, and it will still be true that the union of the generic filter will be a function from $\omega$ to $X$, thereby making $X$ countable in $V[G]$. But what will no longer be true in this class forcing case is that $V[G]\models ZFC$. Indeed, if $X$ is a proper class, then the corresponding forcing extension will definitely not satisfy ZFC, since $X$ will contain members of arbitrarily large ordinal rank (appearing unboundedly high in the $V_\alpha$ hierarchy), the forcing extension will have a function from $\omega$ unbounded in the ordinals, in violation of the Replacement axiom. So the price you pay for making a proper class $X$ countable is that you must give up ZFC in the forcing extension. Indeed, if $X$ is a proper class in $V$, then $X$ contains elements of unboundedly high rank, and this will remain true in any extension of $V$ to a model $W$ with the same ordinals; thus, $X$ will not be a set in any such $W$.

So, to answer your question: every set can be countable in a forcing extension by set forcing (and thus while preserving ZFC), but no proper class can become a set in any extension of the universe to a model of ZF with the same ordinals.


Edit. I see now that you want to consider upward extensions of the models, not just forcing extensions. There are a few observations to make about this. I have been interested in this topic for some time.

The question is the extent to which a model $M$ of set theory can become the $(V_\theta)^N$, for some taller of a taller model of set theory.

  • First, I claim that every model $M$ of set theory can be elementarily embedded into the $V_\theta$ of another model of set theory $N$, and you can even arrange that $(V_\theta)^N\prec N$. This is a sense in which $M$ is continued as you desire to a taller model. You can prove this by a simple compactness argument using the elementary diagram of $M$, augmenting the theory with a new constant symbol $\dot\theta$ and the assertions that the right theory holds. This theory is finitely consistent by an application of the Reflection theorem.

  • Second, some models $M$ of set theory cannot be realized directly as the $V_\theta$ of a taller model. For example, this is true in any pointwise definable model (a model in which every element is definable without parameters), since the larger model would then see that every element of $V_\theta$ is definable in $V_\theta$, which would contradict that the larger model thinks it is uncountable.

  • Third, every countable computably saturated model $M$ of set theory is realized as the $V_\theta$ of a taller model. Indeed, every countable computably saturated model of ZFC is isomorphic to one of its own $V_\theta$'s, and by looking at things from the perspective of that copy, the result obtains. Harvey Friedman was the first to prove results of this kind, concentrating at first on nonstandard models of PA. You can find an account of the argument in my recent paper with Victoria Gitman in "A natural model of the multiverse axioms," Notre Dame Journal of Formal logic, vol 51, 2010. One of the things we prove is that if $M$ is a countable computably saturated model of ZFC, then $M$ is also isomorphic to an element of $M$ that thinks it is an $\omega$-nonstandard model of set theory. Thus, every countable computably saturated model of ZFC exists as a nonstandard model inside another better model.

  • Fourth, it is possible to show that every countable transitive model of set theory can be end-extended to a (possibly nonstandard) model of V=L, which is well-founded to any desired countable height. The reason is that this assertion is true in $L$ itself, and the statement has complexity $\Pi^1_2$ in a code for the structure, so it is true in $V$ by Shoenfield absolutenss.

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Joel, thanks for the detailed answer. I think you may have misunderstood my question though (and I might just be asking it in a confusing way). I know that forcing a class to be countable (or any other cardinality) one violates Replacement. Why is it so important to preserve the ordinals though? Can't we just guarantee they remain an initial segment of the new model's ordinals instead? –  Asaf Karagila Feb 15 '11 at 8:01
    
@Asaf, I assume you want to end up with a transitive (set) model of set theory, right? Suppose that $V=L$ and that there is a unique (necessarily countable) $\alpha$ such that $L_\alpha\models$ZF (this is a consistent scenario). Then if you force over $L_\alpha$ to make a class countable (or a set) as you want, and if you end up with a transitive model of set theory $M$, then $L^M=L_\beta$ for some $\beta$ (note that $\alpha<\beta=ORD^M$), and $L_\beta\models$ZF. But this is impossible, as the only $\gamma$ with $L_\gamma\models$ZF is $\alpha$. –  Andres Caicedo Feb 16 '11 at 1:05
    
@Andres, it doesn't have to be a set. I had a hunch that $V=L$ might be a source of some problems. –  Asaf Karagila Feb 16 '11 at 7:24
    
@Joel, Thanks for the great answer! –  Asaf Karagila Feb 16 '11 at 7:24
    
@Asaf: The point is that if you begin with $M$ and add $g$ and $M[g]$ is not a model of ZF, you may not have enough names inside $M$ to make sense of an extension where you have both $M$ and $g$ and ZF holds, as the smallest such extension may well be the proper class $L[M,g]$. But then you are not doing forcing at all, as you have no way to describe within $M$ what the properties of this extension are (not even under a very loose notion of "describing"). –  Andres Caicedo Feb 16 '11 at 7:32

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