5
$\begingroup$

Here I read that:

enter image description here

Trying to understand why this is true I have studied Kenneth Kunen's "Set Theory: An Introduction to Independence Proofs" (1st ed.), but I'm still confused.

I partially understand (2): A countable transitive model of ZFC can always be extended to a model where CH holds by adding bijections between $\aleph_0$ and any cardinals in between $\aleph_0$ and the cardinality of $2^{\aleph_0}$. But that is just because everything in a countable transitive model is countable when "seen from the outside". How does it work for an arbitrary model? Pretty much all of Kunen's theorems are about countable models.

I do not understand (1) at all. If no cardinals are collapsed, then $\aleph_1^V=\aleph_1^{V[G]}$, and if no new reals are added, then $(2^{\aleph_0})^V=(2^{\aleph_0})^{V[G]}$, and the bijection between the two in $V$ is also in $V[G]$, so CH should still hold. What am I misunderstanding? (I can see how you can extend any countable transitive model to one where CH fails by adding reals because there must always be some that are not already included. But the slide says that you can do it without adding reals and that you can do it with any model.)

$\endgroup$
7
$\begingroup$

Re: (2), we can make sense of forcing extensions over arbitrary models by thinking about Boolean-valued models; this treatment is found in Jech. Alternatively, if we want to talk about models in the usual sense, then we should restrict attention to countable models (at least without further set-theoretic hypotheses - e.g, if we assume MA+$\neg$CH in the "surrounding universe," then models of size $\omega_1$ are as malleable as countable models.)

Incidentally, it's worth pointing out that restricting to transitive models is unnecessary: while forcing is easiest to develop over transitive models, we can force over ill-founded models just as well.


Re (1): Yes, the slide is incorrect, and that's a huge mistake that makes the slide quite misleading.

Maybe the author was trying to write contrasting descriptions:

  • To get $\neg$CH, you collapse no cardinals, but add some reals.

  • To get CH, you collapse some cardinals, but add no reals.

The description at the end, then, should be: even stages satisfy CH, odd stages satisfy $\neg$CH., and stage $2n+1$ and $2n+2$ have the same reals. But the way the end is written makes it sound like all stages have the same reals.

$\endgroup$
  • $\begingroup$ Thank you very much! Just to clarify: Are you saying that any model can be extended to one where CH fails by adding reals? That is, there is no such thing as a model already containing all possible reals? $\endgroup$ – Casper Jun 17 '17 at 14:45
  • $\begingroup$ @Casper: Of course, if your model is $V_\kappa$ for some sufficiently large cardinal $\kappa$, then it contains all the reals. This is some tacit understanding (which is indeed very confusing) that we really care about countable models, at least most of the time, and that anything we do can be ultimately be done in a model which is countable from the meta-theoretic point of view. $\endgroup$ – Asaf Karagila Jun 17 '17 at 14:46
  • $\begingroup$ @AsafKaragile: So no, not every model can be extended to one where CH fails? Or every model can, but not necessarily by adding reals? $\endgroup$ – Casper Jun 17 '17 at 14:48
  • $\begingroup$ @Casper The problem here is that there are a lot of contrasting frameworks for forcing here. The simplest one is the convention that by "model," we mean countable transitive model. Then yes, every countable model has forcing extensions where CH fails by adding new reals - and this isn't surprising, since no countable model can contain all the reals, since there are uncountably many of those. On the other side of things, when we work with Boolean-valued models, our "objects" are really a kind of "potential" object, in a precise sense. (cont'd) $\endgroup$ – Noah Schweber Jun 17 '17 at 14:57
  • $\begingroup$ This approach is ultimately more flexible, but much harder to work with (in my opinion) in the beginning. So my advice would be to always work with countable transitive models. Here, in this framework, are the relevant precise statements: (i) If $M$ is a c.t.m., then $M$ has forcing extensions $M[G]$ and $M[H]$ satisfying CH and $\neg$CH respectively, where $M[G]$ and $M$ have the same reals and $M[H]$ and $M$ have the same cardinals. (ii) If $M$ satisfies CH, then no generic extension of $M$ with the same reals satisfies $\neg$ CH. (cont'd) $\endgroup$ – Noah Schweber Jun 17 '17 at 14:59
3
$\begingroup$

It is true, not every model can be extended in the "normal sense of the word" to a model of $\sf CH$ or a model of $\lnot\sf CH$. If your model is some $V_\kappa$ for an inaccessible $\kappa$, then it contains all the reals, sets of reals, and so on.

Any forcing which changes the truth value of $\sf CH$ adds either reals or sets of reals, so $V_\kappa$ cannot be extended to a larger model where $\sf CH$ changed its truth value.

One can use Boolean-valued models (or a similar, and a perhaps more direct approach using $\Bbb P$-names), but these would generate ill-founded models with just an embedding of your original model into the "generic extension". So this is not the sense in which you want your model to be extended.

However, in set theory, especially when espousing the multiversial view, one can work with the implicit assumption that even the universe is countable in some larger meta-universe. So one can force over the universe to add the necessary generics, perhaps by adding new reals to the universe.

Another way to solve this problem is to recall that anything that you are trying to do, might as well be done over a countable model (simply take a countable elementary submodel of whatever you wanted to force over). This has the same effect as the other approach, as it implies that you're always talking about countable models, in one way or another.

(I do agree, however, that this understanding is not expressed explicitly enough in a lot of places, and it can be confusing for beginners. In my first paper, however, I did talk about countable models etc. etc. and nowadays it feels to me as a very crude way of talking about forcing, and that it can be cleaner and simpler to "just do it". So one has to get used to it, and then move along.)

$\endgroup$
0
$\begingroup$

As for the question about countable transitive models vs. extending $V$: my guess would be that the author of the slides is using an alternate formulation of forcing. Recall that even in the case of countable transitive models $M$, and forcing order $\mathbb{P} \in M$, it is essential in forcing arguments to use the fact that the forcing relation is definable within $M$. So, some authors will take this definition, and view it as being applied directly to the original universe $V$ of $ZFC$ and an arbitrary partial order $\mathbb{P} \in V$. Then for every formula $\phi$, if $ZFC \vdash \phi$, then $ZFC \vdash (\Vdash \phi)$. So, if $ZFC \vdash (\Vdash \psi)$ also, then $Con(ZFC) \Rightarrow Con(ZFC + \psi)$.

Informally, this is close to defining a conservative extension $V[G]$ of $V$. However, as far as I know, it's not possible to actually define a model of $ZFC + \psi$ in this way: for example, for general $p \in \mathbb{P}$, it is not true that either $\Vdash \hat p \in \dot G$ or $\Vdash \hat p \notin \dot G$, even though of course $ZFC \vdash (\Vdash (\hat p \in \dot G \vee \hat p \notin \dot G))$. On the other hand, for any finite number of statements, it would be possible to find $p \in \mathbb{P}$ such that $p$ forces either each statement or its negation - so in situations where you could apply logical compactness, you might be able to formalize an argument in that way (possibly refining $p$ to decide intermediate statements in a proof as well).

You might also be able to prove consistency of $ZFC$ with language extended by proper class $V$, and sets $\mathbb{P}$, $G$, with additional axioms $V$ transitive; $V \models ZFC$; $\mathbb{P} \in V$; $G \subseteq \mathbb{P}$; and $G \notin V$. Plus, if $\mathbb{P}$ is definable without parameters in $V$, you could add an axiom that $\mathbb{P}$ must be exactly the element of $V$ defined in this same way. (I don't recall for sure if the interpretation function $\cdot^G$ is definable as a proper class in $M[G]$ for the countable transitive model case; but if so, you could probably also add a symbol for that, and an axiom that it satisfies the recursive definition of the interpretation function, and an axiom that it's surjective.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.