Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In an answer and comment to

it was suggested that given a forcing argument using a c.t.m., one could always translate the same argument into a setting without c.t.m.'s. If this is the case, then what is one to make of the following argument by Paul Cohen in his paper "The Discovery of Forcing" (pp. 1090-91):

There was another negative result, equally simple, that remained unnoticed until after my proof was completed. This says one cannot prove the existence of any uncountable standard model in which AC holds, and CH is false (this does not mean that in the universe CH is true, merely that one cannot prove the existence of such a model even granting the existence of standard models, or even any of the higher axioms of infinity). The proof is as follows: If $M$ is an uncountable standard model in which AC holds, it is easy to see that $M$ contains all countable ordinals. If the axiom of constructibility is assumed, this means that all the real numbers are in $M$ and constructible in $M$. Hence CH holds. I only saw this after I was asked at a lecture why I only worked with countable models, whereupon the above proof occurred to me.

The same proof can be used to show that one cannot prove the existence of a uncountable standard model in which AC holds, and there exists a nonconstructible real.

If one was to use Boolean-valued models or a Boolean ultrapower approach to 'constructing' models in which CH was false or there existed a nonconstructible real, does this mean that one cannot prove that the models so constructed (assuming the models so constructed were standard models) are uncountable, even if the proof using these two methods make no mention of the models' 'countability'?

share|improve this question
    
Notice that Cohen uses a phrase that should set of blinking red lights: "one cannot prove". We know that in fact one can prove anything that is true by simply assuming that fact as an axiom... So the real question is: what theory is Cohen talking about when he says "one cannot prove". Once that is clarified, these sorts of comments make much more sense. In other words, the issue with countable transitive models is that we are specifically limiting the methods of proof we allow, and with those limited methods it should be unsurprising if we need stronger hypotheses. –  Carl Mummert Nov 27 '13 at 14:41
    
@Carl: Since a more detailed version of the proof (and theorem) is found on pp 108-9 of his book "Set Theory and the Continuum Hypothesis", I will quote the theorem found therein verbatim: "From ZF+SM [SM = "there exists a standard model for ZF"--my comment] or indeed from any axiom system containing ZF which is consistent with V=L, one cannot prove the existence of an uncountable standard model in which AC is true and CH is false, nor even one in which AC holds and which contains non-constructible real numbers." This tells you what theory (or class of theories) he is working in. –  Thomas Benjamin Nov 27 '13 at 15:48

2 Answers 2

up vote 2 down vote accepted

Note that Cohen is specifically talking about standard models. The obstacles to using these methods to contradict Cohen's assertions lie not in demonstrating the uncountability of the models you construct, but rather ensuring that they are standard.

  • Note that Boolean-valued models cannot be standard (or isomorphic to a standard model) unless the underlying Boolean algebra is the two-element algebra, and in this case there is little point even considering them as a separate class of models.

  • The ultrapower $M^{\mathcal{U}}$ of a (standard) model $M$ will not be well-founded unless $\mathcal{U}$ is $\sigma$-complete. The existence of such ultrafilters requires the existence of a measurable cardinal, which itself implies $\mathbf{V} \neq \mathbf{L}$, and is also more suspect than the existence of standard models. At any rate, you are transcending $\mathsf{ZFC}$.

(Cohen's argument is essentially that if one could demonstrate the existence of such an uncountable standard model in $\mathsf{ZFC}$, then you would also get such a model in $\mathsf{ZFC} + \mathbf{V}=\mathbf{L}$. It is in this stronger theory that the problems arise.)

share|improve this answer

To add on Arthur's answer, the use of Boolean-valued models is not in order to construct some object in any meaningful way. It is a way to circumvent the need for a generic filer.

The point of Boolean-valued models is that one can show that they inherit inference rules from first-order logic, and that all the axioms of $\sf ZFC$ have Boolean value of $1$ and are therefore "true" in a very good sense. On the other hands, statements whose value is not $1$ can become consistently false if one takes a generic ultrafilter containing their negation.

Therefore we have that by taking suitable Boolean algebra we can prove that the statement "There is a non-constructible real" is consistent with $\sf ZFC$.

Cohen, as Arthur points out, says that if one works with standard models then when talking about uncountable models one must already include all the countable ordinals and therefore if $V=L$ it will have decided $\sf CH$.

If we want to prove the consistency of $V\neq L$, then we don't want to assume that the universe satisfies that. And working with uncountable models is inadequate for this sort of result.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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