[Math] Why is Cohen’s result insufficient to settle CH

lo.logicset-theory

OK, Cohen has constructed a model in which both ZFC and ~CH are true. Isn't this model an answer to the continuum problem? Hasn't he showed that it is indeed possible to construct a set with cardinality between that of the integers and that of the reals? Why is it still not considered sufficient to settle CH? Why is one model not enough? Why for all models? In other words, why do we have to answer whether "ZFC |- CH" instead of just "CH" itself?

Best Answer

I am sympathetic to this question, which often arises for those first learning of Cohen's theorem, and I don't think it is an idle question. I recall my sophomore undergradatue self being confused about it when I first studied the set-theoretic independence phenomenon. And I think that Carl is right, that this particular issue is not addressed on the other CH questions.

I view the question as arising from the following line of thought: Cohen proved that it is possible that CH fails. Thus, it is possible that there is a set of reals whose cardinality is strictly between the integers and the continuum. But if we can decribe how such a set can exist, then haven't we actually described a set of such intermediate cardinality? That is, doesn't this mean that CH is simply false?

This line of thinking may be alluring, but it is wrong. The reason it is wrong, as Gerhard explains in his comment, is that it doesn't appreciate the role of models, or what might be called the set-theoretic background. What Cohen did was to show that if ZFC is consistent, then so is ZFC + ¬CH. (In contrast, Goedel proved that if ZFC is consistent, then so is ZFC + CH.) Thus, Cohen's intermediate-cardinality set has the property that it is intermediate in cardinality in the model that Cohen describes, with respect to that set-theoretic background, but it will not be intermediate-in-cardinality with respect to other set-theoretic backgrounds. The property of being intermediate in cardinality is dependent on the set-theoretic background in which this property is considered. For example, a set $X$ is uncountable if there is no function from the natural numbers onto it. But perhaps there is no such function mapping onto $X$ in a model of set theory $M$, but there is a larger model of set theory $N$, still having $X$, but for which now there IS a function mapping the natural numbers onto $X$. In fact, this very situation follows from Cohen's forcing method: any set can be made countable in a forcing extension.

Thus, whether a set forms a counterexample to CH cannot be observed looking only at that set---one must consider the set-theoretic universe in which the set is considered, and the possible bijective functions that might witness its countability or not. The very same set of reals can be countable in some models of set theory and bijective with the continuum in others.

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