Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm not sure if my question make sense, but it would also be interesting to know if it didn't, so I will ask anyway:

There exist a countable model of ZF. This means (if I understand it correctly) that we can find a model of ZF in ZF, such that the set of elements in the model is countable seen from the outer ZF.

My question is: Can every model of ZF be "seen from the outside" in a way that makes it countable?

It seems to me, that if we have a model A of ZF, the model will be a element in a countable model B of ZF. Now, if you look at B from "the outside" A is countable.

share|improve this question
add comment

2 Answers 2

up vote 7 down vote accepted

Technically, a model of ZF consists of a set with some relation, representing "being an element of". So the literal answer is no. An uncountable model is simply uncountable. Your argument that one can see every model as a model in a model which is countable doesn't work. Cardinality in a model depends on whether there exists certain bijections in the model. This has nothing to do with "outside cardinality".

However, the first-order statements true in some model of ZF are true in some countable model by the Löwenheim-Skolem theorem. Since we are usually not interested in differences between models that cannot be formulated in the first order theory of sets, we can say that every model of ZF has an, for practical purposes, equivalent countable model.

share|improve this answer
add comment

Michael Greinecker's answer is correct. But there is another subtler weaker sense in which what you asked for could be true.

Namely, the method of forcing shows that every set that exists in any model of set theory can become countable in another larger model of set theory, the forcing extension obtained by collapsing the cardinality of that set to ω. Thus, the concept of countability loses its absolute meaning; whether a set is countable or not depends on the set theoretic background.

So if X is any set, then in some forcing extension of the universe, the set X becomes countable.

In particular, this will be true when X is itself a model of set theory.

There are various ways of viewing the nature of existence of these forcing extensions, among them Boolean-valued models and their quotients, the Boolean ultrapowers, and I refer you to the set-theoretic literature. If one uses the Boolean ultrapower, then stating the theorem from inside V, one attractive way to describe the situation is as follows:

Theorem. If V is the universe of sets and X is any particular set, then there is a class model (W,E) of set theory, and an elementary embedding of V into a submodel W0 of W, such that the image of X in W is thought by W to be countable.

Basically, the model W0 is the Boolean ultrapower of V by the forcing B to collapse X, using any ultrafilter on B, and W is the quotient of the Boolean valued model VB. The elementary embedding maps every object y to the equivalence class of the check name of y.

share|improve this answer
add comment

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.