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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If I have a model of ZFC and a proper class in that model, is there always an extension to another bigger model where this proper class becomes a set? I know that this is possible in particular cases, but I have no idea if this can be done in general.

Thanks in advance

share|cite|improve this question
A class is just a formula with one free variable. Some formulas define sets, and the ones that do not define sets are called proper classes. Hence, think about your question when you consider the formula "$x=x$"; does it define a set or always a proper class? – boumol Aug 26 '11 at 12:25
But for example if I have an inaccessible cardinal $\kappa$, then $V_\kappa$ is a model of ZFC. Now, $(\{x :x=x \})^{V_\kappa} = V_\kappa$, which is not a set viewed from $V_\kappa$, but it's in fact a set (in V). This is the kind of things I mean. Maybe I should ask: given a model M of ZFC is there always an extension where M is a set? – Charlie Aug 26 '11 at 12:31
Sure $V_\kappa$ is a set, but it's not an element of the model $V_\kappa$. If I compute the "proper class" corresponding to the formula x=x IN $V_\kappa$, I get $V_\kappa$, which is a set in V. My question is (one more attempt to make sense of it): any model M of ZFC can be realized as a submodel of a larger model N in such a way that $M \in N$ ? – Charlie Aug 26 '11 at 12:59
If I may contribute yet another phrasing. Suppose that M is a model of ZFC. Must there exist a model N of ZFC such that there is a set $m \in N$ that is isomorphic to M? – Tanner Swett Aug 26 '11 at 13:13
@Charlie:… while I was asking in particular about forcing extensions. However after some discussion JDH's answered regarding end extensions as well, which seems to me to be something you are looking for (at least partially). – Asaf Karagila Aug 26 '11 at 13:49
up vote 6 down vote accepted

This question cannot be answered within ZFC, and the answer depends somewhat on the answerer's philosophical position. It's somewhat related to the question "Are proper classes objects?" on MathOverflow, and the answer someone gives to this question is going to be related to the answer she gives to that question. Unfortunately, as you will see by looking there, both the answer "Yes, proper classes are objects" and the answer "No, proper classes are not objects" have proponents. Similarly, some people would answer this question "Yes", although most set theorists would answer it "No".

For those who accept there is a well-defined, objective meaning to the term "all sets", the answer to this question is "No". Under this viewpoint, the collection of "all sets" is a class model of ZFC which cannot be a set in any other (transitive) model of ZFC, because the collection of "all sets" is itself not a set by Russell's paradox. This is by far the most common position on your question among set theorists.

To argue that every class model of ZFC is a set in a larger (transitive) model of ZFC seems to require giving up the idea that "all sets" has an objective meaning apart from a model of ZFC. In that case, it is consistent that every class model of ZFC is embeddable as a set in a different model of ZFC, and in fact it's consistent that every class model of ZFC is countable in another model of ZFC. By "consistent" I mean that this situation is relatively consistent with ZFC itself, as shown by Gitman and Hamkins in their recent paper "A natural model of the multiverse axioms". However, few set theorists would agree that the multiverse axioms are true.

share|cite|improve this answer

Your Answer


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.