# Proper classes and models of set theory

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.

• 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: math.stackexchange.com/questions/22066/… 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