A few questions about class and set. I find the concepts of proper class and set are quite puzzling. Here are some questions about them.

*

*Is it possible for a collection of something to be a proper class in one model/theory but a set in another?


*In the wiki page of Inner model, it's said that

If M is a model for S, and N is an $L$-structure such that
1.$N$ is a substructure of $M$, i.e. the interpretation $\in_N$ of $\in$ is $\in_M\bigcap N^2$
2.N is a model for T
3.the domain of $N$ is a transitive class of $M$
4.$N$ contains all ordinals of $M$
then we say that $N$ is an inner model of $T$ (in $M$).

Why does it use 'transitive class' rather than 'transitive set'? Is it possible for a transitive class of a model to be small enough to be a set?


*If $Z$ is a proper class all of whose elements are transitive sets, are $\bigcup Z$ and $Z\cup \bigcup Z$ both necessarily proper class?
 A: *

*Yes. For example, consider the minimal model of ZFC (call it $X$). By the Lowenheim-Skolem theorem, $X$ is countable, which by replacement implies that $X$ is a set and not a proper class. Then, try to reason about $X$ itself within $X$: from $X\vDash\textrm{ZFC}$, $X$ satisfies the theorem of ZFC that every set has a rank, which is a specific ordinal assigned to all sets as the supremum of the successors of each of its member's ranks. You can also think of rank as the stage at which a set is a subset of the V-hierarchy. In particular the rank of every ordinal is its own successor. But working in $X$, we cannot define the rank of $X$: $X$ satisfies "for any ordinal $\alpha$, the set $\alpha+1$ has $\textrm{rank}(x)>\alpha$", so $X$ believes that $X$ has no ordinal as its rank. In other words, if working in $X$ we were to assume "$X$ is a set", we would contradict this result of regularity, since $X\vDash``\textrm{rank}(x)\textrm{ is not defined}"$. So from $X$'s perspective, $X$ isn't a set! Similar phenomena happen for other sets like $X\cap\textrm{Ord}$.

*No. From a similar argument to #1's, if we had an inner model $M\vDash\textrm{ZFC}$ that was set-sized, $\textrm{rank}(M)$ would be defined and it would be an ordinal. Then $M$ would not contain the ordinal $\textrm{rank}(M)+1$, so $M$ couldn't have been an inner model.

*For the rest of this answer it's a helpful fact that a superclass of a proper class $Z$ is proper, this is because the superclass will contain all the members of $Z$ including ones of arbitrarily large rank. If $Z$ is a proper class, $Z$ contains members of unboundedly large rank. Going "down a level of $\in$", in order for $Z$ itself to have these high-rank members, we must have that for every $\alpha$ there is a $z\in Z$ such that $z$ itself has a member of rank $>\alpha$. Since $\bigcup Z$ is the class of all the members of $z\in Z$, $\bigcup Z$ has members of arbitrarily large rank and therefore $\bigcup Z$ is proper. $Z\cup\bigcup Z$ is much easier to show proper since it contains all the members of $Z$.

