[Math] Is the class of all ordinals independent of set theory

cardinalslarge-cardinalsordinalsset-theory

I am little confused with this. In any model of set theory (or is only in ZFC?), we start with an initial set, and then by transfinite recursion we build the universe of sets, which is a proper class. The universe of sets can differ, for instance, we don't know if $V=L$, or we at least can assume it either true or false and get different models? (or is it different theories?). I guess there are many other different universes beyond $V$ and $L$, either by transfinite recursion or by some other (unknown to me) method.

The specific question is: Is the set of ordinals in any of those models the same?. I know that there are properties of ordinals (or cardinals) that are not absolute. For instance, there are no ordinals that are measurable cardinals in $L$, but an ordinal that is a measurable cardinal in some other model can still belong to $L$, only that in $L$ it doesn't have that property. So, finally my question: do all the universe of sets of set theory have the same ordinals on every model? In other words, is the universe of sets of the same "size" on all models, so that the proper class or ordinals is well defined, and its size the same? (if that is not the case, do $V$ and $L$ have the same number of ordinals?

So in the end, what I want to know is if the proper class of all ordinals is well defined and always the same, or if it is dependent of the particular set theory or model we are considering in a specific case.

Best Answer

First note that since $\mathsf{ZFC}$ is just a formal first-order theory, it doesn't pick out exactly what form a model's interpretation of the $\in$ symbol takes. So there are models of $\mathsf{ZFC}$ where the $\in$ relation is something other than the real membership relation. Actually, this point is crucial in the development of Boolean-valued models. In such circumstances, what $M$ thinks of as an ordinal will not be what $\mathbf{V}$ thinks of as an ordinal.

But I assume that you are thinking of models where $\in$ is interpreted as the real membership relation. But we do have the following facts.

  • If $\kappa$ is an inaccessible cardinal, then $V_\kappa \models \mathsf{ZFC}$, and $V_\kappa$ clearly does not contain all ordinals.
  • Since the underlying language of $\mathsf{ZFC}$ is countable, by the Löwenheim–Skolem Theorem (if it is consistent) $\mathsf{ZFC}$ has countable models. Taking any countable model $\mathfrak{M} = ( M , E )$ of $\mathsf{ZFC}$ there must be ordinals of $\mathbf{V}$ which are not in $M$. It might be that what $\mathfrak{M}$ thinks of as ordinals are not the ordinals of $\mathbf{V}$, but even if we are "lucky" $\mathfrak{M}$'s version of ordinals is "correct" (which will occur if $M$ is a transitive set and $E$ is the real membership relation, see below) there must be ordinals of $\mathbf{V}$ not appearing in $M$.

In the second point above I mentioned transitive set models of $\mathsf{ZFC}$ (where $\in$ is interpreted as real membership). The existence of models of this kind is an even stronger assumption than the simple consistency of $\mathsf{ZFC}$, then we cannot show from $\mathsf{ZFC}$ (or even $\mathsf{ZFC}+\mathrm{Con}(\mathsf{ZFC})$) that such exist.

But transitive (possibly proper class) models of $\mathsf{ZFC}$ must agree with $\mathbf{V}$ about what ordinals are (if not "how many" there are):

If $M$ is a transitive model of $\mathsf{ZFC}$, and $a \in M$, then $M \models \text{"}a\text{ is an ordinal"}$ iff $a$ really is an ordinal (in $\mathbf{V}$).

This is due to the fact that there is a $\Delta_0$-formula $\phi(x)$ (i.e., a formula in which all quantifiers can be expressed in the form $(\exists x \in y)$ for $(\forall x \in y)$) which in $\mathsf{ZFC}$ defines the property of being an ordinal, and it is known that $\Delta_0$-formulas are absolute for transitive models.

A transitive model $\mathbf{M}$ of $\mathsf{ZFC}$ which contains all the ordinals (from $\mathbf{V}$) is called an inner model.

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