I know that we can assume that formulas are objects in $V_\omega$, and that notions such as formula and satisfiability for a standard model (when the universe is a set) are definable and absolute. Because of that it appears to me that we can define an absolute (metamathematical) formula that "decides" when a formula (as an object in the universe) belongs in ZFC and furthermore through it we can write a formula $S(A)$ that states $(A,\in)\models ZFC$ , which again will be absolute. Is this correct?

Furthermore, given sets $A$ and $E\subset A\times A$ it looks to me that the satisfiability of the model $(A,E)$ can be described with a $\Delta_1$ formula $S(A,E,\ulcorner\phi\urcorner)$ and thus is absolute. Is this true? And does this mean that "$(A,E)\models ZFC$" is absolute as well?

Thanks in advance.

  • 4
    $\begingroup$ Yes, this is absolute, for the reasons you state. There is a small subtlety, in that a model $(M,E)$ of ZFC may actually think that ZFC is inconsistent. Another subtlety is that a satisfaction predicate is easy to define for set models, but is in general not definable for proper classes. $\endgroup$ Commented Feb 26, 2011 at 2:23
  • $\begingroup$ @Andres: Thank you for your answer. $\endgroup$
    – Apostolos
    Commented Feb 26, 2011 at 16:07

1 Answer 1


Here is a way to see directly that the satisfaction relation $M\models\varphi[a]$ has complexity $\Delta_1$ in set theory. This relation is defined (by Tarski) by induction on the complexity of $\varphi$. Thus, we can say that $M\models\varphi[a]$ if and only if there is a function mapping formulas and points in $M$ to the set {satisfied, not satisfied}, which satisfies the inductive properties of Tarski's definition (so that atomic formulas are correctly satisifed, the satisfaction relation on negation inverts the answer, the relation on Boolean combinations works correctly and the relation on quantifiers works correctly). These requirements on the satisfaction relation S have complexity $\Delta_0$, since we need only quantifiy over $S$, over $M$ and over the natural numbers (codes for formulas). Furthermore, one can prove that there is a unique satisfaction relation, so $M\models\varphi[a]$ if and only if there is a satisfaction relation showing this, if and only if all satisfaction relations show this. So the satisfaction relation is $\Delta_1$.

It works the same for any theory, such as ZFC as in your question, since this just adds an additional quantifier over the natural numbers (for every axiom, it is satisfied), which is bounded and therefore doesn't increase the complexity beyond $\Delta_1$.

Thus, being a model of any particular assertion or theory is $\Delta_1$ and therefore absolute in the sense you may have had in mind. For example, any model of set theory will agree with all its forcing extensions and inner models about whether a given set structure is a model of ZFC.

Addendum. But there is another subtle sense in which being a model of ZFC is not absolute. For example, suppose that $M$ is a model of $ZFC+\neg Con(ZFC)$, so that $M$ thinks there are no models of ZFC at all. So this model $M$ must have nonstandard natural numbers, since otherwise it would have the same proofs from ZFC as we do and so it would realize that ZFC is consistent. Inside $M$, we may enumerate the axioms of what $M$ thinks of as ZFC (this includes nonstandard instances of the axioms). In $M$, consider the largest initial segment of the axioms that $M$ thinks is true in some rank initial segment $(V_\alpha)^M$. By the Reflection theorem, this includes all the standard initial segments. Thus, what $M$ thinks is the longest such realizable initial segment of the axioms has nonstandard length. And so the corresponding set $(V_\alpha)^M$ satisfies all the standard instances of axioms of ZFC. That is to say, it really is a model of ZFC, but $M$ doesn't think it is, because $M$ thinks that some of the nonstandard axioms are not true there. So this is an example of non-absoluteness, where two universes of set-theory can disagree about whether a given structure is a model of ZFC or not.

  • 1
    $\begingroup$ Thank you, this was extremely informative. $\endgroup$
    – Apostolos
    Commented Feb 26, 2011 at 16:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .