I fail to understeand how a step in this particular proof of the theorem is permitted in ZFC.

The proof I've seen starts by considering models to all finite subcollections of sentences of a collection $K$, then applies the axiom of choice to construct a choice function that select a single model for every finite subcollection of sentences from the set of models of such subcollection.

Now my problem is: the axiom of choice is obviously applicable only on sets, so how do we know that the set of all models of a given collection of sentences is a set?

In that case we would be good with assuming that the collection of all the sets of models of finite subcollections (that is, the domain of the choice function) is a set of non void sets and the proof goes along.

What am I missing?


1 Answer 1


In fact, there is a general tactic we can use here to "reduce" classes to sets.

Rather than consider the class(es) of all models, look at the sets of models of minimal rank in the cumulative hierarchy - that is, for a theory $T$ let $$Mod^-(T)=\{M: M\models T\wedge\forall N,\forall\alpha\in\mathit{Ord}(N\models T\wedge N\in V_\alpha\implies M\in V_\alpha)\}.$$ Given a theory $T$, let $\mathbb{T}$ be the set of finite subtheories of $T$ and apply choice to $\{Mod^-(S): S\in\mathbb{T}\}$.

This "cut-off" technique is known as Scott's trick. It has a number of cute applications, including the definition of "cardinality" in the absence of the axiom of choice. Note that the axiom of foundation (or regularity) is crucial here.

Another approach is to use collection instead of replacement. The axiom scheme of collection says that, if $A$ is any set and $\varphi$ is any definable relation such that each $a\in A$ is $\varphi$-related to something, then there is a set $B$ such that $\forall a\in A\exists b\in B\varphi(a,b)$. Now take $A=\mathbb{T}$ as above and $\varphi(x,y)\equiv x\models y$. If you're given collection in your axiomatization then groovy! Otherwise, though, this isn't really different from the approach above since you need foundation to prove collection from replacement.

That said, in this particular case there's a different trick we can use to avoid appeal to foundation:

Downward Lowenheim-Skolem. Look at the set of models of $T$ with underlying set $\subseteq \aleph_0\cdot T$.

  • $\begingroup$ Thank you very much, I didn't know about Scott's trick, and it is really a clever method to deal with this kind of problems related to classes' size. I am then right about the fact that the reasoning I've seen is incomplete? $\endgroup$ Commented Jun 12 at 13:44
  • 2
    $\begingroup$ @LorenzoVanni Technically yes? But only if one really cares about how exactly this is done in ZFC specifically. If we treat model theory just like, say, ring theory (math without an explicit foundational system), then I'd call this a complete argument. Also there's another trick we can use in this specific case, which I've just added to my answer. $\endgroup$ Commented Jun 12 at 13:50

You must log in to answer this question.

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