An $\text{interpretation}$ of a theory consists of

A domain of discourse $\mathcal U$, usually required to be non-empty.

For every constant symbol, an element of $\mathcal U$ as its interpretation.

For every $n$-ary function symbol $f$, an $n$-ary function $f :\mathcal U^n\to\mathcal U $.

For every $n$-ary predicate symbol, an $n$-ary relation on $\mathcal U$ as its interpretation (that is, a subset of $\mathcal U^n$).

My question is:

What does a model of $\sf ZFC$ look like? Given that (via $\sf ZFC$) we're trying to axiomatize set theory, how can we describe what a "domain of discourse", a "function", a "relation" mean (as this are (naive?) set theory concepts)?

Could someone explain this in relatively simple terms?


The question seems to come down to a common confusion: how can a model of ZFC be a set, if we want to use ZFC to study sets? Here we are looking at ZFC as an "object theory" - a theory that we are studying, and whose models we are interested in. There is a concept that many students have that we should study ZFC starting with some theory weaker than ZFC.

The answer is, essentially, that things don't work that way. To study "models", we need to work in a metatheory that already has some concept of set or collection. The "metatheory" here is the theory that we use, as a tool, to study the object theory.

There are many options for such a metatheory. One option would be ZFC itself, except that (by the second incompleteness theorem), ZFC can't prove that there is a model of ZFC. We could use ZFC + "there is a model of ZFC" as our metatheory, or we could use a stronger set theory like Morse-Kelley set theory. These are strong enough to study models of ZFC. There are other options, as well.

If we want to work with a "weak" metatheory, which is not strong enough to talk about models, we can only study provability in ZFC, not models per se.

In much of the actual study of set theory, and many set theory textbooks, the metatheory is left unspecified, which tends to perpetuate the confusion about what metatheory is being used. Sometimes the metatheory is left intentionally informal, based on our naive conceptions of sets and collections. Sometimes, models of ZFC are explicitly studied in the metatheory of ZFC + "there is a model of ZFC".

This is true throughout mathematical logic, though. In the study mathematical logic, we apply mathematical techniques to study logic. This means that we apply the same kind of reasoning - including sets, functions, etc. - that we use to study all other kinds of mathematics like algebra, analysis, etc. Occasionally, we do work in weaker metatheories, but in most cases the metatheory we use is ZFC or stronger.

The general idea is that, by studying ZFC as an object theory, we learn more about the nature of sets. This informs us more about our set-theoretic metatheory. We can then use this new insight about the nature of sets to learn more about ZFC, in a kind of cyclical process that has some analogues in textual analysis.

  • $\begingroup$ Could you elaborate a bit on the part about using $\sf ZFC$ or $\sf ZFC + Exists\, a\, model\, of\, ZFC$ as the metatheory? I don't really understad how we can "talk about $\sf ZFC$ from $\sf ZFC$ itself". $\endgroup$ – YoTengoUnLCD Mar 29 '16 at 1:07
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    $\begingroup$ @YoTengoUnLCD It's really no different from proving that if PA is consistent, then it's incomplete . . . from inside PA. One of the major points of logic is that there's no general reason you can't talk about a theory inside that theory (of course, there may be specific reasons, depending on the theory, but the point stands). You can't get something from nothing: in order to use model-theoretic ideas rigorously, you need to first be working inside some theory in which those ideas make sense (or, be a Platonist). $\endgroup$ – Noah Schweber Mar 29 '16 at 1:46
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    $\begingroup$ Suppose we are using ZFC as our metatheory. We can use this to define "models" of first-order theories. And we can identify a particular first-order theory, ZFC. So, we can study models that happen to be models of ZFC, just like we would study models of any other theory. The only small catch is that, if we are using ZFC as our tool to study models, we won't be able to prove there are any models of ZFC, so we will need to include an additional assumption in the metatheory that there are models of ZFC (that is, the assumption that ZFC is consistent). $\endgroup$ – Carl Mummert Mar 29 '16 at 10:10
  • $\begingroup$ @YoTengoUnLCD Notice that we can talk, inside ZFC, of formulas and set of formulas (by using sets to code formulas). Since ZFC is just a set of formulas, we can talk, inside ZFC, about ZFC. $\endgroup$ – Nagase Apr 3 '16 at 18:37

A model of $\mathsf{ZFC}$ is an appropriate structure that satisfies the axioms of $\mathsf{ZFC}$.

The "appropriate structure" is essentially what you quote: but in the case of the language of set theory, which just has one relation ($\in$, i.e., belonging), it simplifies a lot: a model of the language of set theory is just a set $M$ (the "universe of discourse") together with a binary relation $\epsilon$ (destined to correspond to the $\in$ relation of set theory).

However, a model of $\mathsf{ZFC}$ is much more than just a model of the language of set theory: for this, the relation $\epsilon$ has to satisfy a lot of extra constraints, namely the axioms of $\mathsf{ZFC}$. In other words, take every axiom of $\mathsf{ZFC}$, replace every $\forall x$ by $\forall x\in M$ and every $\exists x$ by $\exists x\in M$ and every $\in$ by $\epsilon$, and require them all to hold (for example, the axiom of the empty set of $\mathsf{ZFC}$ says that $\exists x \forall y \neg y\in x$ so we require that there exists $x\in M$ such that for no $y\in M$ does $y\mathbin{\epsilon} x$ hold).

Set theorists are usually interested in "transitive" models of $\mathsf{ZFC}$. A transitive model of $\mathsf{ZFC}$ is a set $M$ which is transitive (i.e., if $x\in M$ and $z\in x$ then $z\in M$) such that the $\in$ relation on $M$ satisfies all the axioms of $\mathsf{ZFC}$. What this last part means is that if you take all the axioms of $\mathsf{ZFC}$ and replace every $\forall x$ by $\forall x\in M$ and every $\exists x$ by $\exists x\in M$, they are still true (we say that the axioms "hold in $M$"). In other words, the structure is $M$ with the relation ${\in}|_{M\times M}$ (i.e., the set of $(z,x)\in M^2$ such that $z\in x$), and we again require the axioms of $\mathsf{ZFC}$ hold.

In practice, being a transitive model of $\mathsf{ZFC}$ means intuitively that $M$ is a set "large enough" and "complete enough" so that all the operations of $\mathsf{ZFC}$ (e.g., taking unions, taking power sets) can be done in $M$. This is a bit simplified, because some operations are not "absolute" (e.g., the powerset in $M$ might not be the real power set) whereas others are (e.g., ordered pairs in $M$ will be real ordered pairs), but it should at least give some kind of idea.

Now $\mathsf{ZFC}$ cannot show that models exist (whether transitive or not), so examples of them can't be exhibited (unless we assume some extra axioms like the existence of inaccessible cardinals), but we can get some intuition of what they would look like by considering near-models. The simplest one consists of taking the empty set and repeatedly taking power sets through the ordinals until we get "far enough" that it looks like the universe of all sets (which can be obtained by doing this through all the ordinals, but this doesn't give a set). For example, if we do this just $\omega$ times we get a model of Zermelo set theory (i.e., $\mathsf{ZFC}$ without replacement).

  • $\begingroup$ Hmm, the thing I don't understand is really what the "universe of discourse" is: Is it a "set"? Isn't that what we're trying to define (in an axiomatic manner) when we talk about $\sf ZFC$? $\endgroup$ – YoTengoUnLCD Mar 28 '16 at 23:57
  • $\begingroup$ The "universe of discourse" (nobody says that, really) is the set of all elements of the model (normally it's called something like $M$ and that's the notation I used), i.e., the set-of-those-guys-that-are-trying-to-act-as-sets. So, yes, it's a set. And so, yes, the models of $\mathsf{ZFC}$ are studied from within $\mathsf{ZFC}$ (indeed, all of mathematics is thought to be done within $\mathsf{ZFC}$). But $\mathsf{ZFC}$ doesn't try to "define" what a set is (that would be futile), just lay out rules on how to manipulate them. (contd.) $\endgroup$ – Gro-Tsen Mar 29 '16 at 11:27
  • $\begingroup$ @YoTengoUnLCD (contd.) I think the problem is that you didn't tell us how much background in set theory you have. A model of $\mathsf{ZFC}$ is a complex set-theoretical structure: before you try to understand that, the first thing is to understand more basic structures like ordinals, transitive sets, models of simpler theories, etc., and get a good working knowledge of $\mathsf{ZFC}$. Understanding $\mathsf{ZFC}$ comes before, not after understanding models of $\mathsf{ZFC}$. So I implicitly, perhaps wrongly, assumed that you knew all that. $\endgroup$ – Gro-Tsen Mar 29 '16 at 11:31

To make model theory go, you have to start out by assuming you already have some fixed notion of what a set is. You can then consider models of a theory, and in particular one thing you can do is plug in the axioms of set theory itself and see what happens. But you can't quite consider your underlying notion of sets as being "a model of ZFC," because as you've noticed that would be circular. (It does induce a model of ZFC, where a "set" is just a set and a "function" is just a function and so forth.)

However, there's a redeeming feature here in the form of Godel's completeness theorem: any statement true in every model of a theory can be deduced from the axioms of the theory by a step-by-step proof. So, assuming the ZFC axioms are consistent, it doesn't matter what notion of sets were fix.


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