18
$\begingroup$

When I am reading through higher Set Theory books I am frequently met with statements such as '$V$ is a model of ZFC' or '$L$ is a model of ZFC' where $V$ is the Von Neumann Universe, and $L$ the Constructible Universe. For instance, in Jech's 'Set Theory' pg 176, in order to prove the consistency of the Axiom of Choice with ZF, he constructs $L$ and shows that it models the ZF axioms plus AC.

However isn't this strictly inaccurate as $V$ and $L$ are proper classes? For instance, by this very method we might as well take it as a $Theorem$ in ZFC that ZFC is consistent since $V$ models ZFC. However this is obviously impossible as ZFC cannot prove its own consistency. I highly doubt that Jech would make a mistake in such classic textbook, so I must be missing something.

How could we, for instance, show Con(ZF) $\implies$ Con(ZF + AC) without invoking the use of proper classes? I imagine, for instance, that we would start with some (set sized) model $M$ of ZFC and apply some sort of 'constructible universe' construction to $M$.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ In ZFC, the (formal) satisfaction relation $\models$ is only defined for set-models. Moreover, in order to deduce the consistency of a theory from the existence of a model, you need to have a set-model. For this reason, the fact that "V/L is a model of ZFC", which cannot be expressed in ZFC with a single statement, does not say anything about the consistency of ZFC. Here is a related question. $\endgroup$
    – Burak
    Commented Feb 15, 2018 at 21:13

3 Answers 3

Reset to default
21
$\begingroup$

What is shown in the cases you mention is not that the model is a model of ZFC, made as a single statement, but rather the scheme of statements that the model satisfies every individual axiom of ZFC, as a separate statement for each axiom.

The difference is between asserting "$L$ is a model of ZFC" and the scheme of statements "$L$ satisfies $\phi$" for every axiom $\phi$ of ZFC.

This difference means that from the scheme, you cannot deduce Con(ZFC).

For the proof that Con(ZF) implies Con(ZFC), one assumes Con(ZF), and so there is a set model $M$ of ZF. The $L$ of this model, which is a class in $M$ but a set for us in the meta-theory, is a model of ZFC, since it satisfies every individual axiom of ZFC. So we've got a model of ZFC, and thus Con(ZFC).

Improve this answer
$\endgroup$
12
  • 3
    $\begingroup$ I somewhat disagree with your last paragraph. The point of the proof is that you get a stronger result. You don't just assume there is a model of ZF and produce a model of ZFC. You actually prove a stronger result, that ZF itself proves any finite number of axioms of ZFC hold in L. Of course, that requires a meta-theoretic jump to conclude from Con(ZF) to Con(ZFC), but it is in fact more than saying "Assume there is a set model, produce another set model". $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 15, 2018 at 18:27
  • 1
    $\begingroup$ @Elie: Probably the old Kunen book; you might to check Felgner's Models of ZF set theory; and since judging by your name there's a good chance you can read Hebrew, you might want to see if Matti Rubin's notes on axiomatic set theory from the early 2000s are still online. $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 15, 2018 at 18:37
  • 1
    $\begingroup$ @AsafKaragila Well, of course the set models are closely related, in that the ZFC model is the $L$ of the ZF model, so whatever extra you are getting is gotten that way, because of the closeness of the models. I don't agree that your way of talking about it is stronger, since you are talking about a scheme of proofs, and this takes place in the meta-theory. $\endgroup$
    – Joel David Hamkins
    Commented Feb 15, 2018 at 18:38
  • 4
    $\begingroup$ @ElieBen-Shlomo I don't view Jech's argument as unrigorous, since I interpret him to be saying the same as what I said. $\endgroup$
    – Joel David Hamkins
    Commented Feb 15, 2018 at 18:39
  • 1
    $\begingroup$ @SimonHenry There is a little more to it than that, since after all, GBC is finitely axiomatizable, but we don't say that GBC implies Con(GBC) just because in GBC we can prove that the universe itself is a model of GBC. The point is that the argument that exists-a-model implies consistency requires having a satisfaction class or truth predicate for the model (since one proves by induction on proofs that the proofs are truth-preserving with respect to the satisfaction class, and contradictions are never true with respect to a truth predicate). $\endgroup$
    – Joel David Hamkins
    Commented Feb 15, 2018 at 21:34
11
$\begingroup$

Yes, that is true. But note that in its nature statements like $\operatorname{Con}(T)$ are meta-theoretic statements. So when we say that $V$ is a model of $\sf ZF$, we mean that in the meta-theory it is a model of $\sf ZF$.

This is often something which is not stressed enough in introductions to $V$ and relative consistency results: when we prove that $L$ is a model of $\sf ZFC$, we do not "just prove a meta-theoretic result", we in fact prove a stronger statement:

There is a formula $L$ in the language of set theory which defines a class that is provably transitive and contains all the ordinals, and for every axiom $\varphi$ of $\sf ZFC$, $\sf ZF\vdash\varphi^\it L$.

So not only you have this model, but in fact $\sf ZF$ itself prove that each axiom of $\sf ZFC$ holds in $L$.

Let me also share, in my first course on axiomatic set theory, which was given by the late Mati Rubin, we had proved that $\sf ZF-Reg$ and $\sf ZF$ are equiconsistent by practically proving that $\sf PRA$ proves that if there is a contradiction in $\sf ZF$, then there is one in $\sf ZF-Reg$.

Of course, the same can be done with $\sf ZF$ and $\sf ZFC$. And it is much more annoying than using the model theoretic approach. Sometimes with impunity when it comes to class models.

Improve this answer
$\endgroup$
3
  • $\begingroup$ One might say that Mati was overly pedantic in his teaching methods, by the way. I mean, I would say that, and I know a few other people who would say otherwise. $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 15, 2018 at 18:25
  • $\begingroup$ Can I ask to clarify something? Did you actually obtain a proof within PRA itself that Con(ZF−F) implies Con(ZF)? Or is it that you prove in ZFC that a proof exists in PRA (as seemingly stated in your answer)? Similarly, what is the weakest well-known system over which we have an actual proof for "Con(Z) implies Con(ZFC)", or at least can write a program to generate the proof? Thanks! $\endgroup$
    – user21820
    Commented Feb 16, 2018 at 14:32
  • $\begingroup$ The former. The point is that the process of moving from the code of $\varphi$ to the code of the relativization of $\varphi$ to WF (or V) is itself a primitive recursive process. So if you assume in PRA that ZF proves a contradiction, the process of transforming this to a proof of contradiction from ZF-Reg is recursive: just apply the relativization function finitely many times. $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 16, 2018 at 14:39
-7
$\begingroup$

The class $V$ of all sets is not a model of $ZFC$, because it is a proper class, not a set.

A model of $ZFC$ is a set (or small class) $U\in V$ which satisfies all the axioms of $ZFC$ when these axioms are restricted or relativized to $U$, even though $U$ does not include all the sets of the $ZFC$ universe.

If a model $U$ of $ZFC$ exists within the universe $V$ of $ZFC$, then its cardinality is "inaccessible" with respect to the universe $V$. Conversely, if an inaccessible cardinal exists in the universe $V$ of $ZFC$, then a small class, or set $U\in V$ exists, which is a model of $ZFC$.

The existence of a model $U$ within the universe $V$ of $ZFC$ implies that $ZFC$ is consistent. $ZFC$ would be inconsistent if its axioms could prove the existence of a model of itself within itself.

I fear that the meta-theorems, meta-theory, and other meta-language are unnecesssary and ill-defined, unless in the universe $V$ of $ZFC$ we are speaking of the properties of a possible model $U, U\in V, U\subsetneqq V$ where the original axioms of $ZFC$ have been restricted to a $U$.

Improve this answer
$\endgroup$
15
  • 5
    $\begingroup$ This does not add to the answers already appearing, with the exception that it adds mistakes (e.g. inaccessibility is only obtained when one restricts to second-order ZFC; being a model of ZFC need not be using the real $\in$ of $V$, but rather it can be using some arbitrary binary relation on the set $U$, etc.) $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 13, 2018 at 18:40
  • 2
    $\begingroup$ A model $(U,\epsilon)$ is isomorphic to one of the form $(U',\in)$ (a "standard model") iff $\epsilon$ is a well-founded relation. This is not necessarily satisfied for an arbitrary model. (And that's true even though the model believes itself to be well-founded. It can be mistaken about that!) Even worse: if ZF(C) is consistent, then $Con(ZFC)+$"There is no standard model" is also consistent, i.e. it is perfectly possible that, while there are models of ZFC, all of them are ill-founded. $\endgroup$
    – Johannes Hahn
    Commented Mar 13, 2018 at 19:49
  • 2
    $\begingroup$ A proof of the fact Johannes mentioned: let $M$ be a model of ZFC + Con(ZFC) + "There is a standard model of ZFC," let $\alpha$ be the thing $M$ thinks is the least height of a standard model of ZFC, and consider $L_\alpha$ ... $\endgroup$
    – Noah Schweber
    Commented Mar 13, 2018 at 20:21
  • 3
    $\begingroup$ Meanwhile, your claim "If a model U of ZFC exists within the universe V of ZFC, then its cardinality is "inaccessible" with respect to the universe V" is false as observed above. It's not even true for models of the form $V_\alpha$ - not every worldly cardinal is inaccessible (assuming they exist). I believe you're thinking of universes in the sense of category theory, but those (as Asaf says) are really about second-order ZFC in a precise (e.g. in a universe, powersets need to be computed correctly). $\endgroup$
    – Noah Schweber
    Commented Mar 13, 2018 at 20:24
  • 3
    $\begingroup$ Unfortunately, the latest comment, claiming that the membership relation of a model of ZFC has to be well-founded, is another error. I agree, though, with @JohannesHahn that the post doesn't need to be deleted. I've seen the same errors elsewhere, so it may be useful for some readers to see them here along with the comments and downvotes, which (I think) suffice to make it clear that they are errors. $\endgroup$
    – Andreas Blass
    Commented Mar 13, 2018 at 21:19

You must log in to answer this question.

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