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I've run across the statement, "The existence of a strongly inaccessible cardinal implies the consistency of ZFC" in several places (Cohen's Set Theory and the Continuum Hypothesis p. 80, for one). His argument is that the set of all sets of rank less than that large cardinal form a model. It seems to me that since we have access to all of those sets without the large cardinal, we could make a model of ZFC without the large cardinal axiom, and thus prove ZFC consistent. This, of course, is false. So, how does the large cardinal provide a model of ZFC that doesn't exist without the large cardinal?

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Without the large cardinal, how do you build your model? Use "the set of all sets"? That doesn't exist. –  George Lowther Nov 8 '10 at 20:32

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A model of ZFC is a set $E$ together with a binary relation $R$ satisfying the axioms of ZFC (and we can always if we can suppose that $E$ is transitive and that $R=\in|_{E\times E}$ then the model is said standard). So we not only need to find a class of sets satisfying the axioms of ZFC, but this class must be itself a set.

If you have a strongly inaccessible cardinal $\kappa$, then there is a set $V_\kappa$ which is a (standard) model of ZFC, hence ZFC is consistent. But without the existence of $\kappa$, even if you could consider the class of all sets of rank “less than $\kappa$”, this would not be a set and the theorem ”there is a model implies the consistency of ZFC” would not be applicable (because there is no model).

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I think my problem is that the large cardinal axiom gives you a new set, but then we don't really use it in the model. Thinking about it more, it does imply the existence of more than one set, some of them smaller than $\kappa$. –  Tim Lewandowski Nov 8 '10 at 20:56
    
In order to build the set of all sets of rank $<\kappa$, you will need transfinite induction on the ordinal $\kappa$ (or something equivalent), so $\kappa$ needs to be a set. You can also see that if $V_\kappa$ is a set, $\kappa$ must also be a set, because it is the set of all ordinals belonging to $V_\kappa$. –  Guillaume Brunerie Nov 8 '10 at 21:32
    
@Guillaume: You are correct, but you need to be a bit careful here. Even if there are no inaccessible cardinals, there may be many $\lambda$ such that $V_\lambda$ is a model of set theory. –  Andres Caicedo Nov 8 '10 at 23:38
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@Tim: no, the single new set given by the large cardinal itself really is key! We don't use it in the model; we use it as the underlying set of the model. –  Peter LeFanu Lumsdaine Nov 9 '10 at 14:51

Just one clarification to Guillaume's answer (his answer has been edited by now): Yes, a model of ZFC is a set $E$ together with a binary relation $R$ on $E$ such that $(E,R)$ satisfies ZFC. The relation $R$ is extensional, that is every $x\in E$ is determined by the collection of $y\in E$ with $yRx$, because $(E,R)$ satisfies the axiom of extensionality.
However, even though $(E,R)$ satisfies the axiom of regularity, the relation $R$ need not be well-founded. $(E,R)$ thinks that $R$ is well-founded, but the real world knows an infinite $R$-decreasing chain in $E$. In this situation, $(E,R)$ is a non-standard model of ZFC which is clearly not isomorphic to a transitive model with the $\in$-relation.

However, an inaccessible cardinal $\kappa$ actually gives you a transitive model of ZFC, and this is more than just having any model of ZFC. (And most approaches to forcing would like to have a transitive set that is a model of ZFC (with the binary relation being the usual $\in$-relation). But there are reasons why we can pretend that we have a transitive set model of ZFC even if we don't really have one.) Now, it is not the case that the inaccessibility of $\kappa$ somehow miraculously gives you a new set that is a model of ZFC. It is more the case that $\kappa$ is so large that if you cut off the universe at $\kappa$, this initial part of the full set-theoretic universe already satisfies ZFC.
I.e., from the point of view of $V_\kappa$, $\kappa$ already looks like the class of all ordinals.

What makes this work is the fact that an inaccessible cardinal has excellent closure properties. $\forall\lambda<\kappa(2^\lambda<\kappa)$ guarantees that every element of $V_\kappa$ is of size $<\kappa$ and the regularity of $\kappa$ now gives you the axiom of replacement. The other axioms are rather easily satisfied in this context.

Note, however, that an inaccesible cardinal can cease to be inaccessible (or even a cardinal) after enlarging the universe (for example by forcing). The $V_\kappa$ of the ground model (the small original universe) is still a model of ZFC, but the $V_\kappa$ of the enlarge universe (where $\kappa$ is not inaccessible anymore) need not be a model of ZFC.

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Yes, not every model of ZFC is standard, sorry for this mistake (see for example [this MO question][1]). I have corrected my post, thank you. [1] : mathoverflow.net/questions/14622/… –  Guillaume Brunerie Nov 8 '10 at 23:03

I'd like to put this as a comment, but I don't have sufficient rights: It might be interesting to point out that the existence of a transitive model (like the one we get using an inaccessible, mentioned above) does not just imply $Con(ZF)$ but even $Con(ZF+Con(ZF))$.

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Why not wait a bit more until you have the proper reputation to comment instead? You're not that far away... –  Asaf Karagila Aug 15 '11 at 11:42
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Yes, this has been pointed out a few times in answers to other questions here at MO: The existence of an $\omega$-model, i.e., one whose $\omega$ is isomorphic to the $\omega$ of the underlying universe, implies Con(ZFC), Con(ZFC+Con(ZFC)), Con(ZFC+Con(...)), etc. In turn, the existence of a transitive model implies the existence of an $\omega$-model, and an omega-model of "there is an $\omega$-model" etc. The existence of a transitive model is strictly weaker than the existence of a $\lambda$ such that $V_\lambda$ is a model, and this is strictly weaker than the existence of an inaccessible. –  Andres Caicedo Aug 15 '11 at 11:47
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Asaf, it is fine for ftonti to post a comment like this, when lacking sufficient rep to post a proper comment; my understanding is that our MO policy specifically allows this, and when the comment is worthwhile, as here, then this is often a good way to earn the necessary reputation. –  Joel David Hamkins Aug 15 '11 at 11:58
    
I agree with Joel. So, +1 to ftonti. –  Todd Trimble Aug 15 '11 at 12:08
    
Joel, Todd: Very well. I was not aware of this social convention, and the post begins with "I wanted to post a comment, but I don't have enough reputation.", now ftonti has enough reputation so the next time he could choose to write an answer or leave a comment. :-) –  Asaf Karagila Aug 15 '11 at 12:29

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