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(1) must ZFC have an infinite model?

(2) if so, why?

(3) is it because of the replacement schema?

(4) if so, is it because we have a finite language and so we can only satisfy or describe countably many instances of replacement?

(5) assuming "yes" to question (1), am I right to say that by Skolem's Theorem, ZFC must have at least one countable model?

pichael
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    Yes, this follows by Löwenheim-Skolem because there are only countably many instances of replacement. – Qiaochu Yuan May 25 '12 at 00:57
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    Yes to all the questions? Sorry...I know there were a lot. One more: so standard formulations of ZFC canNOT have finite models? – pichael May 25 '12 at 01:04
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    Finite models can't possibly satisfy powerset. – Steven Stadnicki May 25 '12 at 01:06
  • Ah. Duh, so there are many reasons ZFC can't have a finite model. Thanks. – pichael May 25 '12 at 01:11
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    The title and the body do not quite match. The body asks whether, if ZFC has a model, must it have a countable model. The title asks whether any model of ZFC must be at least countable. – Arturo Magidin May 25 '12 at 02:37
  • I need to ask clearer questions. The title is what I want to know. The rest is my reasoning and related tangent questions. (1) must ZFC have an infinite model? (2) if so, why? (3) is it because of the replacement schema? (4) if so, is it because we have a finite language and so we can only satisfy or describe countably many instances of replacement? (5) assuming "yes" to question (1), am I right to say that by Skolem's Theorem, ZFC must have at least one countable model? Steven answered (2) nicely. – pichael May 25 '12 at 03:29
  • Please edit the question to make it clearer, instead of relegating the explanation to a comment :) – Mariano Suárez-Álvarez May 25 '12 at 03:37
  • @MarianoSuárez-Alvarez: Done. – pichael May 25 '12 at 03:55
  • For 5), don't you need to have a model $M$ within a model $V$ (a universe), to be able to show that there is a countable model $M^\prime$ in $V$, meaning that merely assuming that $ZFC$ has a model is not enough? – Egbert May 25 '12 at 09:33
  • @Egbert: According to my phil of math class (where I'm studying Skolem's Paradox) I just learned that if a countable first-order collection of sentences (theory) has an infinite model, then it has a countable model. Since ZFC in standard form is such a theory, it must have a countable model. I'd be happy to be corrected... – pichael May 25 '12 at 09:42
  • @pichael: A model is an element of a model for set theory, hence my comment. – Egbert May 25 '12 at 09:48
  • @Egbert: I see. Otherwise, how else could you say that the model M was countable? There needs to be a larger set that contains a bijection from the domain of M and what the universe takes to be w. Is that what you were saying? – pichael May 31 '12 at 01:34
  • You do not need another model $V$, at least not in the usual sense of "model" (a set with an interpretation of the symbol $\in$). It's true that the language and axioms of set theory can be interpreted in models, and then one can speak of other models inside these. But the same language and axioms can also be (and in fact usually are) understood as statements about all sets, not just those in some model. With this understanding, there is no need for a set to serve as the "universe" $V$. – Andreas Blass Oct 11 '15 at 22:43

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The result (existence of a countably infinite model) has absolutely nothing to do with the fact that ZFC is not finitely axiomatizable. Precisely the same result holds for any theory (over an at most countable language) that has an infinite model. In particular, precisely the same result holds for NBG.

There is indeed a countably infinite number of instances of the axiom scheme of replacement. That has no direct connection with the existence of a countably infinite model.

And ZFC can only have infinite models. One needs very little of ZFC for this, not the Axiom of Infinity, not even Powerset.

André Nicolas
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  • Can it be relevant in the following way: if ZFC is not finitely axiomatizable, then by LS Theorem, ZFC must have a countable model? But I think I see what you're getting at. That ZFC is not finitely axiomatizable (that fact) has nothing to do with proving the LS Theorem. – pichael May 25 '12 at 03:37
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    @pichael: If theory $T$ (over a first-order language with equality) has no models of cardinality $\gt N$, for some integer $N$, then $T$ is finitely axiomatizable. But the converse is (very) false. Finite axiomatizability does not imply a finite absolute bound on the size of models. Many finitely axiomatized theories, such as the theory of densely ordered sets, to take a simple example, have only infinite models. – André Nicolas May 25 '12 at 03:42
  • I'm new to this. Can I summarize as follows? If T has only finite models, then T is finitely axiomatizable. However, if T is finitely axiomatizable, this says nothing about what size models it can have. E.g. NBG is finitely axiomatizable, but it surely has infinite models (as you said). Thanks. – pichael May 25 '12 at 04:04
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    @pichael: If $T$ has only finite models, then in fact it cannot have arbitrarily large finite models. (It is easy to show using the Compactness Theorem that if $T$ has arbitrarily large finite models, then $T$ has an infinite model.) So if $T$ has only finite models, there is an absolute (finite) bound on the size of the models. And yes, finite axiomatizability says nothing about the size of the models. – André Nicolas May 25 '12 at 04:11
  • You wrote: "There is indeed a countably infinite number of instances of the axiom scheme of replacement. That has no direct connection with the existence of a countably infinite model." Wow, I didn't know that. What would it mean to satisfy replacement? For some reason I thought you'd have to satisfy all instances of it. But in writing this, I can see that it's not quite right, though I don't know why. Thanks! Sorry for all the questions, but it's very helpful. – pichael May 25 '12 at 04:22
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    @pichael: Replacement "says" that for any formula $\phi(x,y)$, if $\phi$ is "functional," then for any set $z$, the "range" of $\phi$ as $x$ ranges over $z$ is a set. (I am being more than a little vague, typing math in comments is unpleasant, no feedback, one only finds TeX errors after posting and getting a horrible mess.) So there is an instance of Replacement for every formula $\phi(x,y)$. There are infinitely many such formulas. – André Nicolas May 25 '12 at 04:30
  • what is minimally required to get only infinite models of ZFC? Empty-set and pairing? You generate all the naturals that way (though without declaring the existence of N)... – pichael May 25 '12 at 08:27
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    @pichael: That (plus extensionality) forces any model to be infinite, though it is far short of forcing the existence of an infinite set. Parenthetically, you get something fun by having almost the usual axioms, but the negation of Axiom of Infinity. You get a theory which in a technical sense is essentially equivalent to Peano Arithmetic. – André Nicolas May 25 '12 at 09:30
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If ZFC has a model it would have to be infinite. This can follow, as said from power set or the infinity axiom. Furthermore the language of set theory has only one binary relation $\in$, so any theory would be countable and therefore if there is an infinite model there would have to be a countably infinite model.

All this was said before, but I would like to add on an important point:

Even if $\frak M$ is a countable model of ZFC, internally it is a proper class. That is to say, there is no $f\in\frak M$ such that $f$ is a bijection between $\omega$ and $\frak M$.

This model, along with a function witnessing its countability live in a larger model of some strong-enough-theory (this larger model may be a class model).

Note that this has nothing to do with countability. Every set-model of ZFC would think of itself as a proper class, but we "know" (externally) that it is only a set, and if this set happens to live in a universe of ZFC then there is some function from an ordinal (which may be an element of this set-model) onto that model. This should be a hint of how complicated and convoluted infinite objects can get.

Asaf Karagila
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  • What is a class model? A model whose domain is not a set of any other model (or any other set)? – pichael May 25 '12 at 09:31
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    @pichael: Classes are collections which may not be sets. In ZFC given $V$ is a model of ZFC, if $V$ would think itself as a set, it would have to be an element of itself. This would lead a contradiction. Therefore $V$ do not know that it is a set, but rather a class. Class models are simply models which are not sets. – Asaf Karagila May 25 '12 at 09:33
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    @pichael: You may want to read this too. – Asaf Karagila May 25 '12 at 09:40
  • Man, that is so helpful. I've been on forums for weeks before I found this site, and you (and André Nicolas) have made everything so much clearer in just a few comments. Thanks a lot. – pichael May 25 '12 at 09:55
  • I was going to ask what the contradiction was, but can I put it like this: if V "saw itself" as a set, then the powerset axiom would apply (given our intuitions of set in axiomatic form). But then P (V) would be in V. This would contradict Cantor's Theorem (no function surjects a set S onto its powerset). – pichael May 25 '12 at 09:59
  • @pichael: This answer is also about the internal/external "set"-ness of a model. Your argument is true, you can also use other considerations, for example if $V$ was a set then using the axiom of regularity we can see that $V={x\mid x\notin x}$ which is not a set by Russell's paradox. – Asaf Karagila May 25 '12 at 10:01
  • I see. In the linked post, you wrote, "a model of set theory always sees itself as a proper class. The way it sees other models, and they see it can be either as a set or as a class." Is this the general resolution to Skolem's Paradox? -- From an internal perspective, a model might see a set A as uncountable; whereas from another (larger?) model's perspective, A is countable. (I'm hope I'm not reasoning hastily.) I should probably make another question for this... – pichael May 25 '12 at 10:21
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    @pichael: Skolem paradox is resolved by leaving the bijection of the countable model with $\omega$ outside the model. If we have a model of ZFC it might recognize a subclass of itself which is a model - but too large to be a set (if it contains all ordinals, for example). On the other hand, it might know about sets which are models of ZFC, and itself it might be a set in some larger universe. – Asaf Karagila May 25 '12 at 10:28
  • If I said that $M$ sees itself as a proper class, then could I say that it seems itself as being "more" than just merely uncountable? Or that it sees itself as being larger than all the uncountable sets it contains? – pichael Jun 11 '12 at 19:02
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    @pichael: Yes. Proper classes do not have cardinality. Their size is beyond that of any set which they contain. $\frak M$ thinks of itself as a proper class, so it is not only uncountable, it's a whole other size - much like we jump from finite (albeit very very large finite) numbers to the infinite. – Asaf Karagila Jun 11 '12 at 19:07