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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?

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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 0:57
    
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 1:04
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Finite models can't possibly satisfy powerset. –  Steven Stadnicki May 25 '12 at 1:06
    
Ah. Duh, so there are many reasons ZFC can't have a finite model. Thanks. –  pichael May 25 '12 at 1:11
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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 2:37

2 Answers 2

up vote 2 down vote accepted

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.

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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 9:31
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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 9:33
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@pichael: You may want to read this too. –  Asaf Karagila May 25 '12 at 9:40
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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
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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

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.

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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 3:37
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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 3:42
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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 4:11
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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 4:30
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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 9:30

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