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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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    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" $V$.2015-10-11

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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.

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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.2012-06-11
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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.

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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.2012-05-25