# How many countable models of ZFC are there?

If we were looking at just an arbitrary binary relation on a countable set then I guess we would be looking at infinite graphs and those are uncountable. However, ZFC places an extra structure on our 'graphs', is this enough to reduce the number of models to countable (sort of how demanding that our functions from $\mathbb{R}$ to $\mathbb{R}$ are continuous reduces their number to $2^{\aleph_0}$)?

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I am assuming you are asking for the number of isomorphism classes of countable models of ZFC. There are no more than $2^{\aleph_0}$ binary relations on a countable set, and hence there are at most $2^{\aleph_0}$ isomorphism classes of countable models of ZFC.

On the other hand, even though I firmly believe in the consistency of ZFC, we could be living in a model of ZFC that believes that ZFC is inconsistent. In this case there would be no countable model of ZFC at all.

Hence we have to assume something in order to have countable models of ZFC at all. A slightly stronger assumption than the consistency of ZFC is that we have a countable transitive model $M$ of ZFC. We can extend countable transitive models of ZFC by forcing and also, because of the uniqueness of the Mostowski collapse, two transitive models are the isomorphic iff they are the same.

Now, if $\mathbb P\in M$ is any sufficiently non-trivial forcing notion (like Cohen forcing, for example), then we can construct $2^{\aleph_0}$ distinct $\mathbb P$-generic filters over $M$.
Each filter generates a countable transitive forcing extension of $M$. Now, this forcing extension might contain some of the other generic filters that we have constructed, but only countably many.
It follows that in total we get $2^{\aleph_0}$ different forcing extensions of $M$.

In other words, if there is one countable transitive model of ZFC, then there are exactly $2^{\aleph_0}$ pairwise non-isomorphic countable transitive models of ZFC.

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When you say you "believe in the consistency of ZFC", isn't that implicitly a belief about the world we live in? To me, saying without technical qualifications that ZFC is consistent means exactly that the world we do our mathematics in contains no finite proof of a contradiction from ZFC, and therefore the world we live in must contain a model of ZFC. – Henning Makholm Nov 24 '11 at 12:21
The question is, what do you base mathematics on? The infinite seems to be counterintuitive at times and hence we say what the axioms are that we use in our proofs. I like to use ZFC as the basis for mathematics, and on that foundation I can't prove that there is a countable model of ZFC. I think it is naive to claim that there is a world that we are doing mathematics in and we somehow intuitively know what a proof or a correct statement is. So what I believe is that I won't ever derive a contradiction from ZFC, but I don't usually want to use ZFC+"ZFC is consistent" to prove something. – Stefan Geschke Nov 24 '11 at 13:25