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Why can't a (standard?) model of ZFC "say of itself" that it is countable?

That is, why is there no bijection $f$ ∈ between and $\omega^$?

(I've read that it fails regularity, or even without regularity we get Cantor's paradox. But a direct answer to the question would be most helpful.)

Thanks.

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    If ZFC is inconsistent, one could prove that that there is such an $f$. Conversely, if one could prove there is such an $f$, then ZFC would be inconsistent.2012-06-08

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If $\frak M$ is a [standard] model of ZFC then we know several things:

  1. $\frak M$ thinks that $\{x\mid x\notin x\}=\frak M$ is not a set.
  2. If $f\in\frak M$ and $\frak M$ thinks that $f$ is a function, then the range of $f$ is a set in $\frak M$. (This is an instance of the axiom schema of replacement)
  3. $\omega^\frak M$ is a set in $\frak M$.

These combined tell us that if $\frak M$ knew about a function from its own $\omega$ onto its entire universe it would violate the second thing in the list above, and will not be a model of ZFC.

In the case of a standard model, we can also have the contradiction from the fact that if such $f$ was in $\frak M$ then we would have $\frak M\in\frak M$ and that, as you said, would contradict the axiom of regularity (both in the universe and in $\frak M$) but this is in addition to the above argument.

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    @Andrew: I'm not quite sure how to help here. I think most set theorists just developed their intuion through use, rather than reading it somewhere. The key point is that you just need to grok the idea of what it means to prove something, and what it means from$a$semantic point of view. I'm afraid that I can't help much more than that.2018-07-08
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If you can prove the existence of uncountable sets in ZFC, and if you can also prove a proposition saying the model is countable, then you have a contradiction in ZFC. With countable models of ZFC, the statement that a set is uncountable is true in the model if the set is not "internally" countable, i.e. no enumeration of the set is a member of the model.

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    Yes, I agree. It does happen that models in which CH holds. But you say that "It could happen that CH is true" and not "It does happen that CH is true". I honestly believe that we are waaaaaay over the point that we both fully understand what the other meant and now simply continue to comment out of some inertia. I'll stop now.2012-06-09