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The well-known Skolem construction yields a countable model of ZFC,elemetarily equivalent to the universe of sets $V$.Why this construction is not a proof of existence of models of ZFC,as such proofs cannot exist by Godel's theorems?

user122424
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    Gödel's theorem only says that models of ZFC cannot be constructed within ZFC. The Skolem construction can be viewed in several ways, but if it is applied to $V$ then it cannot be formalized in ZFC, as ZFC does not have enough strength to construct the elementary diagram of $V$, which is needed to begin that construction. – Carl Mummert Sep 05 '15 at 12:12
  • Closely related: http://math.stackexchange.com/q/1368436/630 . I think that, if this question was clarified, it would likely be a duplicate of that question. – Carl Mummert Sep 05 '15 at 14:30

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We can have a model $V$ of ZFC, which is uncountable, but which does not think there exist any countable models of ZFC: Externally, Lowenheim-Skolem gives us a countable elementary submodel $M\subseteq V$, but there's no reason to believe $M\in V$, so $V$ might not "see" that such an $M$ exists at all.

Note also that $V$ may "think" that $V\not\models ZFC$! This is because if $V$ has nonstandard natural numbers, this will yield nonstandard axioms of what $V$ thinks is $ZFC$, and it's possible that one of these might be false in $V$ (actually, this phrase doesn't make sense - the sentence in question will be nonstandard, so infinite in length, so it's not even clear what "false in $V$" means - but this can be made precise and true with some effort). So even ignoring the fact that $V$ can't construct its own theory, $V$ might think "$ZFC$ is false," even though $V$ satisfies every actual axiom of $ZFC$. (Lest you think this is a dodge, by the way, note that nonstandard natural numbers are crucial to Godel's theorem, so this is really very much in the right spirit.)

Noah Schweber
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