Skolem's paradox has been explained by the proposition that the notion of countability is not absolute in first-order logic. Intuitively, that makes sense to me, as a smaller model of ZFC might not be rich enough to talk about a bijection that a larger model understands. I'm wondering, however, if there are models of ZFC that disagree on the notion of being finite/infinite, so that a given set is thought of as infinite in one model, but finite in another? I suspect the answer is no, but I'm not sure how to prove it. Does anyone have any insight?

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    $\begingroup$ pretty sure a "standard model of ZFC" doesn't exist. $\endgroup$ – Dustan Levenstein Dec 14 '11 at 20:43
  • $\begingroup$ What do you call the one we like to work with? $\endgroup$ – Elchanan Solomon Dec 14 '11 at 20:44
  • $\begingroup$ There isn't one, as I understand it. There are many models of ZFC, some of which we have immediate reason to reject right away, and others for which it's not so clear. $\endgroup$ – Dustan Levenstein Dec 14 '11 at 20:50
  • $\begingroup$ I think usually one likes to assume when doing various relative consistency results that ZF is consistent (has a model). One usually arbitrarily fixes one model. $\endgroup$ – William Dec 14 '11 at 20:50
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    $\begingroup$ In ZFC, "infinite" (as in "not finite") is equivalent to "bijectable with a proper subset of itself". For a model of ZFC to "think" that a set is infinite, it has to produce a bijection between the set and a proper subset of itself. And to think that a set is finite, it has to have a bijection from the set to an $n\in\mathbb{N}$. $\endgroup$ – Arturo Magidin Dec 14 '11 at 20:51

The finite/infinite distinction is not absolute. Indeed, different models of set theory can think vastly different things about the sizes of a set whose elements they have in common.

For example, there can be models of set theory $M$ ad $N$, both satisfying all the ZFC axioms (assuming that ZFC is consistent), such that a set $x$ is thought to be finite in $M$, but $N$ thinks it has uncountably many elements.

To construct such an example, let $M$ be any model of ZFC, and let $N$ be the model of ZFC that arises from an internal utrapower of $M$ by a nonprincipal ultrafilter on $\mathbb{N}$. Thus, $N$ is a class of $M$, and for any set $x\in N$, the collection of objects that $N$ thinks are elements of $x$ is a set in $M$. So this is a sense in which the set exists in both models. Since the ultrafilter was nonprincipal, it is not difficult to see that $N$ will have a nonstandard collection of integers. So let $n$ be one of the nonstandard integers of $N$, and consider the set in $N$ consisting of the predecessors of $n$ in $N$. Since $N$ thinks that $n$ is a natural number, and doesn't see that it is nonstandard, it follows that $n$ thinks that $n$ has only finitely many predecessors. But $M$, on the other hand, can see that $n$ is nonstandard, and in fact arises from the ultrapower construction, and so $M$ can see that $n$ has continuum many predecessors. So the set of predecessors of $n$ is finite in $N$ and size continuum in $M$.

One can make much worsely behaved examples: for any $M\models$ZFC and any cardinal $\kappa$ of $M$, there is a model $N\models$ZFC constructible inside $M$, such that $N$ has a set that it thinks is finite, but which $M$ thinks has infinite size $\kappa$.

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  • $\begingroup$ Wow. Thanks for the excellent answer. To be honest, I'm rather surprised that this is true... $\endgroup$ – Elchanan Solomon Dec 14 '11 at 23:44
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    $\begingroup$ It shouldn't really be surprising, since we know already from the compactness theorem that "finite" is not first order expressible. For example, there is no first-order theory whose models are exactly the finite groups, or the finite fields, of the fields of non-zero characteristic. And the example I give above can be seen in that light. $\endgroup$ – JDH Dec 14 '11 at 23:47
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    $\begingroup$ Meanwhile, there are numerous positive results also: if the two models have the same ordinals, for example, or even just the same $\omega$, then they agree on the notion of finite for the sets they have in common. The method of forcing, in contrast, shows that they needn't agree on the notion of countability, even when they have all the same ordinals, since any set can become countable in a forcing extension. $\endgroup$ – JDH Dec 14 '11 at 23:49

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