Question regarding Skolem's Paradox

Question

I am asking this question because I am presently confused as to the main, seemingly paradoxical aspect of Skolem's Paradox.
As I understand it, the paradox is formulated thus: The (Downward) Löwenheim-Skolem Theorem says (roughly) that any set of sentences expressed in a language of cardinality $\kappa \geq \aleph_0$ with a model of cardinality $\lambda > \kappa$ has a model for any cardinality $\kappa \leq \kappa' < \lambda$. In particular, any set of sentences (expressed in a countable language) has a countable model. Thus, the set of sentences which constitute the axioms of ZF Set Theory possess a countable model, and this (supposedly) induces a paradoxical state of affairs as one such axiom within the theory claims the existence of an uncountable set.
But, and forgive me if I am excessively thick, why is this taken to be paradoxical? What is countable is the model's universe, not the sets contained therein, and the axiom declaring the existence of an uncountable set refers only to those elements within the universe, no? The argument for the seemingly paradoxical character of this model's existence appears to claim thus: as this model (let us call it $\mathcal M$) is countable (i.e., it only possesses countably many elements), there are only countably many $m \in |\mathcal M|$ such that $\mathcal M \models \bar m \in \mathrm{\textbf m}$ (here $\mathrm{\textbf m}^\mathcal M$ is the element in $|\mathcal M|$ satisfying the sentence $\phi(x)$ being interpreted as '$x$ is uncountable'), which would seem to suggest $\mathrm{\textbf m}^\mathcal M$ is countable.
But, why is that suggestion so? As in, why does the fact that there are countably many elements satisfying belonging to that highlighted set within the considered universe imply —so it is claimed— the set itself must be countable? Clearly, the set itself, when interpreted, can have a 'life' or existence of its own, with actual elements outside the considered universe, no? That there are only finitely many $n \in |\mathcal N|$ satisfying $\mathcal N \models \bar n \in \bar{\mathbb R}$ given $|\mathcal N| = \{\mathbb R, 0, 1, 2\}$ does not mean $\mathbb R$ itself is finite, no?
Now I know the above given set clearly does not satisfy the ZF axioms, but my question is whether I am simply missing some essential fact that renders the before given argument obvious, because presently I simply do not understand why the existence of only countably many terms belonging to the 'uncountable' set within the countable model's unvierse should indicate anything about the properties and cardinality of the actual, interpreted highlighted set.
I thank everyone in advance for all answers and help.

Answer 1

The issue is not that a countable model shows that $\mathbf{m}^\mathscr{M}$ "really is" countable - as you say it doesn't. The issue is that (depending on your philosophical presuppositions) the existence of a countable model for ZFC suggests that someone who accepts the literal truth of ZFC could do so while maintaining that there are in fact only countably many mathematical objects, which seems paradoxical since intuitively it seems ZFC has statements pretty explicitly stating that there are uncountably many such objects. Or to phrase the issue as a question, we generally take it that the 'standard model' of ZFC is the "correct" interpretation and involves uncountably many objects, but what makes this interpretation correct, when there are models which validate all the statments of ZFC which don't involve uncountably many objects?

Answer 2

There are a few issues that need to be clarified. The explanation you offer (that some sets of the model are countable in the model only because many of their elements lie "outside the model") doesn't actually resolve the paradox. The usual explanation is different.

Transitive models: The proof of Skolem's theorem seems to show that there is a countable transitive standard model of ZFC. This is a model $(M,\in)$ with $M$ is a countable set, where $\in$ is the same as the $\in$ relation out in the "real universe" $V$. (That's the "standard" part.) Also, if $x\in y\in M$, then $x\in M$. (That's the "transitive part.) So a set $y$ in $M$ doesn't have any elements that are "outside of $M$", and all the sets of $M$ are countable "in the real universe $V$".

Usual explanation: Say that $y$ is a set that "$M$ thinks" is uncountable, i.e., $$M\models \neg\exists f(f\text{ is a bijection between }\omega\text{ and }y)$$ The part in English can be expressed formally in the language of ZFC. So there is no bijection inside $M$, but since $y$ is countable in $V$, there is a bijection in $V$.

This is the real issue. Even if the model weren't transitive, and $y\cap M$ were countable while $y$ wasn't, $M$ would still have sets it "thinks" are uncountable, although they are countable in the "real world". (Here, $y\cap M$ would be such a set.) Nowadays one says that countability is not absolute. (Gödel introduced this notion.)

Axiom SM: Axiom SM (SM=Standard Model) is exactly what I said the Skolem proof "seems to show". The argument cannot be carried out formally inside ZFC, because the notion of satisfaction (or truth) in $V$ cannot be formalized in the language of ZFC. But informally, the Skolem is rather convincing. If we believe that "$V\models\varphi$" always has meaning, then it does seems like you ought to able to carry out the Skolem construction to produce a countable standard transitive model of ZFC.

Workaround: Although the Skolem proof cannot be formally carried out inside ZFC, you can carry it out for any finite fragment of ZFC (i.e., any finite subset of the axioms). The proof that the power set of $\omega$ is uncountable uses only finitely many axioms (like any proof), and so there is a standard transitive model of this ZFC fragment. Skolem's paradox applies to this fragment, as does the usual explanation.

There's a lot more say: look up absoluteness, and also Mostowski collapse, for more insight into this issue.