#notes on this
As a consequence of Cantor’s theorem, there exists a set of uncountable cardinality
Löwenheim-Skolem theorem: if a first-order theory has an infinite model, then it has a countable model
Seems to be a paradox!
Let $T$ denote some standard axiomatization of set theory
Assuming $T$ has a model, Löwenheim-Skolem theorem asserts existence of some countable model $\mathbf M$ for $T$
Let $\Omega(x)$ be the assertion, expressed using first-order logic, that $x$ is uncountable
By Cantor’s theorem we have that $T \vdash \exists m\ \Omega(m)$ and so there is a $m \in \mathbf M$ where $\mathbf M \vDash \Omega(m)$
But since $\mathbf M$ is countable then there are only countably many $n \in \mathbf M$ such that $n \in m$!
Resolution
When evaluating the semantics of a first-order sentence such as $\Omega(m)$, we must recognize that the meaning of first-order language is relative to the model in which we interpret it
For instance, $\forall$- and $\exists$- bindings quantify exactly over the domain of $\mathbf M$ rather than any other domain that we had in mind. Likewise, the inclusion relation $\in$ need not represent “genuine” inclusion but need only be some binary relation between elements of $\mathbf M$ which satisfies the appropriate properties
Though we have some intention for what first-order sentences “ought” to mean, $\mathbf M$ has no mandate to respect that intention.
As humans, we have a certain sense and intuition around what it means for a set to be uncountable. Guided by that sense and intuition, we write the definition of $\Omega$ as a first-order expression of uncountability
But this definition was predicated on a certain understanding of first-order logic which $\mathbf M$ need not follow
As such, the fact that $\mathbf M \vDash \Omega(m)$ may have a meaning far differing from $m$ being uncountable.
(For what it’s worth, I think the article does a better job explaining this than me. Go read it! §2.2)
Details on how $\mathbf M$ has “strange interpretations”
Transitive models
A model $\mathbf X$ is called transitive if (1) every member of $\mathbf X$ is a set and any every member of a member of $\mathbf X$ is also a member of $\mathbf X$; and (2) the relation $\in$ in $\mathbf X$ is the same as $\in$ in the language being used to describe $\mathbf X$
Stronger, if $\mathbf X$ is a transitive model and if $f, m \in \mathbf X$ then $\mathbf X \vDash f : \omega \to m \text{ is bijective}"$ if and only if $f$ is a bijection in the metalanguage
Conceptually, in a transitive model the interpretations of membership and bijectivity are “correct” (relative to the metalanguage)
An example of how a transitive model can “misinterpret” axioms
Take $\mathbf Z$ a transitive model of ZFC
Powerset axiom is $\forall x\ \exists y\ \forall z \mid z \subseteq x \leftrightarrow z \in y$
But $\forall$ only quantifies over members of $\mathbf Z$
So for a given $x$ its powerset $y$ only contains sets $z$ where $z \subseteq x \land z \in \mathbf Z$
If $\mathbf Z$ is countable, then $x$ contains at most a countable number of sets
(In §2.4¶4 they say that the source Resnik 1966 follows “this phenomenon through the case of the real numbers”. Seems interesting! #onwards)
In non-transitive models we get even more ways to misinterpret!
Assuming is $\mathbf M$ is transitive, where can it differ from the metalanguage for Skolem’s paradox to arise?
$\mathbf M$ can differ on $\exists$
The form of $\Omega$ is $\Omega(m) = \neg \exists f : \omega \to m \text{ bijective}$
Since they agree on “bijective”, then they must differ on "$\exists$"
This is exactly the case. Since $\mathbf M$ is countable, then $m = \{ n \mid \mathbf M \vDash n \in m \}$ is countable, so there exists at least one bijection $\tilde f : \omega \to m$. But any such bijection is not a member of $\mathbf M$, so $\mathbf M$ considers $m$ to be uncountable. That is, whereas $\exists$ in the metalanguage can “see” $\tilde f$, $\exists$ in $\mathbf M$ cannot.
Skolem’s paradox is intimately tied to first-order logic
Does not arise in second-order settings
Standard proofs fail in constructivist settings
Personally, it seems that this situation suggests that the semantics of first-order logic are not strong enough. If we are confident that we know what uncountability means, and we are confident that $\Omega$ is a proper expression of uncountability in first-order logic, then it seems that any appropriately-faithful interpretation of $\Omega$ would demand an uncountable model.
The source also includes some discussion of philosophical consequences of Skolem’s paradox (§3) which I chose to skip/skim for now (#onwards?)

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