Morley's categoricity theorem states:
If a first-order theory in a countable language is categorical in some uncountable cardinality, then it is categorical in all uncountable cardinalities.
As per the definition this only applies to uncountable cardinalities. But why doesn't the Löwenheim–Skolem theorem (assuming a countable language $L$) imply categoricity in all infinite cardinalities if a theory $T$ is $\kappa$-categorical, for $\kappa$ is some infinite cardinality?
An isomorphism $\pi\colon A\to B$ can be 'dumbed down' to 3 main points
- $A$ and $B$ are elementary equivalent (for $\varphi\in L$ then $M\models \varphi \iff N\models \varphi$)
- $A$ and $B$ have the same 'structure' ($n$-ary functions and $n$-ary relations are preserved)
- |$A$| = |$B$|, they have the same cardinality
For a model $M\models T$, the Löwenheim–Skolem theorem guarantees there are elementary extensions of $M$, $M\preceq N$ and $M\preceq U$. If $|M| =|N|$, then doesn't this imply $M$ and $N$ are isomorphic as it satisfies the 3 checkboxes for isomorphic structures? So if $T$ has $1$ isomorphism class in an infinite cardinality, then it should have $1$ for all infinite cardinalities (otherwise, a contradiction will occur where it isn't categorical in any infinite cardinality)?
This link seems to support this: Spectrum of a theory.
The Löwenheim-Skolem theorem shows that if $I(T,\kappa)$ is nonzero for one infinite cardinal then it is nonzero for all of them.
Where $I(T,K)$ is the number of isomorphism classes.
This application of the Löwenheim–Skolem theorem must be wrong in some way, otherwise we wouldn't have a need for Morley's categoricity theorem and we wouldn't have theories that are $\omega$-categorical. My intuition tell me that $M\preceq N$ and $M\preceq U$ isn't enough to guarantee an isomorphism, even if $|M| =|N|$. Thus, we require the categoricity theorem (which I believe does some stuff with saturated models?)
Best Answer
You write:
This "dumbed down" interpretation is imprecise and incorrect. I think it's giving you entirely the wrong intuition.
First: I don't know what is meant by "$A$ and $B$ are bijective". Bijectivity is a property of a function, not a pair of sets. If you mean that $A$ and $B$ have the same cardinality, that's true whenever $A\cong B$. But in order for $\pi\colon A\to B$ to be an isomorphism, you need to know that $\pi$ itself is a bijection, not just that there exists some bijection between $A$ and $B$.
Similarly, the fact that $A$ and $B$ are elementarily equivalent is a consequence of $A\cong B$. But it is not part of the definition of isomorphism, and it is certainly not sufficient for $A\cong B$. It is possible to have structures $A$ and $B$ with $A\equiv B$ (elementarily equivalent) and $|A| = |B|$ (same cardinality), but $A\not\cong B$.
The correct definition is that $\pi\colon A\to B$ is an isomorphism if $\pi$ is a bijection $A\to B$ which preserves the function symbols in $L$ and preserves and reflects the relation symbols in $L$. This is what you need to check to show that two structures are isomorphic, not the "check boxes" you listed.
I have no idea why you think the quote from Wikipedia supports your idea. It just says that if $T$ (in a countable language!) has some infinite model, then it has infinite models of all infinite cardinalities. It does not say anything about isomorphisms between these models. (Put another way: It's saying $I(T,\kappa)\geq 1$, not $I(T,\kappa) = 1$.)
Moral: Don't try to "dumb down" mathematical concepts. Without precise definitions, you won't get anywhere.