Ryll-Nardzewski theorems states that if $T$ is a countable complete theory, then $T$ is $\aleph\_0$-categorical if and only if for every $n<\omega$ there are only finitely many formulas $\varphi(x\_1,\ldots,x\_n)$ up to equivalence relative to $T$.


$T$ is a countable theory if it can be built in a countable language.


My question is: Is there a complete uncountable theory which is $\aleph\_0$-categorical, but that for some $n<\omega$ there are infinitely many formulas $\varphi(x\_1,\ldots,x\_n)$ up to equivalence relative to $T$?


Thanks



Why Ryll-Nardzewski theorem fails for uncountable theories?
Let $T$ be the complete theory of $\mathbb N$, with a binary predicate $<$ for the standard ordering, unary predicates for all subsets of $\mathbb N$, and constants for all the elements of $\mathbb N$. This clearly has uncountably many inequivalent unary formulas. I claim that its standard model is, up to isomorphism, the only countable model. The main point in the proof is that there exist continuum many infinite subsets $A\_i$ of $\mathbb N$ any two of which have a finite intersection. If $M$ were a nonstandard model, it would have an element $c$ greater (in the sense of $M$) than all the standard numbers; each $A\_i$ woujld have an element $a\_i$ greater than $c$ (because the sentence saying $A\_i$ has arbitrarily large elements is in $T$); and these $a\_i$'s would all be distinct (because each $A\_i\cap A\_j$ has a standard upper bound).


