# Why Ryll-Nardzewski theorem fails for uncountable theories?

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

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Welcome to MO! A local custom is that every question should normally bear at least one arXiv-style tag, e.g. "lo.logic" or "ct.category-theory". I've just added one to yours. – Tom Leinster Dec 13 '12 at 22:15
Ah yes. Another custom is that you don't ask the same question simultaneously here and at math.stackexchange, as this potentially wastes people's time. Your question here may get closed for this reason. – Tom Leinster Dec 13 '12 at 23:03
Yet another custom is to accompany links with text so you know why you may or may not want to click through. – François G. Dorais Dec 13 '12 at 23:32

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).