Why Ryll-Nardzewski theorem fails for uncountable theories? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:21:24Z http://mathoverflow.net/feeds/question/116326 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116326/why-ryll-nardzewski-theorem-fails-for-uncountable-theories Why Ryll-Nardzewski theorem fails for uncountable theories? Camilo Arosemena 2012-12-13T22:05:53Z 2012-12-14T14:02:19Z <p>Ryll-Nardzewski theorems states that if $T$ is a countable complete theory, then $T$ is $\aleph_0$-categorical if and only if for every $n&lt;\omega$ there are only finitely many formulas $\varphi(x_1,\ldots,x_n)$ up to equivalence relative to $T$.</p> <p>$T$ is a countable theory if it can be built in a countable language.</p> <p>My question is: Is there a complete uncountable theory which is $\aleph_0$-categorical, but that for some $n&lt;\omega$ there are infinitely many formulas $\varphi(x_1,\ldots,x_n)$ up to equivalence relative to $T$?</p> <p>Thanks</p> http://mathoverflow.net/questions/116326/why-ryll-nardzewski-theorem-fails-for-uncountable-theories/116327#116327 Answer by Andreas Blass for Why Ryll-Nardzewski theorem fails for uncountable theories? Andreas Blass 2012-12-13T22:29:37Z 2012-12-13T22:29:37Z <p>Let $T$ be the complete theory of $\mathbb N$, with a binary predicate <code>$&lt;$</code> 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).</p>