Is there a complete theory that is not $\kappa$ categorical for any $\kappa$?

Question

One of the theorem I have studied states that:

Let $\kappa \ge |L|$, If $\Sigma$ is a $\kappa$-categorical theory of $L$, and every model of $\Sigma$ is infinite, $\Sigma$ is complete.

I wondered why this is not if and only if theorem, but couldn't find a counter-example.

Later on, in a summary I found that $Th(\mathbb N)$ is not categorical for any $\kappa \ge \aleph_0$, I tried proving it but to no avail.

I am looking for a proof on why $Th(\mathbb N)$ is not categorical for any $\kappa$ or a complete theory $\mathbb T$ that is not categorical for any $\kappa$ with proof.

Answer 1

There's a simple recipe for producing such theories. Given two countable complete theories $T_0,T_1$, we can form their "side-by-side theory" $SBS(T_0,T_1)$; this theory basically says that the structure consists of a model of $T_0$ disjoint union a model of $T_1$.

Formally, suppose for simplicity that $T_0$ and $T_1$ have disjoint relational languages $L_0$ and $L_1$. Then $SBS(T_0,T_1)$ has language $L_0\cup L_1\cup \{U\}$, where $U$ is a new unary relation symbol, and axioms saying (i) the $L_0$-reduct of $U$ is a model of $T_0$, (ii) the $L_1$-reduct of $\neg U$ is a model of $T_1$, and (iii) no relation from $L_0$ (resp. $L_1$) holds of any tuple not entirely from $U$ (resp. $\neg U$).

Now if $T_0$ is not $\aleph_0$-categorical and $T_1$ is not categorical in any uncountable cardinality, $SBS(T_0,T_1)$ is not categorical in any cardinality.

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In the case of $Th(\mathbb{N})$ specifically, remember the Baldwin-Lachlan theorem; can you show that $Th(\mathbb{N})$ has uncountably many countable models?

HINT: How many types are there over $\mathbb{N}$? How many can be realized in a single countable model?

Answer 2

You could take the theory of dense linear orders with no first or last element that extend $\mathbb R$ -- or in other words, the elementary diagram of $(\mathbb R,{<}).$

It is well known that the theory of dense linear orders with no first or last element is complete. Unfortunately it is $\aleph_0$-categorical -- but adding new constants for each real (and axioms that say how they compare) prevents that without disturbing the completeness. Every sentence $\varphi$ still mentions only finitely many reals, so it will have a proof or disproof according to whether $\forall x_1\ldots x_n(x_1<x_2<\cdots<x_n \to \varphi)$ is true in $\mathbb Q$.

For each cardinality $\ge\mathfrak c$ there's now both a model where each element has uncountably many predecessors, and a model where some elements have countably many predecessors. (You can construct them by taking an arbitrary model and sticking either $\mathbb Q$ or $\mathbb R$ in front).

Answer 3

The theory of real closed fields is complete, as we know from quantifier elimination argument.

It is not hard to show there are two countable models which are not isomorphic, the real closure of $\Bbb Q$ and the closure of $\Bbb Q(\pi)$ (any transcendental would work). It also has two non isomorphic models of size continuum: the reals and the hyperreals. By Morley's theorem it has no uncountable size where it is categorical.