Suppose $T$ is a complete theory in a countable language with infinite models.

  1. How we could say there are at most $2^{\aleph_0}$ non-isomorphic countable models of T?

  2. What we could say about number of finite models which are non-isomorphic?


closed as off-topic by user21820, Namaste, C. Falcon, Aweygan, user91500 Jan 21 '17 at 8:28

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  • 1
    $\begingroup$ Hint for part 1: Any countably infinite model is isomorphic to one whose carrier set is $\mathbb N$. $\endgroup$ – Henning Makholm Jan 20 '17 at 15:52

Every countable model is isomorphic to a model whose base set is $\mathbb{N}$, so we can assume that our models have a base set $\mathbb{N}$. Now let $S$ denote the set of all possible interpretations of the language $L$. The set $S$ can be mapped one-one in a set of cardinality $2^{\aleph_0}$, since the language is countable.

Indeed, for simplicity, let's assume there are only function symbols (the relational symbols being not much harder to treat), and let $\{f_n \mid n\in \mathbb{N}\}$ be an enumeration of the language, and for all $n$, let $a(n)$ denote the arity of $n$. Then $S$ can be sent one-one into $\displaystyle \prod_{n\in \mathbb{N}} F(\mathbb{N}^{a(n)}, \mathbb{N})$ (where $F(X,Y)$ denotes the set of functions from $X$ to $Y$).Now this is of cardinality $2^{\aleph_0}$ so there are at most $2^\aleph_0$ $L$-structures so at most $2^{\aleph_0}$ models. QED.

Since $T$ is complete and has infinite models, it cannot have any finite model. Indeed let for all $n$ $F_n$ be the sentence $\exists x_1,.... \exists x_n, (x_1\neq x_2 \land... \land x_{n-1} \neq x_n)$ (where the non written part are existentital quantifiers, and $x_i\neq x_j$ for $i\neq j$). Then if $T$ disproved any of the $F_n$, it couldn't have any infinite model, so out of completeness, for any $n$, $T\vdash F_n$, so that any model of $T$ must be infinite.

  • 4
    $\begingroup$ A more economical way to state that the universe has at least $n$ elements would be $$\forall x_1\forall x_2\ldots \forall x_{n-1}\exists y(y\ne x_1\land y\ne x_2\land\cdots\land y\ne x_{n-1})$$ $\endgroup$ – Henning Makholm Jan 20 '17 at 16:01

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