# A theory with exactly $n$ countable models, for each $n>1$

For each $n>1$ we shall construct a first-order theory $T_n$ with exactly n countable models. Let $n>1$, consider the language $L_n=\left\{{R,c_1,...,c_n}\right\}$, where $R$ is a binary relation symbol, and each $c_n$ is a distinct constant symbol. Put $$\phi=\forall{x}\forall{y}(x\neq{c_1}\wedge{y\neq{c_2,...c_n}}\longrightarrow{¬Rxy})$$ and $$\psi=\forall{x}\forall{y}\forall{z}(x\neq{y}\wedge{x\neq{z}\wedge{y\neq{z}\wedge{Rxy}}}\longrightarrow{¬Rxz}).$$ Set $$T_n=\left\{{c_i\neq{c_j}:i\neq{j}}\right\}\cup\left\{{\phi,\psi}\right\}.$$ It is clear that $T_n$ is consistent, and thus by Lowenheim-Skolem theorem $T_n$ has a countable model. Let $\mathcal{M}$ be a countable model of $T_n$, then since $\mathcal{M}\models{\phi}$ we must have that either $R^{\mathcal{M}}=\emptyset$ or $R^{\mathcal{M}}=\left\{{(c_1,c_{n_1}),...,(c_1,c_{n_m})}\right\}$, but since $\mathcal{M}\models{\psi}$ we get that if $R^{\mathcal{M}}\neq{\emptyset}$, then $\left |{R^{\mathcal{M}}}\right |=1$.

Therefore we must have that either $R^{\mathcal{M}}=\emptyset$ or $R^{\mathcal{M}}=\left\{{(c_1,c_m)}\right\}$ for some $m\leq{n}$. It is clear that if $\mathcal{M}$ and $\mathcal{N}$ are models of $T_n$ with $R^{\mathcal{M}}=\emptyset$ and $R^{\mathcal{N}}=\emptyset$, then $\mathcal{M}\simeq{\mathcal{N}}$.

And also if $R^{\mathcal{M}}=\left\{{(c_1,c_i)}\right\}$ and $R^{\mathcal{N}}=\left\{{(c_1,c_j)}\right\}$, then $\mathcal{M}\simeq{\mathcal{N}}$ if and only if $i=j$, for $\mathcal{M}\models{c_i\neq{c_j}}$ for $i,j\in{\left\{{2,...,n}\right\}}$ with $i\neq{j}$

By Lowenheim-Skolem theorem each case has a countable model.

Thus $T_n$ has exactly $n$ countable models.

I would like to see others examples.

Thanks

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 Hi Camilo. You say that $R^{\mathcal{M}} = \{ (c_{1},c_{m}) \}$ and $R^{\mathcal{N}} = \{ (c_{1},c_{m}) \}$ imply $\mathcal{M} \cong \mathcal{N}$. However, one can have $R^{\mathcal{M}} = \{ (c_{1},c_{i}) \}$ and $R^{\mathcal{N}} = \{ (c_{1},c_{j}) \}$, where $i,j \in \{ 2,\ldots,n \}$ are distinct. After all, $\mathcal{M}$ and $\mathcal{N}$ are different models, so there is no need to stick to a single $c_{m}$ for both. – Haskell Curry Oct 14 '12 at 17:17 Isn't that exactly Camilo's argument? There are exactly $n-1$ countable models with nonempty rlation and another countable model with empty relation. – Hagen von Eitzen Oct 14 '12 at 17:26 Hi @HaskellCurry, I just modified my post justifying the doubt you have, thanks for noticing – Camilo Arosemena Oct 14 '12 at 21:29

Let $L_n$ be the language with $n$ nullary predicate symbols (aka primitive propositions) $P_1, \ldots, P_n$, and $T_n$ be the theory with $\forall x \forall y (x=y)$ as one axiom and another axiom asserting that exactly one of the $P_i$ is true. Clearly a model is just a choice of $i$, so there are $n$ models.

If you don't like nullary predicates, use unary predicates instead, and have the second axiom assert that exactly one of the $n$ sentences $\forall x(P_i(x))$ is true.

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Our theory $T_n$ is over the predicate calculus with equality. The language has a unary predicate symbol $Q$, a binary predicate symbol $\lt$, and no other non-logical symbols. We now describe the axioms of $T_n$.

(i) If $Q$ fails at $x$ or at $y$, then neither $x\lt y$ nor $y\lt x$ holds.

(ii) Under $\lt$, the objects that satisfy $Q$ form a (non-empty) densely ordered set with no first or last element.

(iii) There are at most $n-1$ objects that satisfy $\lnot Q$.

Since the theory of densely ordered sets with no first or last element is $\omega$-categorical, we are done.

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