Let $\mathcal L_3 = \{<,c_o,c_1,\ldots\}$, and $T_3$ be the theory of DLO with sentences added stating $c_o < c_1 < \ldots$. Now let $\mathcal L_4 = \mathcal L_3 \cup \{P\}$, where $P$ is a unary predicate. Let $T_4 = T_3$ unioned with the sentences:

$$ \forall x \forall y(x < y \rightarrow \exists z \exists w (x < z < y \wedge x < w < y \wedge P(z) \wedge \neg P(w)))$$

In other words, P is a dense-codense set. Show that $T_4$ is complete theory with exactly four countable models.

And how would you generalize the above construction to give theories with exactly 5,6,... models?


$T_4$ is not complete because there is no proof as to whether there is a least or greatest element.

Also, note $T_4$ is not $\kappa$-categorical. For uncountable models, one of $P$ or $P^c$ could be countable (eg in $\mathbb{R}$ let $P(x)$ mean $x$ is rational/irrational) or both could be uncountable (eg in $\mathbb{R}^2$ with lexicographic order let $P(\langle x,y \rangle ))$ mean $y$ is rational).

There are also more than 4 countable models. There could be a lowest element, a greatest element, both, or neither. Then any combination of lowest and greatest element could be either $P$ or not $P$...

The constants in the language and the criteria that they are in order do nothing to lower the number of models because the $c_n$'s could be mapped to any countable subset that may or may not include a maximal or least element.

I just thought -- if there were greatest and lowest elements, then there would be four countable models. Both elements could be in $P$, both could be in $P^c$, one in $P$ and one in $P^c$, and vice versa. We could then add one extra model by adding another predicate $Q$ and adding enough to the theory to have a proof that $Q$ is the same everywhere, but could be either true or false at say the least element in only one of the four models -- basically we split one of the 4 models into 2 to get 5 models.


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