Can the number of countable models of a complete first-order theory decrease after adding constants? If $T$ is a countable complete first-order theory with infinite models, the number of countable models it has, $I(T,\omega)$, must be an element of $N=\{1,3,4,5,6,7,\dots,\omega,\omega_1,2^\omega\}$ (although we don't know if $\omega_1$ can happen). For which pairs $n,m\in N$ does there exist a countable complete theory $T$ with $n$ countable models but $m$ countable models after adding finitely many constants to the theory? Countably many new constants? In particular can we have $m<n$? EDIT: By 'adding constants,' I mean adding constants whose type is completely specified, i.e. expanding by constants and then passing to a complete theory in the expanded language.
Let $n\rightarrow m$ denote the statement "There exists a complete countable theory $T$ and a finite tuple of constants $\overline{a}$ such that $I(T,\omega)=n$ and $I(T_\overline{a},\omega)=m$." And let $n\rightarrow_\omega m$ denote the statement "There exists a complete countable theory $T$ and a countable set of distinct constants $A$ such that $I(T,\omega)=n$ and $I(T_A,\omega)=m$."
 Some easy results and relevant observations:


*

*If $n\rightarrow m$ (resp. $n\rightarrow_\omega m$) and $k \rightarrow \ell$ (resp. $k \rightarrow_\omega \ell$), then $nk \rightarrow m\ell$ (resp. $nk \rightarrow_\omega m\ell$). (Take the disjoint union of the relevant theories.)

*$n \rightarrow n$ for every $n\in N-\{\omega_1\}$. (This is obvious for $n=1$. There are easy examples for $n=\omega,2^\omega$ and the standard examples for $n=3,4,5\dots$ all have constants which do not increase the number of countable models.)

*$n^2+n\rightarrow (n+1)^2$ for any $1<n<\omega$. (DLO with $n-1$ colors and a countable set of constants of order type $\omega + \omega^\ast$. By itself this theory has $n^2 + 1$ countable models. Adding a constant in between $\omega$ and $\omega^\ast$ makes the theory have $(n+1)^2$ models.)

*$1\not\rightarrow n$ and $n\not\rightarrow 1$ for any $n\in N - \{1\}$.

*$1\rightarrow_\omega 2^\omega$ (For example: DLO.)

*$1\rightarrow_\omega \omega$ (For example: A structureless set.)

*$n\not\rightarrow_\omega 1$ for any $n\in N$.

*$1 \rightarrow_\omega n$ for every $2<n<\omega$. (The standard examples of Ehrenfeucht theories are $\omega$-categorical theories with countably many constants added.)

*If a theory is not small, then it will have $2^\omega$ countable models after adding any countable set of constants.

 A: I believe that the number of models can indeed decrease. The following seems to be an example of $5 \to 3$:
Start with $T$ the model companion of the theory of "valued trees": the language has two sorts $M$ and $\Gamma$, $\Gamma$ is equipped with a linear order $\leq_{\Gamma}$ and is a model of DLO. The sort $M$ has a tree structure, say in the language $\{\leq ,\wedge \}$, so $\leq $ is a partial order such that the set of predecessors of any point is a chain and $\wedge$ maps two points to their infimum. There is also a valuation map $v:M \to \Gamma$ which is increasing and such that for any $a\in M$, the restriction of $v$ to $\{x\in M:x<a\}$ is an injection onto $\{\gamma\in \Gamma:\gamma<v(a)\}$. This theory is $\aleph_0$-categorical.
Add a countable set of constants $\{c_0,c_1,\ldots,\}$ interpreted such that $c_0<c_1<\cdots$. Call $T_1$ the resulting theory. Then $T_1$ has 5 countable models:


*

*the prime model where the $c_i$'s are cofinal;

*two models with no element above all the $c_i$'s (one with a minimal valuation larger that all $v(c_i)$ and one without);

*two models with an element above all the $c_i$'s (same as the previous case).


If we now add another constant which is above all the $c_i$'s, then we have only 3 models (one where the constant is minimal above the $c_i$'s, one where it isn't, but there is such a minimal point, one where there is no such minimal point).
Edit: Example of $2^{\omega}\to_{\omega} 3$:
We modify a little bit the example above. Take the constants $c_i$ to name a subtree that has exactly $\aleph_0$ branches, all going all the way up in the tree. This has $2^{\aleph_0}$ many countable models, since we can independently add or not elements at the top of each branch.
Now add $\aleph_0$ constants, one above each branch of the named tree and impose that they have the same valuation. If I am not mistaken, this has now only 3 countable models similarly as above.
A: See the paper Nonnessential extensions of complete theories(B. Omarov)
Translated from Algebra i Logika, Vol. 22, No. 5, pp. 542-550, September-October, 1983]
Original article submitted June ii, 1982.
By adding constants (that may realize new types) Omarov write `A. D. Taimanov posed the following questions: "Is it possible to lower
the number of countable models from continuum to countable , and from continuum to the finite number k ?"
These questions are also answered affirmatively.
A: Edit: Emil Jeřábek confiems that this doesn’t answer the question; the OP is using “adding constants” to mean something different from what I would understand by it.  However, I’m leaving this here for others who understand the question the same way I did.
The number of countable models cannot decrease.
Let $T$ be any theory, $T^+$ any expansion of $T$ by new constant symbols (and possibly new function symbols too).  Taking reducts gives a map from models of $T^+$ to models of $T$. This map doesn’t change cardinality of models, respects isomorphism, and is surjective except possibly on the empty model.  So on isomorphism classes of models of any nonzero cardinality $\kappa$, it induces a surjection, showing $I(T^+,\kappa) \geq I(T,\kappa)$.
(I am slightly worried I’m misunderstanding something in your terminology conventions, since this answer seems a bit too easy?)
