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.



Can the number of countable models of a complete first-order theory decrease after adding constants?
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.


