Suppose we have a first order theory $T$ in a language which contains the identity symbol $=$, constants $a\_1$, $a\_2$, $a\_3$, ... $b\_1$, $b\_2$, $b\_3$, ... and no relation or function symbols. For natural numbers $i \neq j$, suppose that $a\_i \neq a\_j$ is contained in $T$, and let us suppose that $T$ has no models whose domain consists only of $a\_1, a\_2, a\_3, ...$ - that is to say, let us suppose that in every model of $T$, at least one of the $b$s is distinct from all of the $a$s. 


Must there be some number $n$ such that in every model of $T$, one of $b\_1$, $b\_2$, ... , $b\_n$ is distinct from all the $a$s? 



a theory of identity with constants
The answer is no.


Here is a simplified counterexample (my earlier post was more
complicated).


Let $T$ be the theory asserting $a\_i\neq a\_j$ and also the
assertions, for every particular $k$ and $j$, that if $b\_1=a\_k$, then $b\_{k+1}\neq a\_j$.


This theory is consistent, since we can let $b\_1\neq a\_k$ for any $k$, which vacuously satisfies all the implications. Also, in any model of $T$, if $b\_1=a\_k$, then $b\_{k+1}\neq a\_j$ for any $j$. So no model of $T$ consists only of the $a\_i$'s.


But meanwhile, there is no $n$ as you request, since for any $n$,
we can let $b\_1=\cdots=b\_n=a\_n$, and $b\_{n+1}\neq a\_j$ any $j$.
This will be a model of $T$ where the $b\_1,\ldots,b\_n$ are among
the $a\_i$'s, and so the requested property is violated.


