A theory of identity with constants

Question

Suppose we have a first order theory $T$ in a language which contains the identity symbol $=$, constants $a_1, a_2, a_3, \dotsc, b_1, b_2, b_3, \dotsc$ 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 $a_1, a_2, a_3, \dotsc$ — 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 $b_1, b_2, \dotsc, b_n$ is distinct from all the $a$s?

Answer 1

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.