The question is straightforward as expressed in the title: If a theory $T$ has a model that satisfies sentence $s$, does every model of $T$ satisfy $s$? I think I know the answer as yes, but why this is the case, I cannot find or remember. What would be such proof?
If a theory $T$ has a model that satisfies sentence $s$, does every model of $T$ satisfy $s$
logicmodel-theory
Best Answer
The answer, as stated above, is no (unless we assume that $T$ is complete).
If you know a bit of abstract algebra, the simplest natural counterexample to your guess is probably to take $T$ to be the group axioms and $s$ to be the sentence "$\forall x\forall y(x*y=y*x)$." Some but not all groups are abelian, which is to say that some but not all models of $T$ satisfy $s$.
Leaving abstract algebra aside, here's a "purely combinatorial" counterexample:
Let $T$ be the empty theory, $T=\emptyset$. Then every structure satisfies $T$, vacuously: for $\mathcal{M}$ an arbitrary structure, we have "$\mathcal{M}\models\varphi$ for every $\varphi\in T$" since there aren't any $\varphi\in T$ to begin with.
Now consider the sentence $s:=$ "$\exists x\exists y(x\not=y)$." The structures satisfying this sentence are exactlyt he structures with more than one element. Since there are structures which do have more than one element, and structures which don't have more than one element, that means there are models of $T$ which disagree about whether $s$ is true or false.