In a previous question, Does the empty model satisfy a contradiction?, I asked whether the empty model satisfies a contradiction. I used an open formula, $x=x$. Now, I am asking, is there a sentence $S$ (that is, a closed formula) such that the empty model satisfies both $S$ and $\neg S$? I think there isn't, but what is the proof?
The empty model does not satisfy both a sentence and its negation.
model-theory
Related Solutions
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.
The standard way to make sense of $\mathfrak{M}\models\varphi$ when $\varphi$ is a formula with free variables is to demand that $\mathfrak{M}$ satisfy every instantiation of $\varphi$; this is the same as $\mathfrak{M}$ satisfying the universal closure of $\varphi$. Under this definition, if we admit the empty structure we do indeed have $\emptyset\models(x=x)\wedge \neg(x=x)$.
However, it is no longer the case that a contradiction (in the sense of a formula gotten by substituting first-order formulas for propositional variables in an unsatisfiable propositional sentence) is explosive, as this example demonstrates. Variable-free contradictions are still explosive, of course. Basically, since we've changed our semantics we have to re-evaluate how our various notions relate to each other. And if we interpret "contradiction" as "unsatisfiable formula," then $(x=x)\wedge \neg(x=x)$ is no longer a contradiction ... as, again, this example demonstrates.
So there's no tension here at all, just a disconnect between two previously-equivalent properties arising because of a slightly broader semantics.
And if that's not how you're defining "$\mathfrak{M}\models\varphi$" when $\varphi$ has free variables, how are you defining it?
Best Answer
Think about how $\models$ is defined in the first place: by definition, $\mathcal{A}\models\neg\varphi$ iff $\mathcal{A}\not\models\varphi$. So excluded middle in the metatheory prevents the situation you're asking about: to get $\mathcal{A}\models\varphi$ and $\mathcal{A}\models\neg\varphi$ we'd need to have both $\mathcal{A}\models\varphi$ and $\neg(\mathcal{A}\models\varphi) $, which is impossible.
(And if that's not how you're defining $\models$, then how are you defining it?)