The empty model does not satisfy both a sentence and its negation.

Question

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?

Answer 1

Here's another way to prove that the empty model does not satisfy both $\varphi$ and $\lnot \varphi$ when $\text{FV}(\varphi) = \varnothing$. Empty models are tricky, it is worth running back to the definition of $\models$ when working with them.

In particular, I am using the following definition:

$$ M \models \varphi \;\;\text{if and only if}\;\; M, v \models \varphi \; \text{for all valuations $v$ of the free variables of $\varphi$} $$

Let $M$ be a model; it can be empty or not.

Let $\varphi$ be a wff; it can be closed or not.

Let $[\varphi]$ be the set of all assignments from $\mathrm{FV}(\varphi)$ to the domain of $M$ in which $\varphi$ holds.

$\varphi$ is a tautology if and only if $[\varphi]$ is equal to $(\mathrm{FV}(\varphi) \to M)$, the set of all assignments over its free variables.

Suppose $M$ is empty.

If $\varphi$ contains any free variables whatsoever, then $[\varphi]$ will be empty, but $(\text{FV}(\varphi) \to M)$ will also be empty, so $\varphi$ will be true. Note that $\text{FV}(\lnot \varphi) = \text{FV}(\varphi)$.

Suppose $\varphi$ is closed. This means that there is a single map, the empty assignment, in $(\text{FV}(\varphi) \to M)$. $\varphi$ either holds or fails at $(M, \varepsilon)$, with $\varepsilon$ being the empty assignment.

Since it can't both hold and fail, it is impossible for $M, \varepsilon \models \varphi$ and $M, \varepsilon \models \lnot \varphi$ to both be true and hence impossible for $M \models \varphi$ and $M \models \lnot \varphi$ to both be true.

Answer 2

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?)