If the sets defined by a formula are all trivial then what can we conclude from that? Let $\mathcal L$ be a language, $n$ a positive integer and $\phi=\phi(x_1,\dots,x_n)$ an $\mathcal L$-formula having $x_1,\dots,x_n$ as free variables.
Then for every $\mathcal L$-structure $\mathfrak A$ there is a set $\phi^{\mathfrak A}$ defined by $\phi$ as: $$\phi^{\mathfrak A}:=\{(a_1,\dots,a_n)\in |\mathfrak A|^n: \mathfrak A\vDash\phi[a_1,\dots,a_n]\}$$where $|\mathfrak A|$ denotes the domain of structure $\mathfrak A$.
Now let it be that for every $\mathcal L$-structure $\mathfrak A$ we have: $$\phi^{\mathfrak A}=\varnothing\text{ or }\phi^{\mathfrak A}=|\mathfrak A|^n$$
Can we conclude from this that one of the following statements must be true?

*

*$\phi^{\mathfrak A}=\varnothing$ for every $\mathcal L$-structure $\mathfrak A$.

*$\phi^{\mathfrak A}=|\mathfrak A|^n$ for every $\mathcal L$-structure $\mathfrak A$.


No, we cannot; let $\phi'$ be any $\mathcal{L}$-sentence (so $\phi'$ contains no free variables), and let $\phi$ be the $\mathcal{L}$-formula $\phi'\wedge\bigwedge_{i=1}^nx_i=x_i$. (Note that $\phi$ does indeed contain every $x_i$ as a free variable, albeit in a vacuous way.) Then if $\mathfrak{A}\models\phi'$ we have $\phi^\mathfrak{A}=|\mathfrak{A}|^n$, and if $\mathfrak{A}\models\neg\phi'$ we have $\phi^\mathfrak{A}=\emptyset$. Since every $\mathcal{L}$-sentence either holds or does not hold in an $\mathcal{L}$-structure, one of these two cases must apply. Thus, chosing $\phi'$ so that there are some structures in which it holds and some structures in which it does not provides a countexample.
