# 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. It basically only means that $\phi$ is independent of its variables but not necessarily of the models, e g. think about $\phi(x,y)=\exists z_1,z_2:z_1\ne z_2$, it is false in a model with one element but true in other models, independently of the values of $x,y$. – Berci Nov 23 '20 at 18:56

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.