Definability in pure-second order logic

Question

Are the universe and the empty set the only definable sets without parameters in the language of pure second-order logic? TIA

Answer 1

Let me try to help by explicating the argument Noah mentioned. I think this is part of logic folklore—it amounts at bottom to the facts that every permutation of a pure set is an isomorphism, and isomorphisms preserve truth.

Specifically, in the language of pure second-order logic, you have a a domain $D$ of individuals, the domain of discourse, which exhibits no additional structure—no relations (except $=$), no named constants, no functions. The first order quantifiers $\exists x$ and $\forall x$ range over the elements of $D$, and the second-order quantifiers $\exists X$ range over the subsets and relations on $D$.

Since the domain $D$ has no atomic structure, it follows immediately that every permutation of the domain $\pi:D\to D$ is an automorphism, an isomorphism of the structure with itself. From this, it follows that $D\models\varphi[a,b,c,R,S,T]$, where $a,b,c$ are individuals from the domain and $R,S,T$ are relations on $D$, if and only if $D\models\varphi[\pi(a),\pi(b),\pi(c),\pi(R),\pi(S),\pi(T)]$ for any formula $\varphi$, whether in first-order or second-order logic (or higher-order).

This can be proved by induction on formulas. It is true for atomic formulas, since $\pi$ respects $=$ and we have transformed the second-order parameters to their images under $\pi$, and it combines inductively via Boolean combinations and first-order quantifiers. The second-order quantifiers are also preserved, for essentially the same reason—if there is a relation $R$ on $D$ fulfilling $\exists R\ \varphi[a,b,c]$, then $D$ satisfies $\varphi[a,b,c,R]$ and so by induction it satisfies $\varphi[\pi(a),\pi(b),\pi(c),\pi(R)]$, which means that $D$ fulfills $\exists R\varphi[\pi(a),\pi(b),\pi(c)]$.

Basically, any permutation is an isomorphism of the pure set $D$ with itself, trivially so, and isomorphisms preserve truth, regardless of the order.

It now follows that if you have a definable class $\{\ d\in D\mid\varphi(d)\ \}$ in $D$, with no parameters, then if it contains any point at all, it will contain all points, since there are permutations moving any $d$ to any other $d'$. So the only definable classes without parameters are the empty set or the universal class.