Propositional Logic: find sets $\emptyset \neq K_1 \subseteq K_2 \subseteq K_3 \subsetneq A$ such that $K_1, K_3$ are definable, and $K_2$ isn't

Edit: An assignment $$v$$ is a funcion $$v : \{p_i : i \in \Bbb N\} \to \{true, false\}$$ ($$p_i$$ are atomic variables).

A set of Assignments $$K$$ is definable if there exists $$\Gamma \subseteq WFF$$ such that $$K= \{v : v \vDash \Gamma \} := Asgn(\Gamma)$$.

The question: Let $$A$$ be the set of all assignments. Find sets $$\emptyset \neq K_1 \subseteq K_2 \subseteq K_3 \subsetneq A$$ such that $$K_1, K_3$$ are definable, and $$K_2$$ isn't.

One undefinable set I know is the set of all assinments assiging true to a finite number of variables. I tried to look for a (non trivial) set that contains it, and is definable, but I can't find any.

• It might be worth editing into the question a definition of assignments. – J.G. Dec 23 '18 at 11:01
• I added some definitions – user401516 Dec 23 '18 at 11:06
• By definable I mean that there exists any set $\Gamma \subseteq WFF$ that defines it, as I wrote in my post. It does not have to be a single formula (or even a finite set). I will edit my post so it is clear I am working in propositional logic, thank you for your comment. – user401516 Dec 23 '18 at 11:08

Hint: let $$P_1, P_2, \ldots$$ enumerate all the propositional variables and take $$K_1$$ to be defined by $$\Gamma_1 = \{P_1, \lnot P_2, \lnot P_3, \lnot P_4,\ldots\}$$ and $$K_3$$ to be defined by $$\Gamma_3 = \{P_1\}$$. Now see if you can adapt your idea to find a suitable $$K_2$$.