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

Question

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.

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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.

Answer 1

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 $K_2$.