I have the following question: If I have two models which are defined as followed - $$ v_1(p_i)=T \leftrightarrow 2|i $$ $$ v_2(p_i)=T \leftrightarrow \exists k \in \mathbb{N}. i=2k+1 $$
Is there a Set X such that $Ass(X)= \{ v_1, v_2 \}$
If not - how can I show that?
I would also like to generalize that question - is a set with any two models definable?
Thanks!
CPL = classical propositional calculus
Let $\mathcal{L}$ be a set of all formulas of CPL. By a complete set $T$ of formulas of CPL I mean a subset of $\mathcal{L}$ such that there exists a model $\nu$ such that $$T = \big\{\phi\,\big|\,\nu\Vdash \phi\big\}$$ Note that such $T$ has the following properties:
1. For every formula $\phi$ either $\phi\in T$ or $\neg \phi \in T$ 2. If $\phi\in T$, then for every formula $\psi$ we have $\phi\vee\psi \in T$. 3. If $\phi_1,...,\phi_n\in T$, then $\phi_1\wedge ...\wedge \phi_n\in T$.
We will use these properties below.
Fact. Let $T, T_1,...,T_n$ be complete sets of formulas for CPL. If $$T_1\cap T_2\cap ...\cap T_n\subseteq T$$ then there exists $k$ such that $T=T_k$.
Proof. We show first that $T_k\subseteq T$ for some $k$. Suppose to the contrary that for each $1\leq k\leq n$ there exists $\phi_k\in T_k$ such that $\phi_k\in T_k\setminus T$. Then $\neg \phi_k\in T$ according to 1. Thus also $(\neg \phi_1)\wedge (\neg \phi_2)\wedge...\wedge (\neg \phi_k)\in T$ according to 3. Hence $$\phi_1\vee \phi \vee...\vee \phi_k\not \in T$$ On the other hand since $\phi_k\in T_k$, we derive that $\phi_1\vee\phi_2\vee ...\vee\phi_n\in T_k$ by 2. Thus $$\phi_1\vee\phi_2\vee ...\vee\phi_n\in T_1\cap T_2\cap ...\cap T_n \subseteq T$$ This is a contradiction and hence we proved that $T_k\subseteq T$ for some $1\leq k\leq n$. This implies that $T_k = T$, because by definition complete sets of formulas are maximal among consistent sets of formulas.
Now consider $T_1, T_2$ complete sets of formulas corresponding to $\nu_1,\nu_2$, respectively. Let $X= T_1\cap T_2$. Clearly $\nu_1\Vdash X$ and $\nu_2\Vdash X$. On the other hand suppose that $\nu\Vdash X$. Let $T$ be a complete set corresponding to $\nu$. Then $X = T_1\cap T_2\subseteq T$. By Fact we have $T_1 = T$ or $T_2 = T$. This means that $\nu = \nu_1$ or $\nu = \nu_2$. Hence $\mathrm{Ass}(X) = \{\nu_1,\nu_2\}$.
The general case. Consider the set of all models $M$ of CPL. We define a topology $\tau$ on $M$. Basis of $\tau$ consists of sets $$M(\phi) = \{\nu \in M\,|\,\nu \Vdash \phi\}$$ One can easily verify that this is a topology on $M$. This topology (as it is intuitively clear from its definition) is closely related to metalogical properties of CPL. Closed subsets in the topology are of the form $$V(X) = \{\nu \in M\,|\,\forall_{\phi \in X} \nu \not \Vdash \phi\} $$ From this description you can prove that the compactness theorem for $\Vdash$ is equivalent to the compactness of $(M,\tau)$.
It also turns out that closed subsets of $\tau$ are precisely definable sets. The Fact above really says that finite sets of models are closed in $\tau$.