So when reading section 6.2 of Hodges' A shorter model theory I came across this example:
Now, what does it mean precisely "by symmetry"? What is the argument he is trying to make?
I tried to prove on my own that for each $s\subseteq\omega$ there exists a countable model of $T$ that omits $\Phi_s$, but the only "straightforward" way I came up to do it is by constructing explicit models with $\omega$ as the domain, starting from a "ground" model in which every element (natural number) satisfy the $i$-th relation $P_i$ according to the $i$-th digit of its binary expansion and then adjusting it according to the type I want to omit.
But of course I have the feeling that there is a more succinct and elegant way of proving this. Any hint?
Thanks!
The symmetry Hodges is referring to is the symmetry between $P_i$ and $\lnot P_i$ in the axioms of $T$, for all $i\in \omega$.
Let $M$ be a model of $T$. Now pick some $X\subseteq \omega$, and define a structure $M_X$ by starting with $M$ and, for each $i\in X$, replacing the interpretation $P_i^M$ of the relation symbol $P_i$ by its complement $M\setminus P_i^M$.
Then $M_X\models T$ "by symmetry". For each axiom $\varphi\in T$, there is a corresponding axiom $\varphi_X\in T$, obtained by replacing $P_i$ by $\lnot P_i$ and replacing $\lnot P_i$ by $P_i$, for each $i\in X$. Since $M\models \varphi_X$, it follows that $M_X\models \varphi$.
Now if we want to omit a given type $\Phi_s$, we pick an arbitrary countable model $M$ and note that $M$ must omit some type $\Phi_t$. Letting $X = s\triangle t$ (the symmetric difference of $s$ and $t$), we have $M_X\models T$ and $M_X$ omits $\Phi_s$.