How can $n\neq 0-$types contain sentences?

Question

I was studying Introduction to Stability Theory by Anand Pillay, where he begins proves the following lemma:

lemma 1.15 let $p(\bar{x})\in S_n(M)$ and A $\supseteq M$. Then $p$ has an extension $q \in S_n(A)$ which is an heir of $p$.

where

• $M$ is a model in some monster model
• $S_n(M)$ (resp. $S_n(A)$) denotes the set of $n$-types from the language $M$ (resp. $A$)
• In this scenario, $q$ is an heir of $p$ if every $\mathcal{L}(M)$ formula $\phi(\bar{x}, \bar{y})$ for which there is some tuple $\bar{a} \in A$ with $\phi(\bar{x}, \bar{a}) \in q$ there is some tuple $\bar{m} \in M$ for which $\phi(\bar{x}, \bar{m}) \in p$ (or using terminology defined in the book, $q$ is an heir of $p$ if every $\mathcal{L}(M)$ formula represented in $q$ is represented in $p$)

His strategy of the proof as follows (with details omited):

1. Let $T(A)$ be the set of $\mathcal{L}(A)$-sentences true in the monster model.
2. Define $\Sigma$ to be $T(A) \cup p \cup \{\neg \phi(\bar{x}, \bar{y}) \in \mathcal{L}(M), \bar{a} \in A, \phi(\bar{x}, \bar{y})$ is not represented in $p\}$.
3. Pillay proves that $\Sigma$ is consistent.
4. $\Sigma$ has an extension $\Sigma'$ which is a complete and consistent set of formulae with free variables $\bar{x}$ in $\mathcal{L}(A)$. As $T(A) \subseteq \Sigma'$, $\Sigma'$ can be considered as a type $q \in S_n(A)$. He then finishes off the proof (irrelevant to my confusion).

My confusion arises that $\Sigma'$ is considered a type, where it clearly contains sentences of $0$ free variables from $T(A)$. Question: what is going on here? Why is $\Sigma'$ considered an $n$-type when it clealy contains formulas of not $n$-free variables? Perhaps there's something obvious I've missed which allows us to kind of ignore the sentences?

On another note, I also believe in step $2$ of the proof, Pillay meant $\{\neg \phi(\bar{x}, \mathbf{\bar{a}}) \in \mathcal{L}(M), \bar{a} \in A, \phi(\bar{x}, \bar{y})$ is not represented in $p\}$ instead of $\{\neg \phi(\bar{x}, \bar{y}) \in \mathcal{L}(M), \bar{a} \in A, \phi(\bar{x}, \bar{y})$ is not represented in $p\}$ ?

Answer 1

We don't actually have to use all the variables. A formula in $x_1,...,x_n$ is a formula whose set of free variables is a subset of $\{x_1,...,x_n\}$, and an $n$-type is a set of such formulas. Remember, the key point is that we need to make sense of the question

Does $\varphi$ hold when we substitute $a_i$ for $x_i$, for $i\in\{1,...,n\}$?

This doesn't need $\varphi$ to actually use each variable, merely that it not use any other variable. That said, if you want to actually use each variable you can always replace your given $\varphi$ with $\varphi\wedge(\bigwedge_{1\le i\le n}x_i=x_i)$ if you want, and this won't change anything.

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This might get easier if you think about types set-theoretically. An $n$-type "is" a collection of definable (from the appropriate parameters) subsets of $\mathcal{M}^n$ with the finite intersection property; both $\mathcal{M}^n$ itself and $\emptyset$ are such sets, and they correspond to sentences true and false in $\mathcal{M}$ respectively. More generally, any definable subset $A$ of $\mathcal{M}^k$ "induces" a definable subset $A_m$ of $\mathcal{M}^{k+m}$ given by $$(a_1,..., a_{k+m})\in A_m\iff (a_1,...,a_k)\in A,$$ and this basically reflects the fact that omitting variables isn't really an issue for us.