Removing parameters from the definition of a type-definable set

Let $M$ be an $\aleph_0$-saturated structure (in a possibly uncountable language). I have a set $A \subseteq M$. Let $F$ be the family of $M$-definable subsets of $M$ containing $A$. I have the following question:

Is there a partial type $\Sigma(x)$ over a finite subset of $M$ such that every $M$-definable subset $X$ of $M$ is a superset of $\Sigma(M)$ iff $X \in F$?

That is, I want to find a set like $\bigcap F$ that is type-definable by fewer (finite) number of parameters.

When is can I find such a set, and when can I not? (For instance, it seems that I can find such a partial type for any subset $A$ in $(\Bbb Q, <)$, but that structure has very tame definable sets.)

• You seem to be asking for a certain one-one map from $F$s to partial 1-types with definitely many parameters. In the example of $(\mathbb{Q}, <)$, each Dedekind cut $\{q \in \mathbb{Q} : q < r\}$ has a different $F$, so there are uncountably many of those; but there are only countably many $\Sigma$ after you eliminate quantifiers. So it seems pigeonhole-impossible in this case. – realdonaldtrump Oct 6 '17 at 6:19
• Finitely many parameters. Sorry for phoneposting – realdonaldtrump Oct 6 '17 at 6:29

For any structure $M$ and any $A\subseteq M$, your set $F$ includes the set $M\setminus \{b\}$ (defined by the formula $x\neq b$) for all $b\notin A$, so $\bigcap F = A$.
Now suppose we had a set $\Sigma$ satisfying the property in your question. Clearly $\Sigma(M)\subseteq \bigcap F = A$. Conversely, suppose $a\in A$. Then $M\setminus \{a\}$ is not in $F$, so $\Sigma(M)\not\subseteq (M\setminus \{a\})$, so $a\in \Sigma(M)$. So also $\Sigma(M) = A$.
So for these rather trivial reasons (we only had to look at the formula $x = y$!), you can find such a set $\Sigma$ if and only if your original set $A$ is itself type-definable over a finite set of parameters.