Non-Forking Types have Non-Forking Completions

Theorem
Let $T$ be a complete $\mathcal{L}$-theory.

Let $\mathfrak{C}$ be a monster model for $T$.

Let $A\subseteq B$ be subsets of the universe of $\mathfrak{C}$.

Let $\pi(\bar x)$ be an $n$-type over $B$.

If $\pi$ does not fork over $A$, then there is a complete $n$-type $p$ over $B$ such that $\pi \subseteq p$ and $p$ does not fork over $A$.

Proof
Suppose $\pi$ does not fork over $A$.

We will use Zorn's Lemma to find a candidate for the needed complete type.

Consider the collection $\Pi$ of all non-forking sets $\pi'$ of $\mathcal{L}$-formulas with parameters from $B$ such that $\pi'$ contains $\pi$.

Order $\Pi$ by subset inclusion.

Since a set forks iff a finite subset forks, the union over any chain is still a non-forking set, and hence is an upper bound for the chain.

Thus, by Zorn's Lemma, there is a maximal (with respect to subset inclusion) $p$ in $\Pi$.

Suppose $p$ is not complete.
 * By definition, for some $\phi(\bar x, \bar b)$, $p$ contains neither $\phi(\bar x, \bar b)$ nor $\neg\phi(\bar x, \bar b)$.


 * Since $p$ is non-forking, by Formula and its Negation can't both Cause Forking, at least one of $p\cup\phi(\bar x, \bar b)$ or $p\cup\neg\phi(\bar x, \bar b)$ is non-forking as well.


 * Hence $p$ is not maximal in $\Pi$, contradicting the choice of $p$.

Thus $p$ is complete.