Cowen-Engeler Lemma

Theorem
Let $X$ and $Y$ be sets.

Let $M$ be a set of mappings from subsets of $X$ to $Y$.

Thus each element of $M$ is a partial mapping from $X$ to $Y$.

Define a mapping $\Phi: X \to \powerset Y$ thus:


 * $\map \Phi x = \set {\map f x: f \in M \text{ and } x \in \Dom f}$

That is:
 * $\map \Phi x = \set {y \in Y: \exists f \in M: \tuple {x, y} \in f}$

Suppose that the following hold:

Then there exists an element of $M$ with domain $X$.

Proof
Let $\map {\operatorname {Fin} } X$ be the set of finite subsets of $X$.

For each $S \in \map {\operatorname {Fin} } X$, let $\Gamma_S = \set {f \in M: S \subseteq \Dom f}$.

By $(2)$, $\Gamma_S$ is non-empty for each $S \in \map {\operatorname {Fin} } X$.

For each $S \in \map {\operatorname {Fin} } X$, $\Gamma_S \cap \Gamma_T = \Gamma_{S \cup T}$:

If $f \in \Gamma_{S \cup T}$, then $S \cup T \subseteq \Dom f$.

Then $S \subseteq \Dom f$ and $T \subseteq \Dom f$, so $f \in \Gamma_S$ and $f \in \Gamma_T$.

Thus $f \in \Gamma_S \cap \Gamma_T$.

If $f \in \Gamma_S \cap \Gamma_T$, then $f \in \Gamma_S$ and $f \in \Gamma_T$.

Thus $S \subseteq \Dom f$ and $T \subseteq \Dom f$.

Thus $S \cup T \subseteq \Dom f$.

So $f \in \Gamma_{S \cup T}$.

Thus $\set {\Gamma_S: S \in \map {\operatorname {Fin} } X}$ has the finite intersection property.

Then by the corollary to the Ultrafilter Lemma, there is an ultrafilter $\mathcal U$ on on $M$ which includes $\set {\Gamma_S: S \in \map {\operatorname {Fin} } X}$ as a subset.

We will now construct a mapping $\phi: X \to Y$.

Fix $x \in X$.

Note that by the definitions of $\Phi$ and $\Gamma$:
 * $\map \Phi x = \set {\map f x: f \in \Gamma_{\set x} }$

For each $y \in \map \Phi x$, let $K_y = \set {f \in \Gamma_{\set x} : \map f x = y}$.

Then $\set {K_y : y \in \map \Phi x}$ is a partition of $\Gamma_{\set x}$.

Because $\set x \in \map {\operatorname {Fin} } X$:
 * $\Gamma_{\set x} \in \mathcal U$

By $(1)$, $\map \Phi x$ is finite.

Thus by Component of Finite Union in Ultrafilter, there is exactly one $y \in \map \Phi x$ such that $K_y \in \mathcal U$.

We define $\map \phi x$ to be that $y$.

By the definition of $K_y$, for each $x \in X$ we have:


 * $\set {f \in \Gamma_{\set x}: \map f x = \map \phi x} \in \mathcal U$

We must show that $\phi \in \mathcal M$.

Let $S$ be a finite subset of $X$.

Let $\displaystyle \Psi = \bigcap_{x \mathop \in S} \set {f \in \Gamma_{\set x}: \map f x = \map \phi x}$.

Since:
 * $\set {f \in \Gamma_{\set x}: \map f x = \map \phi x} \in \mathcal U$ for each $x \in X$
 * $S$ is finite
 * $\mathcal U$ is a filter

it follows that:
 * $\Psi \in \mathcal U$

Thus $\Psi \ne \O$.

Choose any $f \in \Psi$.

Then $\map f x \in \Gamma_{\set x}$ for each $x \in S$.

So $f \in M$ and $S$ is a finite subset of the domain of $f$.

Since $M$ has finite character, $f \restriction S \in M$.

Furthermore, by the definition of $\Psi$:
 * $\map f x = \map \phi x$

for each $x \in S$.

Thus:
 * $\phi \restriction S = f \restriction S \in \phi$

So we see that $\phi \restriction S \in M$ for each finite subset $S$ of $X$.

Since $M$ has finite character, $\phi \in M$.