Theorem: Cowen-Engeler Lemma

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:
{{begin-itemize}}
{{item|(1):|$\map \Phi x$ is finite for each $x \in X$.}}
{{item|(2):|For each finite subset $F$ of $X$, there exists an element $f \in M$ such that the domain of $f$ is $F$.}}
{{item|(3):|$M$ has finite character.
That is, a mapping $f$ from some subset of $X$ to $Y$ is an element of $M$ {{iff}} its restriction to each finite subset of $\Dom f$ is in $M$.
}}
{{end-itemize}}
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 $S \in \map {\operatorname {Fin} } X$.
{{tidy|presentation is ugly and difficult to follow}}
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:
:$f \in \Gamma_S \land f \in \Gamma_T$
Thus:
:$f \in \Gamma_S \cap \Gamma_T$
If:
:$f \in \Gamma_S \cap \Gamma_T$
then:
:$f \in \Gamma_S \land f \in \Gamma_T$
Thus:
:$S \subseteq \Dom f \land T \subseteq \Dom f$
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 {{Corollary|Ultrafilter Lemma}}, there is an ultrafilter $\UU$ 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 $\Gamma_{\set x}$.
{{explain}}
Because $\set x \in \map {\operatorname {Fin} } X$:
:$\Gamma_{\set x} \in \UU$
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:
:$K_y \in \UU$
We define $\map \phi x$ to be that $y$.
By the definition of $K_y$, for each $x \in X$ we have:
:$\set {f \in \Gamma_{\set x}: \map f x = \map \phi x} \in \UU$
We must show that:
:$\phi \in M$
Let $S$ be a finite subset of $X$.
Let:
:$\ds \Psi = \bigcap_{x \mathop \in S} \set {f \in \Gamma_{\set x}: \map f x = \map \phi x}$
Since:
:$\set {f \in \Gamma_{\set x}: \map f x = \map \phi x} \in \UU$ for each $x \in X$
:$S$ is finite
:$\UU$ is a filter
it follows that:
:$\Psi \in \UU$
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, $\phi \in M$.
{{qed}}
{{BPI|Ultrafilter Lemma}}
{{Namedfor|Robert H. Cowen|name2 = Erwin Engeler|cat = Cowen|cat2 = Engeler}}