Cowen-Engeler Lemma

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

Let $M$ be a set of mappings from subsets of $X$ to $Y$. That is, each element of $M$ is a partial mapping from $X$ to $Y$.

Define a mapping $\Phi: X \to \mathcal P(Y)$ thus:


 * $\Phi(x) = \left\{{ f(x): f \in M \text{ and } x \in \operatorname{Dom}(f) }\right\}$

That is, $\Phi(x) = \left\{{ y \in Y \mid \exists f \in M: (x, y) \in f }\right\}$

Suppose that the following hold:

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

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

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

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

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

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

Then $S \subseteq \operatorname{Dom}(f)$ and $T \subseteq \operatorname{Dom}(f)$, 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 $S \subseteq \operatorname{Dom}(f)$ and $T \subseteq \operatorname{Dom}(f)$.

Thus $S \cup T \subseteq \operatorname{Dom}(f)$.

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

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

Then by the corollary to the Ultrafilter Lemma, there is an ultrafilter $\mathcal U$ on on $M$ which includes $\{ \Gamma_S: S \in \operatorname{Fin}(X) \}$ 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 $\Gamma$, $\Phi(x) = \{ f(x) \mid f \in \Gamma_{\{x\}} \}$.

For each $y \in \Phi(x)$, let $K_y = \{ f \in \Gamma_{\{x\}} \mid f(x) = y \}$.

Then $\{ K_y | y \in \Phi(x) \}$ is a partition of $\Gamma_{\{x\}}$.

Since $\{x\} \in \operatorname{Fin}(X)$, $\Gamma_{\{x\}} \in \mathcal U$.

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

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

We define $\phi(x)$ to be that $y$.

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


 * $\{ f \in \Gamma_{\{x\}} \mid f(x) = \phi(x) \} \in \mathcal U$

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

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

Let $\displaystyle \Psi = \bigcap_{x \in S} \{ f \in \Gamma_{\{x\}} \mid f(x) = \phi(x) \}$.

Since $\{ f \in \Gamma_{\{x\}} \mid f(x) = \phi(x) \} \in \mathcal U$ for each $x \in X$, $S$ is finite, and $\mathcal U$ is a filter, $\Psi \in \mathcal U$.

Thus $\Psi \ne \varnothing$.

Choose any $f \in \Psi$.

Then $f(x) \in \Gamma_{\{x\}}$ 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, $f(x) = \phi(x)$ for each $x \in S$ by the definition of $\Psi$.

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, $\phi \in M$.