Finite Character for Sets of Mappings

Theorem
Let $S$ and $T$ be sets.

Let $\FF$ be a set of mappings from subsets of $S$ to $T$.

That is, let $\FF$ be a set of partial mappings from $S$ to $T$.

Then the following are equivalent:

$(1)$ implies $(2)$
Let $\FF$ have finite character in the sense of Definition:Finite Character/Mappings.

That is, suppose that for each partial mapping $f \subseteq S \times T$:


 * $f \in \FF$ for each finite subset $K$ of the domain of $f$, the restriction of $f$ to $K$ is in $\FF$.

Let $q \subseteq S \times T$.

First suppose that $q \in \FF$.

Let $r$ be a finite subset of $q$.

Then $r$ is a partial mapping from its domain into $T$.

Since $r$ is finite, its domain, $\Dom r$, is a finite subset of the domain of $q$.

Since $\FF$ is of finite character as a set of mappings, $q \restriction \Dom r$ is in $\FF$.

But $q \restriction \Dom r = r$.

Therefore $r \in \FF$.

Suppose instead that for each finite subset $r$ of $q$, $r \in \FF$.

Then for each finite subset $r$ of $q$, $r$ is a partial mapping from $S$ to $T$.

Let $x \in \Dom q$.

Let $y_1, y_2 \in T$.

Suppose that $\tuple {x, y_1}, \tuple {x, y_2} \in q$.

Then $\set {\tuple {x, y_1}, \tuple {x, y_2} }$ is a finite subset of $q$.

Therefore it must be a partial mapping.

Thus $y_1 = y_2$.

As this holds for all such $x, y_1, y_2$, $q$ is a partial mapping.

Let $K$ be a finite subset of $\Dom q$.

Then the restriction of $q$ to $K$ is a finite subset of $q$.

Thus $q \restriction K \in \FF$.

As this holds for all finite subsets $K$ of $\Dom q$, $q \in \FF$.

So we see that $\FF$ has finite character in the sense of Definition:Finite Character.

$(2)$ implies $(1)$
Let $\FF$ have finite character in the sense of Definition:Finite Character.