# Finite Character for Sets of Mappings

## Theorem

Let $S$ and $T$ be sets.

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

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

Then the following are equivalent:

\((1)\) | $:$ | $\mathcal F$ has finite character in the sense of Definition:Finite Character/Mappings. | ||||||

\((2)\) | $:$ | $\mathcal F$ has finite character as a set of subsets of $S \times T$ in the sense of Definition:Finite Character. |

## Proof

### $(1)$ implies $(2)$

Let $\mathcal F$ have finite character in the sense of Definition:Finite Character/Mappings.

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

- $f \in \mathcal F$ if and only if for each finite subset $K$ of the domain of $f$, the restriction of $f$ to $K$ is in $\mathcal F$.

Let $q \subseteq S \times T$.

First suppose that $q \in \mathcal F$.

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, $\operatorname{Dom} \left({r}\right)$, is a finite subset of the domain of $q$.

Since $\mathcal F$ is of finite character as a set of mappings, $q \restriction \operatorname{Dom} \left({r}\right)$ is in $\mathcal F$.

But $q \restriction \operatorname{Dom} \left({r}\right) = r$.

Therefore $r \in \mathcal F$.

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

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

Let $x \in \operatorname{Dom} \left({q}\right)$.

Let $y_1, y_2 \in T$.

Suppose that $\left({x, y_1}\right), \left({x, y_2}\right) \in q$.

Then $\left\{ {\left({x, y_1}\right), \left({x, y_2}\right)}\right\}$ 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 $\operatorname{Dom} \left({q}\right)$.

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

Thus $q \restriction K \in \mathcal F$.

As this holds for all finite subsets $K$ of $\operatorname{Dom} \left({q}\right)$, $q \in \mathcal F$.

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

$\Box$

### $(2)$ implies $(1)$

Let $\mathcal F$ have finite character in the sense of Definition:Finite Character.