Finite Character for Sets of Mappings
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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:
\((1)\) | $:$ | $\FF$ has finite character in the sense of Definition:Finite Character/Mappings. | |||||||
\((2)\) | $:$ | $\FF$ has finite character as a set of subsets of $S \times T$ in the sense of Definition:Finite Character. |
Proof
$(1)$ implies $(2)$
Let $\FF$ have finite character in the sense of Definition:Finite Character/Mappings.
That is, suppose that for each partial mapping $f \subseteq S \times T$:
- $f \in \FF$ if and only if 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 $r \in \FF$.
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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.
$\Box$
$(2)$ implies $(1)$
Let $\FF$ have finite character in the sense of Definition:Finite Character.
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