Mappings Partially Ordered by Extension
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Theorem
Let $S$ and $T$ be sets.
Let $F$ be the set of all mappings from $S$ to $T$.
Let $\mathcal R \subseteq F \times F$ be the relation defined as:
- $\left({f, g}\right) \in \mathcal R \iff \operatorname{Dom} \left({f}\right) \subseteq \operatorname{Dom} \left({g}\right) \land \forall x \in \operatorname{Dom} \left({f}\right): f \left({x}\right) = g \left({x}\right)$
That is, $f \mathrel{\mathcal R} g$ if and only if $g$ is an extension of $f$.
Then $\mathcal R$ is an ordering on $F$.
Proof
Let $x \in \operatorname{Dom} \left({f}\right)$ such that $f \left({x}\right) = y$.
Then by definition $x \in \operatorname{Dom} \left({g}\right)$ and $g \left({x}\right) = y$.
Thus by definition of subset, $f \subseteq g$.
We have that Subset Relation is Ordering.
Hence the result.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 14$: Order