Mappings Partially Ordered by Extension

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