Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty/Lemma

Lemma for Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty

Let $S$ be a set.

Let $\struct {T, \preccurlyeq}$ be an ordered set.

Let $\struct {T^S, \preccurlyeq}$ denote the ordered set of all mappings from $S$ to $T$.

Let $S = \O$.

Then $\struct {T^S, \preccurlyeq}$ is a lattice.

Proof

Recall the definition of lattice:

Let $\struct {S, \preceq}$ be an ordered set.

Then $\struct {S, \preceq}$ is a lattice if and only if:

for all $x, y \in S$, the subset $\set {x, y}$ admits both a supremum and an infimum.

Let $S = \O$.

Then there is one mapping from $S$ to $T$, and that is the empty mapping $e: S \to T$.

Thus we have $T^S = \set e$

From Supremum of Singleton and Infimum of Singleton:

$\sup \set e = e = \inf \set e$

Hence $T^S$ (trivially, in the degenerate sense) is a lattice.

$\blacksquare$