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

Theorem
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 $T$.

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

either:
 * $\struct {T, \preccurlyeq}$ is a lattice

or:
 * $S = \O$

Proof
Recall the definition of lattice:

First a lemma to take care of the case where $S = \O$:

Sufficient Condition
Let $\struct {T^S, \preccurlyeq}$ be a lattice.

Let $u, v \in T$ be arbitrary.

We have that as $T^S$ is the set of all mappings from $S$ to $T$.


 * Domain is Non-Empty:

Let $S \ne \O$.

Then there exists $x \in S$ and $f, g \in T^S$ such that:

We have that $\struct {T^S, \preccurlyeq}$ be a lattice.

Then by definition $\set {f, g}$ admits a supremum and an infimum.

Recall that a mapping $c: S \to T$ is the supremum of $\set {f, g}$ in $T^S$ :


 * $(1): \quad c$ is an upper bound of $\set {f, g}$ in $T^S$
 * $(2): \quad c \preccurlyeq d$ for all upper bounds $d$ of $\set {f, g}$ in $T^S$.

Then:

and for all $d \in T^S$ such that $f \preccurlyeq d$ and $g \preccurlyeq d$:

That is:
 * $\map c x = \sup \set {u, v}$

Similarly, recall that a mapping $c: S \to T$ is the infimum of $\set {f, g}$ in $T^S$ :


 * $(1): \quad c$ is a lower bound of $\set {f, g}$ in $T^S$
 * $(2): \quad d \preccurlyeq c$ for all lower bounds $d$ of $\set {f, g}$ in $T^S$.

Then:

and for all $d \in T^S$ such that $d \preccurlyeq f$ and $d \preccurlyeq g$:

That is:
 * $\map c x = \inf \set {u, v}$

As $u, v \in T$ are arbitrary, it follows that $\struct {T, \preccurlyeq}$ is a lattice.


 * Domain is Empty:

Let $S = \O$.

From the lemma:
 * $T^S$ is a lattice.

So, if $T^S$ is a lattice, one of the possibilities is that $S = \O$.

It follows that if $T^S$ is a lattice, then either:
 * $\struct {T, \preccurlyeq}$ is a lattice

or:
 * $S = \O$

Necessary Condition
Let $S = \O$.

From the lemma:
 * $T^S$ is a lattice.

Suppose $S \ne \O$ and $\struct {T, \preccurlyeq}$ is a lattice.

Let $f, g \in T^S$ be arbitrary mappings from $S$ to $T$.

Let $x \in S$ be arbitrary.

Then there exists $u, v \in T$ such that:

We have that $\struct {T, \preccurlyeq}$ is a lattice.

Then by definition $\set {u, v}$ admits a supremum and an infimum.

Hence as $x$ is arbitrary:
 * $\forall x \in S: \set {\map f x, \map g x}$ admits both a supremum and an infimum.

Because $T^S$ is the set of all mappings from $S$ to $T$, there exists a mapping $c \in T^S$ such that:
 * $\forall x \in S: \map c x = \sup \set {\map f x, \map g x}$

That is:
 * $c = \sup \set {f, g}$

Similarly, because $T^S$ is the set of all mappings from $S$ to $T$, there exists a mapping $d \in T^S$ such that:
 * $\forall x \in S: \map d x = \inf \set {\map f x, \map g x}$

That is:
 * $d = \inf \set {f, g}$

So $\set {f, g}$ admits both a supremum and an infimum in $T^S$.

As $f$ and $g$ are arbitrary, it follows by definition that $\struct {T, \preccurlyeq}$ is a lattice.