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 $S = \O$:
=== Lemma ===
=== 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.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.20$
Category:Ordered Sets
Category:Lattice Theory
Category:Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty