# Condition for Ordered Set of All Mappings to be Total Ordering

## 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 totally ordered set
$\card S \le 1$

or:

$\card T \le 1$

## Proof

### Sufficient Condition

Let $\struct {T^S, \preccurlyeq}$ be a totally ordered set.

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

Let $S$ and $T$ be such that:

 $\ds \exists a, b \in S: \,$ $\ds a$ $\ne$ $\ds b$ $\ds \exists x, y \in T: \,$ $\ds x$ $\ne$ $\ds y$

Let $f: S \to T$ be defined as:

 $\ds \map f a$ $=$ $\ds x$ $\ds \map f b$ $=$ $\ds y$ $\ds \map g a$ $=$ $\ds y$ $\ds \map g b$ $=$ $\ds x$

As $\struct {T^S, \preccurlyeq}$ is a totally ordered set, either $f \preccurlyeq g$ or $g \preccurlyeq f$.

Suppose $f \preccurlyeq g$.

Then:

 $\ds \map f a$ $\preccurlyeq$ $\ds \map g a$ $\ds \map f b$ $\preccurlyeq$ $\ds \map g b$ $\ds \leadsto \ \$ $\ds x$ $\preccurlyeq$ $\ds y$ $\ds \leadsto \ \$ $\ds y$ $\preccurlyeq$ $\ds x$

As $x \ne y$ this shows that $\preccurlyeq$ is not antisymmetric on $T$.

This contradicts the assumption that $\struct {T, \preccurlyeq}$ is an ordered set.

Hence if both $\card T > 1$ and $\card S > 1$ it cannot be the case that $\struct {T^S, \preccurlyeq}$ is a totally ordered set.

Thus by the Rule of Transposition, either $\card T \le 1$ or $\card S \le 1$.