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


 * $\card S \le 1$
 * $\card S \le 1$

or:
 * $\card T \le 1$

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:

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

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

Suppose $f \preccurlyeq g$.

Then:

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 $\card S \le 1$.