Condition for Trivial Relation to be Mapping

Theorem
Let $S$ and $T$ be sets.

Let $\RR = S \times T$ be the trivial relation in $S$ to $T$.

Then $\RR$ is a mapping either:
 * $(2): \card S = 0$

or:
 * $(1): \card T = 1$

where $\card {\, \cdot \,}$ denotes cardinality.

Proof
By definition, the trivial relation in $S$ to $T$ is the set:
 * $\RR = \set {\tuple {s, t}: s \in S, t \in T}$

Proof by Cases:


 * $(1)$: Let $\card S = 0$.

That is, $S = \O$.

By definition of the trivial relation, in this case $\RR = \O$.

This is the empty mapping.

From Empty Mapping is Mapping, this is a mapping.


 * $(2)$: Let $\card S \ne 0$.

That is, $S \ne \O$.


 * $(2) \, \text (a)$: Let $\card T = 0$.

That is, $T = \O$.

From Relation to Empty Set is Mapping iff Domain is Empty, $\RR$ cannot be a mapping for $S \ne \O$.


 * $(2) \, \text (b)$: Let $\card T = 1$.

So, let $T = \set t$.

Then the trivial relation in $S$ to $T$ is:


 * $\RR = \set {\tuple {s, t}: s \in S}$

We have that each element of $S$ is associated with exactly one element of $T$.

Hence $\RR$ is a mapping by definition.


 * $(2) \, \text (c)$: Let $\card T > 1$.

Then:
 * $\exists t_1, t_2 \in T: t_1 \ne t_2$

We have that $S \ne \O$, so $\exists s \in S$.

Thus by definition of trivial relation:
 * $\tuple {s, t_1} \in \RR$

and:
 * $\tuple {s, t_2} \in \RR$

but $t_1 \ne t_2$.

Thus $\RR$ is not a mapping.

Hence the result.