Image under Left-Total Relation is Empty iff Subset is Empty

Theorem
Let $\RR \subseteq S \times T$ be a left-total relation.

Let $A \subseteq S$.

Then:
 * $\RR \sqbrk A = \O$ $A = \O$

Necessary Condition
We prove the contrapositive statement:
 * $A \ne \O \implies \RR \sqbrk A \ne \O$

Let $s \in A$.

By definition of left-total relation:
 * $\exists t \in T : \tuple{s, t} \in R$

By definition of image:
 * $\exists t \in T : t \in \RR \sqbrk A$

Hence:
 * $\RR \sqbrk A \ne \O$

The result follows from Rule of Transposition.

Sufficient Condition
Follows immediately from Image of Empty Set is Empty Set.