Image under Left-Total Relation is Empty iff Subset is Empty
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Theorem
Let $\RR \subseteq S \times T$ be a left-total relation.
Let $A \subseteq S$.
Then:
- $\RR \sqbrk A = \O$ if and only if $A = \O$
Proof
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.
$\Box$
Sufficient Condition
Follows immediately from Image of Empty Set is Empty Set.
$\blacksquare$