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

This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code.

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.

$\blacksquare$