Relation is Total iff Union with Inverse is Trivial Relation

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Theorem

Let $\RR$ be a relation on $S$.


Then $\RR$ is a total relation if and only if:

$\RR \cup \RR^{-1} = S \times S$

where:

$\RR^{-1}$ is the inverse of $\RR$.
$S \times S$ is the trivial relation on $S$.


Proof

Necessary Condition

Let $\RR$ be a total relation.

By definition of relation, both $\RR \subseteq S \times S$ and $\RR^{-1} \subseteq S \times S$.

So from Union is Smallest Superset (and indeed, trivially) $\RR \cup \RR^{-1} \subseteq S \times S$.


Let $\tuple {a, b} \in S \times S$.

As $\RR$ is total, either:

$\tuple {a, b} \in \RR$

or:

$\tuple {b, a} \in \RR$


From the definition of inverse relation, this means that either:

$\tuple {a, b} \in \RR$

or:

$\tuple {a, b} \in \RR^{-1}$

That is:

$\tuple {a, b} \in \RR \cup \RR^{-1}$

and so by definition of subset:

$S \times S \subseteq \RR \cup \RR^{-1}$


Hence, by definition of set equality:

$\RR \cup \RR^{-1} = S \times S$

$\Box$


Sufficient Condition

Let $\RR \cup \RR^{-1} = S \times S$.

Let $\tuple {a, b} \in S \times S$.


Then by definition of set union:

$\tuple {a, b} \in \RR$

or:

$\tuple {a, b} \in \RR^{-1}$


That is, by definition of inverse relation:

$\tuple {a, b} \in R$

or:

$\tuple {b, a} \in R$


So by definition $\RR$ is total.

$\blacksquare$


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