Relation is Total iff Union with Inverse is Trivial Relation

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

Then $\RR$ is a total relation :
 * $\RR \cup \RR^{-1} = S \times S$

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

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

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.

Also see

 * Relation is Connected iff Union with Inverse and Diagonal is Trivial Relation