Relation is Total iff Union with Inverse is Trivial Relation

Theorem
Let $\mathcal R$ be a relation on $S$.

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

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

Necessary Condition
Let $\mathcal R$ be a total relation.

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

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

Let $\left({a, b}\right) \in S \times S$.

As $\mathcal R$ is total, either:
 * $\left({a, b}\right) \in \mathcal R$

or:
 * $\left({b, a}\right) \in \mathcal R$

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


 * $\left({a, b}\right) \in \mathcal R$

or:
 * $\left({a, b}\right) \in \mathcal R^{-1}$

That is:
 * $\left({a, b}\right) \in \mathcal R \cup \mathcal R^{-1}$

and so by definition of subset:
 * $S \times S \subseteq \mathcal R \cup \mathcal R^{-1}$

Hence, by definition of set equality:
 * $\mathcal R \cup \mathcal R^{-1} = S \times S$

Sufficient Condition
Let $\mathcal R \cup \mathcal R^{-1} = S \times S$.

Let $\left({a, b}\right) \in S \times S$.

Then by definition of set union:
 * $\left({a, b}\right) \in \mathcal R$

or:
 * $\left({a, b}\right) \in \mathcal R^{-1}$

That is, by definition of inverse relation:
 * $\left({a, b}\right) \in R$

or:
 * $\left({b, a}\right) \in R$

So by definition $\mathcal R$ is total.

Also see

 * Relation is Connected iff Union with Inverse and Diagonal is Complete Set