Relation is Connected and Reflexive iff Total

Theorem
Let $S$ be a set.

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

Then:
 * $\mathcal R$ is both a connected relation and a reflexive relation

iff:
 * $\mathcal R$ is a total relation.

Necessary Condition
Let $\mathcal R$ be a relation on $S$ which is both connected and reflexive.

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

Suppose $a = b$.

Then as $\mathcal R$ is reflexive, $\left({a, b}\right) \in \mathcal R$.

Suppose $a \ne b$.

Then as $\mathcal R$ is connected, $\left({a, b}\right) \in \mathcal R \lor \left({b, a}\right) \in \mathcal R$.

That is:
 * $\forall \left({a, b}\right) \in S \times S: \left({a, b}\right) \in \mathcal R \lor \left({b, a}\right) \in \mathcal R$.

Thus $\mathcal R$ is a total relation.

Sufficient Condition
Let $\mathcal R$ be a total relation on $S$.

Then:
 * $\forall \left({a, b}\right) \in S \times S: \left({a, b}\right) \in \mathcal R \lor \left({b, a}\right) \in \mathcal R$

by definition.

Thus:
 * $\forall \left({a, a}\right) \in S \times S: \left({a, a}\right) \in \mathcal R$

and so $\mathcal R$ is reflexive.

If $a \ne b$ the condition still holds, and so:


 * $\forall \left({a, b}\right) \in S \times S: a \ne b \implies \left({a, b}\right) \in \mathcal R \lor \left({b, a}\right) \in \mathcal R$

and so $\mathcal R$ is connected.