# 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
$\mathcal R$ is a total relation.

## Proof

### Necessary Condition

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

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

Suppose $a = b$.

Then as $\mathcal R$ is reflexive, $\tuple {a, b} \in \mathcal R$.

Suppose $a \ne b$.

Then as $\mathcal R$ is connected, $\tuple {a, b} \in \mathcal R \lor \tuple {b, a} \in \mathcal R$.

That is:

$\forall \tuple {a, b} \in S \times S: \tuple {a, b} \in \mathcal R \lor \tuple {b, a} \in \mathcal R$.

Thus $\mathcal R$ is a total relation.

$\Box$

### Sufficient Condition

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

Then:

$\forall \tuple {a, b} \in S \times S: \tuple {a, b} \in \mathcal R \lor \tuple {b, a} \in \mathcal R$

by definition.

Thus:

$\forall \tuple {a, a} \in S \times S: \tuple {a, a} \in \mathcal R$

and so $\mathcal R$ is reflexive.

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

$\forall \tuple {a, b} \in S \times S: a \ne b \implies \tuple {a, b} \in \mathcal R \lor \tuple {b, a} \in \mathcal R$

and so $\mathcal R$ is connected.

$\blacksquare$