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

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

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

where $\RR^{-1}$ is the inverse of $\RR$ and $\Delta_S$ is the diagonal relation.

Necessary Condition
Let $\RR$ be a connected relation.

By definition of relation:
 * $\RR \subseteq S \times S$
 * $\RR^{-1} \subseteq S \times S$
 * $\Delta_S \subseteq S \times S$

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

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

Suppose $a = b$.

Then $\tuple {a, b} \in \Delta_S$ by definition of the diagonal relation.

Then by Set is Subset of Union:
 * $\tuple {a, b} \in \RR \cup \RR^{-1} \cup \Delta_S$

Suppose otherwise, that is, that $a \ne b$.

As $\RR$ is connected, either:
 * $\tuple {a, b} \in \RR$

or:
 * $\tuple {b, a} \in \RR$

From the definition of inverse relation, either:


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

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

That is:
 * $\tuple {a, b} \in R \cup \RR^{-1}$

Again by Set is Subset of Union:
 * $\tuple {a, b} \in \RR \cup \RR^{-1} \cup \Delta_S$

So, by definition of subset:
 * $S \times S \subseteq \RR \cup \RR^{-1} \cup \Delta_S$

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

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

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

Suppose $a \ne b$.

By definition of diagonal relation:
 * $\tuple {a, b} \notin \Delta_S$

So, by definition of set union:
 * $\tuple {a, b} \in \RR$

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

That is, by definition of inverse relation:
 * $\tuple {a, b} \in \RR$

or:
 * $\tuple {b, a} \in \RR$

So by definition $\RR$ is connected.

Also see

 * Relation is Total iff Union with Inverse is Trivial Relation