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

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

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

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

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

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

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

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

Suppose $a = b$.

Then $\left({a, b}\right) \in \Delta_S$ by definition of the diagonal relation.

Then by Subset of Union it follows that $\left({a, b}\right) \in \mathcal R \cup \mathcal R^{-1} \cup \Delta_S$.

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

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

Again by Subset of Union it follows that $\left({a, b}\right) \in \mathcal R \cup \mathcal R^{-1} \cup \Delta_S$.

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

Hence, by Equality of Sets:
 * $\mathcal R \cup \mathcal R^{-1} \cup \Delta_S = S \times S$

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

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

Suppose $a \ne b$.

By definition of diagonal relation it follows that $\left({a, b}\right) \notin \Delta_S$.

So, 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 connected.

Also see

 * Total Relation Union Inverse is Complete Set