Reflexive Closure of Transitive Relation is Transitive

Theorem
Let $S$ be a set.

Let $\RR$ be a transitive relation.

Let $\RR^=$ be the reflexive closure of $\RR$.

Then $\RR^=$ is also transitive.

Proof
Let $a, b, c \in S$.

Suppose that $a \mathrel {\RR^=} b$ and $b \mathrel {\RR^=} c$.

If $a = b$, then since $b \mathrel {\RR^=} c$, also $a \mathrel {\RR^=} c$.

If $b = c$, then since $a \mathrel {\RR^=} b$, also $a \mathrel {\RR^=} c$.

The only case that remains is that $a \ne b$ and $b \ne c$.

Then by the definition of $\RR^=$, $a \mathrel \RR b$ and $b \mathrel \RR c$.

Since $\RR$ is transitive, it follows that:


 * $a \mathrel \RR c$

and hence also $a \mathrel {\RR^=} c$.

Thus $\RR^=$ is transitive.

Also see

 * Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering