Reflexive Closure of Antisymmetric Relation is Antisymmetric

Theorem

Let $S$ be a set.

Let $RR$ be an antisymmetric relation on $S$.

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

Then $\RR^=$ is also antisymmetric.

Proof

Let $a, b \in S$.

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

By definition of $\RR^=$, this means:

$a \mathrel \RR b$ or $a = b$

$b \mathrel \RR a$ or $b = a$

If $a = b$ or $b = a$ we are done, by definition of an antisymmetric relation.

So suppose $a \mathrel \RR b$ and $b \mathrel \RR a$.

Because $\RR$ is antisymmetric, it follows that $a = b$ in this case as well.

Hence $\RR^=$ is also antisymmetric.

$\blacksquare$