Reflexive Closure of Antisymmetric Relation is Antisymmetric

Theorem
Let $S$ be a set.

Let $\mathcal R$ be an antisymmetric relation on $S$.

Let $\mathcal R^=$ be the reflexive closure of $\mathcal R$.

Then $\mathcal R^=$ is also antisymmetric.

Proof
Let $a, b \in S$.

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

By definition of $\mathcal R^=$, this means:


 * $a \mathrel{\mathcal R} b$ or $a = b$
 * $b \mathrel{\mathcal R} a$ or $b = a$

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

So suppose $a \mathrel{\mathcal R} b$ and $b \mathrel{\mathcal R} a$.

Since $\mathcal R$ is antisymmetric, it follows that $a = b$ in this case as well.

Hence $\mathcal R^=$ is also antisymmetric.