Intersection of Relation with Inverse is Symmetric Relation

Theorem

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

Then $\RR \cap \RR^{-1}$, the intersection of $\RR$ with its inverse, is symmetric.

Proof

Let $\tuple {x, y} \in \RR \cap \RR^{-1}$

By definition of intersection:

$\tuple {x, y} \in \RR$

$\tuple {x, y} \in \RR^{-1}$

By definition of inverse relation:

$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR^{-1}$

$\tuple {x, y} \in \RR^{-1} \implies \tuple {y, x} \in \paren {\RR^{-1} }^{-1}$

By Inverse of Inverse Relation the second statement may be rewritten:

$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR^{-1}$

$\tuple {x, y} \in \RR^{-1} \implies \tuple {y, x} \in \RR$

Then by definition of intersection:

$\tuple {y, x} \in \RR \cap \RR^{-1}$

Hence $\RR \cap \RR^{-1}$ is symmetric.

$\blacksquare$

Sources

• 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.19$: Some Important Properties of Relations: Exercise $6$