Inverse Relation Equal iff Subset

Theorem

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. If $\RR$ is a subset or superset of its inverse, then it equals its inverse.

That is, the following are equivalent:

\(\text {(1)}: \quad\)

\(\ds \RR\)

\(\subseteq\)

\(\ds \RR^{-1}\)

\(\text {(2)}: \quad\)

\(\ds \RR^{-1}\)

\(\subseteq\)

\(\ds \RR\)

\(\text {(3)}: \quad\)

\(\ds \RR\)

\(=\)

\(\ds \RR^{-1}\)

Proof

$(1)$ implies $(2)$

Suppose $\RR \subseteq \RR^{-1}$.

Then:

\(\ds \tuple {a, b}\)

\(\in\)

\(\ds \RR^{-1}\)

\(\ds \leadsto \ \ \)

\(\ds \tuple {b, a}\)

\(\in\)

\(\ds \RR\)

Definition of Inverse Relation

\(\ds \leadsto \ \ \)

\(\ds \tuple {b, a}\)

\(\in\)

\(\ds \RR^{-1}\)

Definition of Subset (as $\RR \subseteq \RR^{-1}$)

\(\ds \leadsto \ \ \)

\(\ds \tuple {a, b}\)

\(\in\)

\(\ds \RR\)

Definition of Inverse Relation

Hence $\RR^{-1} \subseteq \RR$.

$\Box$

$(2)$ implies $(1)$

Since $(1)$ implies $(2)$ was already established above, interpreting it on $\RR^{-1}$ yields:

$\RR^{-1} \subseteq \paren {\RR^{-1} }^{-1}$

By Inverse of Inverse Relation, $\paren {\RR^{-1} }^{-1} = \RR$.

Hence $(2)$ implies $(1)$.

$\Box$

$(3)$ if and only if $(1)$ and $(2)$

This is precisely the definition of set equality.

$\blacksquare$