Equivalence of Definitions of Symmetric Relation

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Theorem

The following definitions of the concept of Symmetric Relation are equivalent:

Definition 1

$\RR$ is symmetric if and only if:

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

Definition 2

$\RR$ is symmetric if and only if it equals its inverse:

$\RR^{-1} = \RR$

Definition 3

$\RR$ is symmetric if and only if it is a subset of its inverse:

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


Proof

Definition 1 implies Definition 3

Let $\RR$ be a relation which fulfils the condition:

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


Then:

\(\ds \) \(\) \(\ds \tuple {x, y} \in \RR\)
\(\ds \) \(\leadsto\) \(\ds \tuple {y, x} \in \RR\) by hypothesis
\(\ds \) \(\leadsto\) \(\ds \tuple {x, y} \in \RR^{-1}\) Definition of Inverse Relation
\(\ds \) \(\leadsto\) \(\ds \RR \subseteq \RR^{-1}\) Definition of Subset


Hence $\RR$ is symmetric by definition 3.

$\Box$


Definition 3 implies Definition 2

Let $\RR$ be a relation which fulfils the condition:

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

Then by Inverse Relation Equal iff Subset:

$\RR = \RR^{-1}$

Hence $\RR$ is symmetric by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $\RR$ be a relation which fulfils the condition:

$\RR^{-1} = \RR$


Then:

\(\ds \) \(\) \(\ds \tuple {x, y} \in \RR\)
\(\ds \) \(\leadsto\) \(\ds \tuple {x, y} \in \RR^{-1}\) as $\RR^{-1} = \RR$
\(\ds \) \(\leadsto\) \(\ds \tuple {y, x} \in \RR\) Definition of Inverse Relation

Hence $\RR$ is symmetric by definition 1.

$\blacksquare$


Sources

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