Reflexive Relation/Examples/Reflexive Relation on Cartesian Plane

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Examples of Use of Symmetric and Transitive Relation is not necessarily Reflexive

The subset of the Cartesian plane defined as:

$\RR := \set {\tuple {x, y} \in \R^2: x \le y \le x + 1}$

determines a relation on $\R^2$ which is reflexive, but neither symmetric nor transitive.


Reflexive Relation

We note that, by definition:

$\forall x \in \R: \tuple {x, y} \in \RR$ such that $x = y$

and so:

$\forall x \in \R: \tuple {x, x} \in \RR$

Hence $\RR$ is reflexive.


Non-Symmetric Relation

Proof by Counterexample:

\(\ds 0 \le 0 + 1\) \(\le\) \(\ds 0 + 1\)
\(\ds \leadsto \ \ \) \(\ds \tuple {0, 1}\) \(\in\) \(\ds \RR\)


$1 > 0$

and so:

$\tuple {1, 0} \notin \RR$

thus demonstrating that $\RR$ is not symmetric.


Non-Transitive Relation

Proof by Counterexample:

\(\ds \tuple {0, 1}\) \(\in\) \(\ds \RR\)
\(\, \ds \land \, \) \(\ds \tuple {1, 2}\) \(\in\) \(\ds \RR\)


$\tuple {0, 2} \notin \RR$

thus demonstrating that $\RR$ is not transitive.


The relation $\RR$ is illustrated below: