Reflexive Relation/Examples/Reflexive Relation on Cartesian Plane

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.

Proof

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.

$\Box$

Non-Symmetric Relation

Proof by Counterexample:

\(\ds 0 \le 0 + 1\)

\(\le\)

\(\ds 0 + 1\)

\(\ds \leadsto \ \ \)

\(\ds \tuple {0, 1}\)

\(\in\)

\(\ds \RR\)

But:

$1 > 0$

and so:

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

thus demonstrating that $\RR$ is not symmetric.

$\Box$

Non-Transitive Relation

Proof by Counterexample:

\(\ds \tuple {0, 1}\)

\(\in\)

\(\ds \RR\)

\(\, \ds \land \, \)

\(\ds \tuple {1, 2}\)

\(\in\)

\(\ds \RR\)

but

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

thus demonstrating that $\RR$ is not transitive.

$\Box$

The relation $\RR$ is illustrated below:

$\blacksquare$

Sources

• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Relations