# Symmetric and Transitive Relation is not necessarily Reflexive/Examples/Subset of Cartesian Plane

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

The subset of the Cartesian plane defined as:

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

determines a relation on $\R^2$ which is symmetric and transitive but not reflexive.

## Proof

### Non-Reflexive Relation

We note that, for example, $\tuple {2, 2} \notin \RR$.

Hence $\RR$ is non-reflexive.

$\Box$

### Symmetric Relation

 $\ds \tuple {x, y}$ $\in$ $\ds \RR$ $\ds \leadsto \ \$ $\ds -1 \le x \le 1$ $\land$ $\ds -1 \le y \le 1$ $\ds \leadsto \ \$ $\ds -1 \le y \le 1$ $\land$ $\ds -1 \le x \le 1$ $\ds \leadsto \ \$ $\ds \tuple {y, x}$ $\in$ $\ds \RR$

thus demonstrating that $\RR$ is symmetric.

$\Box$

### Transitive Relation

 $\ds \tuple {x, y}$ $\in$ $\ds \RR$ $\, \ds \land \,$ $\ds \tuple {y, x}$ $\in$ $\ds \RR$ $\ds \leadsto \ \$ $\ds -1 \le x \le 1$ $\land$ $\ds -1 \le y \le 1$ $\, \ds \land \,$ $\ds -1 \le y \le 1$ $\land$ $\ds -1 \le z \le 1$ $\ds \leadsto \ \$ $\ds -1 \le x \le 1$ $\land$ $\ds -1 \le z \le 1$ $\ds \leadsto \ \$ $\ds \tuple {x, z}$ $\in$ $\ds \RR$

thus demonstrating that $\RR$ is transitive.

$\Box$

The relation $\RR$ is illustrated below:

$\blacksquare$