Symmetric and Transitive Relation is not necessarily Reflexive/Examples/Subset of 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:

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

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:

Symmetric-Transitive-NonReflexive.png

$\blacksquare$


Sources