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