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.
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
\(\ds 0 \le 0 + 1\) | \(\le\) | \(\ds 0 + 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {0, 1}\) | \(\in\) | \(\ds \RR\) |
But:
- $1 > 0$
and so:
- $\tuple {1, 0} \notin \RR$
thus demonstrating that $\RR$ is not symmetric.
$\Box$
Non-Transitive Relation
\(\ds \tuple {0, 1}\) | \(\in\) | \(\ds \RR\) | ||||||||||||
\(\, \ds \land \, \) | \(\ds \tuple {1, 2}\) | \(\in\) | \(\ds \RR\) |
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