Symmetric and Transitive Relation is not necessarily Reflexive/Proof 3

Theorem

Let $S$ be a set.

Let $\alpha \subseteq S \times S$ be a relation on $S$.

Let $\alpha$ be both symmetric and transitive.

Then it is not necessarily the case that $\alpha$ is also reflexive.

Let $S = \set {1, 2}$ be a set.

Let $\RR$ be the relation on $S$ defined as:

$\forall x, y \in S: x \mathrel \RR y \iff x = y = 2$

$\ds x$

$\RR$

$\ds y$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds y = 2$

$\ds \leadsto \ \ $

$\ds y$

$=$

$\ds x = 2$

$\ds \leadsto \ \ $

$\ds y$

$\RR$

$\ds x$

Thus $\alpha$ is symmetric.

Now let $x \mathrel \RR y$ and $y \mathrel \RR z$

Then:

$x = y = 2, y = z = 2$

and so:

$x \mathrel \RR z$

Now let $x = 1$.

Then it is not the case that:

$x \mathrel \RR x$

and so $\RR$ is not reflexive.

Hence $\RR$ is both symmetric and transitive but not reflexive.

$\blacksquare$

1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Exercise $5$