Symmetric and Transitive Relation is not necessarily Reflexive/Proof 3

Proof
Proof by Counterexample:

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

Let $\mathcal R$ be the relation on $S$ defined as:
 * $\forall x, y \in S: x \mathrel {\mathcal R} y \iff x = y = 2$

Thus $\alpha$ is symmetric.

Now let $x \mathrel {\mathcal R} y$ and $y \mathrel {\mathcal R} z$

Then:
 * $x = y = 2, y = z = 2$

and so:


 * $x \mathrel {\mathcal R} z$

Now let $x = 1$.

Then it is not the case that:
 * $x \mathrel {\mathcal R} x$

and so ${\mathcal R}$ is not reflexive.

Hence ${\mathcal R}$ is both symmetric and transitive but not reflexive.