Symmetric and Transitive Relation is not necessarily Reflexive/Proof 2

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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.


Proof by Counterexample:

Let $S = \Z$ be the set of integers.

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

$\forall x, y \in S: x \mathrel \alpha y \iff x = y = 0$

\(\ds x\) \(\alpha\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds y = 0\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds x = 0\)
\(\ds \leadsto \ \ \) \(\ds y\) \(\alpha\) \(\ds x\)

Thus $\alpha$ is symmetric.

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


$x = y = 0, y = z = 0$

and so

$x \mathrel \alpha z$

Now let $x = \Z$ such that $x \ne 0$.

Then it is not the case that:

$x \mathrel \alpha x$

and so $\alpha$ is not reflexive.

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