Reflexive and Symmetric Relation is not necessarily Transitive/Proof 2

Proof
Proof by Counterexample:

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

Let $n \in \Z$ such that $n > 1$.

Let $\alpha$ be the relation on $S$ defined as:
 * $\forall x, y \in S: x \mathrel \alpha y \iff \size {x - y} < n$

where $\size x$ denotes the absolute value of $x$.

It is seen that:
 * $\forall x \in \Z: \size {x - x} = 0 < n$

and so:
 * $\forall x \in \Z: x \mathrel \alpha x$

Thus $\alpha$ is reflexive

Then it is seen that:


 * $\forall x, y \in \Z: \size {x - y} = \size {y - x}$

and so if $\size {x - y} < n$ then so is $\size {y - x} < n$.

That is, if $x \mathrel \alpha y$ then $y \mathrel \alpha x$.

Thus $\alpha$ is symmetric.

Now let $x = -\paren{n - 1}, y = 0, z = n - 1$.

We have:

and so both $x \mathrel \alpha y$ and $y \mathrel \alpha z$.

But:

Thus:
 * $\neg x \mathrel \alpha z$

and so $\alpha$ is not transitive.

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