Preordering is not necessarily Ordering

Theorem
Let $S$ be a set.

Let $\RR$ be a preordering on $S$.

Then it is not necessarily the case that $\RR$ is also an ordering on $S$.

Proof
Consider the relation $\RR$ on the powerset of the natural numbers:


 * $\forall a, b \in \powerset \N: a \mathrel \RR b \iff a \setminus b \text { is finite}$

where $\setminus$ denotes set difference.

It is demonstrated in Preordering Example: Finite Set Difference on Natural Numbers that;
 * $\RR$ is a preordering on $\powerset \N$

but:
 * $\RR$ is not an ordering on $\powerset \N$.