Set of Integers Bounded Above by Integer has Greatest Element/Proof 2

Proof
Since $S$ is bounded above, $\exists M \in \Z: \forall s \in S: s \le M$.

Hence we can define the set $S' = \left\{{-s: s \in S}\right\}$.

$S'$ is bounded below by $-M$.

So from Set of Integers Bounded Below has Smallest Element, $S'$ has a smallest element, $-g_S$, say, where $\forall s \in S: -g_S \le -s$.

Therefore $g_S \in S$ (by definition of $S\,^\prime$) and $\forall s \in S: s \le g_S$.

So $g_S$ is the greatest element of $S$.