Set of Integers Bounded Above by Integer has Greatest Element

Theorem
Let $\Z$ be the set of integers.

Let $\varnothing \subset S \subseteq \Z$ such that $S$ is bounded above.

Then $S$ has a greatest element.

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

Hence $\forall s \in S: 0 \le M - s$.

Thus the set $T = \left\{{M - s: s \in S}\right\} \subseteq \N$.

The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$.

Hence:
 * $\left({\forall s \in S: b_T \le M - s}\right) \land \left({\exists g_S \in S: b_T = M - g_S}\right)$

So:

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

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

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

$S\,^\prime$ is bounded below by $-M$.

So from Integers Bounded Below has Smallest Element, $S\,^\prime$ 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 $S$.

Also see

 * Integers Bounded Below has Smallest Element
 * Well-Ordering Principle