Set of Integers Bounded Above by Integer has Greatest Element/Proof 2
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Theorem
Let $\Z$ be the set of integers.
Let $\le$ be the ordering on the integers.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded above in $\struct {\Z, \le}$.
Then $S$ has a greatest element.
Proof
Since $S$ is bounded above, $\exists M \in \Z: \forall s \in S: s \le M$.
Hence we can define the set $S' = \set {-s: s \in S}$.
$S'$ is bounded below by $-M$.
So from Set of Integers Bounded Below by Integer 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 $S'$) and $\forall s \in S: s \le g_S$.
So $g_S$ is the greatest element of $S$.
$\blacksquare$