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 $S$.

$\blacksquare$