Set of Integers Bounded Above by Integer has Greatest Element

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

Then $$S$$ has a maximal or "greatest" element.

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

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 minimal element, which we can call $$b_T \in T$$.

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

So:

$$ $$ $$ $$

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

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

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 Minimal Element, $$S^\prime$$ has a minimal 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 maximal element of $$S$$.