Supremum of Absolute Value of Difference equals Supremum of Difference

Theorem
Let $S$ be a non-empty real set.

Let $\ds \sup_{x, y \mathop \in S} \paren {x - y}$ exist.

Then $\ds \sup_{x, y \mathop \in S} \size {x - y}$ exists and:


 * $\ds \sup_{x, y \mathop \in S} \size {x - y} = \sup_{x, y \mathop \in S} \paren {x - y}$

Proof
Consider the set $\set {x - y: x, y \in S, x - y \le 0}$.

There is a number $x'$ in $S$ as $S$ is non-empty.

Therefore, $0 \in \set {x - y: x, y \in S, x - y \le 0}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, and $x - y \le 0$.

Also, $0$ is an upper bound for $\set {x - y: x, y \in S, x - y \le 0}$ by definition.

Accordingly:


 * $\ds \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} = 0$

Consider the set $\left\{{x - y: x, y \in S, x - y \ge 0}\right\}$.

There is a number $x'$ in $S$ as $S$ is non-empty.

Therefore, $0 \in \left\{{x - y: x, y \in S, x - y \ge 0}\right\}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, $x - y \ge 0$.

Accordingly:


 * $\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y} \ge 0$