Supremum of Absolute Value of Difference equals Supremum of Difference

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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:

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


Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: I can't immediately think of how it would be done, but it would be good if we could devise a neater and more compact notation that what is used here. All the complicated mathematics is being done in the underscript, which makes it not easy to follow. (Improved Dec. 2016.)
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\(\ds \sup_{x, y \mathop \in S} \paren {x - y}\) \(=\) \(\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \paren {x - y}\) as ($x - y \ge 0$ or $x - y \le 0$) is true
\(\ds \) \(=\) \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} }\) by Supremum of Set Equals Maximum of Suprema of Subsets
\(\ds \) \(=\) \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}, 0}\) as $\ds \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} = 0$
\(\ds \) \(=\) \(\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}\) as $\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y} \ge 0$
\(\ds \) \(=\) \(\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}\) as $\size {x − y} = x − y$ since $x − y \ge 0$
\(\ds \) \(=\) \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y} }\) as the two arguments of max are equal
\(\ds \) \(=\) \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{y, x \mathop \in S, y − x \mathop \ge 0} \size {y - x} }\) by renaming variables $x \leftrightarrow y$
\(\ds \) \(=\) \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \le 0} \size {x - y} }\)
\(\ds \) \(=\) \(\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \size {x - y}\) by Supremum of Set Equals Maximum of Suprema of Subsets
\(\ds \) \(=\) \(\ds \sup_{x, y \mathop \in S} \size {x - y}\) as ($x - y \ge 0$ or $x - y \le 0$) is true

$\blacksquare$

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