Max yields Supremum of Parameters

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Theorem

Let $\left({S, \preceq}\right)$ be a totally ordered set.

Let $x, y \in S$.

Then:

$\max \left({x, y}\right) = \sup \left({\left\{{x, y}\right\}}\right)$

where:

$\max$ denotes the max operation
$\sup$ denotes the supremum.


Proof

As $\left({S, \preceq}\right)$ be a totally ordered set, all elements of $S$ are comparable by $\preceq$.

Therefore there are two cases to consider:

Case 1: $x \preceq y$

In this case:

$\max \left({x, y}\right) = y = \sup \left({\left\{{x, y}\right\}}\right)$


Case 2: $y \preceq x$

In this case:

$\max \left({x, y}\right) = x = \sup \left({\left\{{x, y}\right\}}\right)$

In either case, the result holds.

$\blacksquare$


Comment

Note that it is considered abuse of notation to write

$\max = \sup$

This is because

$\max: S \times S \to S$

while

$\sup: \mathcal P \left({S}\right) \to S$

where $\mathcal P \left({S}\right)$ is the power set of $S$.


Also see