Max Operation Yields Supremum of Parameters

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

Let $x, y \in S$.

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

Proof
There are two cases to consider:

Case 1: $x \preceq y$
In this case $\max(x,y)$ $=$ $y$ $=$ $\sup(\{x,y\})$.

Case 2: $y \preceq x$
In this case $\max(x,y)$ $=$ $x$ $=$ $\sup(\{x,y\})$.

(Note: Equal signs are hypertext.)

In either case, the result holds.

Comment
Notice that it would be incorrect to write
 * $\max$ $=$ $\sup$

since
 * $\max:S \times S \to S$

while
 * $\sup:\mathcal P(S) \to S$

where $\mathcal P(S)$ is the power set of $S$.