Max Operation Yields Supremum of Parameters

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

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

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 $S$.