Max Operation Yields Supremum of Parameters
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Theorem
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $x, y \in S$.
Then:
- $\max \set {x, y} = \sup \set {x, y}$
where:
- $\max$ denotes the max operation
- $\sup$ denotes the supremum.
General Case
Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
Then:
- $\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
Proof
As $\struct {S, \preceq}$ be a totally ordered set, all elements of $S$ are $\preceq$-comparable.
Therefore there are two cases to consider:
Case 1: $x \preceq y$
In this case:
- $\max \set {x, y} = y = \sup \set {x, y}$
Case 2: $y \preceq x$
In this case:
- $\max \set {x, y} = x = \sup \set {x, y}$
In either case, the result holds.
$\blacksquare$
Warning
Note that it is considered abuse of notation to write
- $\max = \sup$
This is because:
- $\max: S \times S \to S$
while:
- $\sup: \powerset S \to S$
where $\powerset S$ is the power set of $S$.