Max Operation Yields Supremum of Parameters/General Case

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
Then:
:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
where:
:$\max$ denotes the max operation
:$\sup$ denotes the supremum.
== Proof ==
We will prove the result by induction on the number of operands $n$.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\max \set {x_1} = \sup \set {x_1}$
By definition of the max operation:
:$\max \set {x_1} = x_1$
From Supremum of Singleton:
:$\sup \set {x_1} = x_1$
So:
:$\max \set {x_1} = \sup \set {x_1} = x_1$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\max \set {x_1, x_2, \dotsc, x_k} = \sup \set {x_1, x_2, \dotsc, x_k}$
from which it is to be shown that:
:$\max \set {x_1, x_2, \dotsc, x_k, x_{k+1}} = \sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$
=== Induction Step ===
This is the induction step.
Now:
As $\struct {S, \preceq}$ is a totally ordered set, all elements of $S$ are comparable by $\preceq$.
Therefore there are two cases to consider:
==== Case 1: $x_{k + 1} \preceq \sup \set {x_1, x_2, \dotsc, x_k}$ ====
Let $x_{k + 1} \preceq \sup \set {x_1, x_2, \dotsc, x_k}$.
By definition of the max operation:
:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \sup \set {x_1, x_2, \dotsc, x_k}$
By definition, $\sup \set {x_1, x_2, \dotsc, x_k}$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k}$.
That is:
:$\forall 1 \le i \le k : x_i \preceq \sup \set {x_1, x_2, \dotsc, x_k}$
Thus:
:$\forall 1 \le i \le k + 1 : x_i \preceq \sup \set {x_1, x_2, \dotsc, x_k}$
So $\sup \set {x_1, x_2, \dotsc, x_k}$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.
Let $y$ be any other upper bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.
Then:
:$y$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k}$.
By definition of the supremum:
:$\sup \set {x_1, x_2, \dotsc, x_k} \preceq y$
It has been shown that the supremum of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$ is:
:$\sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \sup \set {x_1, x_2, \dotsc, x_k}$
Thus it follows:
:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$
==== Case 2: $\sup \set {x_1, x_2, \dotsc, x_k} \preceq x_{k + 1}$ ====
Let $\sup \set {x_1, x_2, \dotsc, x_k} \preceq x_{k + 1}$.
By definition of the max operation:
:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = x_{k + 1}$
By definition, $\sup \set {x_1, x_2, \dotsc, x_k}$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k}$.
That is:
:$\forall 1 \le i \le k : x_i \preceq \sup \set {x_1, x_2, \dotsc, x_k}$
By definition of an ordering:
:$\preceq$ is transitive.
Thus:
:$\forall 1 \le i \le k : x_i \preceq x_{k + 1}$
By definition of an ordering:
:$\preceq$ is reflexive.
Thus:
:$x_{k + 1} \preceq x_{k + 1}$
It follows that $x_{k + 1}$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.
Let $y$ be any other upper bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.
Then:
:$x_{k + 1} \preceq y$.
It has been shown that
:$\sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = x_{k + 1}$.
Thus it follows:
:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1}} = \sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$
In either case, the result holds.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
Category:Max Operation