# 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:

$\max \set {x_1} = x_1$
$\sup \set {x_1} = x_1$

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:

 $\ds \max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$ $=$ $\ds \max \set {\max \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }$ Definition of Max Operation (General Case) $\ds$ $=$ $\ds \max \set {\sup \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }$ Induction hypothesis

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:

$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$

$\blacksquare$