# Max Operation Equals an Operand

## 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:

$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$

where:

$\max$ denotes the max operation.

## Proof

We will prove the result by induction on the number of operands $n$.

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$

### Basis for the Induction

$\map P 1$ is the case:

$\exists i \in \closedint 1 1 : x_i = \max \set {x_1}$

By definition of the max operation:

$\max \set {x_1} = x_1$

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:

$\exists i \in \closedint 1 k : x_i = \max \set {x_1, x_2, \dotsc, x_k}$

from which it is to be shown that:

$\exists i \in \closedint 1 {k + 1} : x_i = \max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$

### Induction Step

This is the induction step.

By the definition of Max Operation:

$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \max \set {\max \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }$

By the induction hypothesis:

$\exists i \in \closedint 1 k : x_i = \max \set {x_1, x_2, \dotsc, x_k}$

So:

$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \max \set {x_i, x_{k + 1} }$

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: $x_{k + 1} \preceq x_i$

By definition of the max operation:

$\max \set {x_i, x_{k + 1} } = x_i$

$\Box$

#### Case 2: $x_i \preceq x_{k + 1}$

By definition of the max operation:

$\max \set {x_i, x_{k + 1} } = x_{k + 1}$

$\Box$

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:

$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$

$\blacksquare$