Non-Empty Finite Set of Natural Numbers has Greatest Element

Theorem
Let $A$ be a non-empty finite set of natural numbers.

Then $A$ has a greatest element.

Proof
The proof proceeds by induction.

By definition, if $A$ is a non-empty finite set, then $A$ has $m$ elements for some $m \in \N$ such that $m > 0$.

So, for all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * If $A$ has $n$ elements, then $A$ has a greatest element.

Basis for the Induction
Let $A$ have one element which we will call $x$

Trivially:
 * $\forall y \in A: y \le x$

and so $A$ has a greatest element $x$, by definition.

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:
 * If $A$ has $k$ elements, then $A$ has a greatest element

from which it is to be shown that:
 * If $A$ has $k + 1$ elements, then $A$ has a greatest element.

Induction Step
This is the induction step:

Let $A$ have $k + 1$ elements:
 * $A = \set {a_1, a_2, \ldots, a_k, a_{k + 1} }$

Then:
 * $A = \set {a_1, a_2, \ldots, a_k} \cup \set {a_{k + 1} }$

By the induction hypothesis, $\set {a_1, a_2, \ldots, a_k}$ has a greatest element, which we will call $m$.

By definition of greatest element, $m$ is one of the elements of $\set {a_1, a_2, \ldots, a_k}$.

As $A$ is a set, all elements of $A$ are distinct.

Hence:
 * $\forall a \in \set {a_1, a_2, \ldots, a_k}: a \ne a_{k + 1}$

It follows that $m \ne a_{k + 1}$.

Suppose $a_{k + 1}$ is such that $a_{k + 1} > m$.

By definition of greatest element:
 * $\forall y \in \set {a_1, a_2, \ldots, a_k}: y \le m$

and so:
 * $\forall y \in \set {a_1, a_2, \ldots, a_k}: y < a_{k + 1}$

It follows that:
 * $\forall y \in A: y \le a_{k + 1}$

and so $A$ has a greatest element, that is: $a_{k + 1}$.

Suppose $a_{k + 1}$ is such that $a_{k + 1} < m$.

Then it follows immediately that:
 * $\forall y \in A: y \le m$

and so $A$ has a greatest element, that is: $m$.

In both cases $A$ has been shown to have a greatest element.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore, for all $n \in \N_{>0}$:
 * If $A$ has $n$ elements, then $A$ has a greatest element.