Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element

Theorem
Let $n \in \N_{>0}$ be a non-zero natural number.

Let $\N^*_n$ denote the Initial segment $\left\{{1, 2, \ldots, n}\right\}$ of the non-zero natural numbers.

Then every non-empty subset of $\N^*_n$ has a greatest element.

Proof
The proof will proceed by the Principle of Finite Induction on $\N_{>0}$.

Let $S$ be the set defined as:
 * $S := \left\{ {n \in \N:}\right.$ Every non-empty subset of $\N^*_n$ has a greatest element$\left.{}\right\}$

Basis for the Induction
We have that:
 * $\N^*_1 = \left\{{1}\right\}$

The only non-empty subset of $\left\{{1}\right\}$ is $\left\{{1}\right\}$ itself.

This has a greatest element $1$.

So $1 \in S$.

This is our basis for the induction.

Induction Hypothesis
It is to be shown that, if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$.

This is the induction hypothesis:
 * Every non-empty subset of $\N^*_k$ has a greatest element.

It is to be demonstrated that it follows that:
 * Every non-empty subset of $\N^*_{k + 1}$ has a greatest element.

Induction Step
This is our induction step:

Let $T \subseteq \N^*_{k + 1}$.

Either $k + 1 \in T$ or $k + 1 \notin T$.

If $k + 1 \notin T$ then $T \subseteq \N^*_k$.

In that case, then from the induction hypothesis:
 * $T$ has a greatest element.

If $k + 1 \in T$ then:
 * $\forall x \in T: x \le k + 1$

and so $k + 1$ is the greatest element of $T$.

So in both cases $T$ has a greatest element and so $k + 1 \in S$.

So $k \in S \implies k + 1 \in S$ and the result follows by the Principle of Finite Induction:


 * $\forall n \in \N$: every non-empty subset of $\N^*_n$ has a greatest element.