# Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements

## Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $T \subseteq S$ be a finite, non-empty subset of $S$.

Then $T$ has a maximal element and a minimal element.

### Corollary

Let $\struct {S, \preceq}$ be a finite ordered set.

Let $x \in S$.

Then there exists a maximal element $M \in S$ and a minimal element $m \in S$ such that:

$m \preceq x \preceq M$

## Proof

We will show that each finite subset of $S$ has a minimal element.

The existence of a maximal element then follows from duality.

Proof by induction:

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:

Every set with $n$ elements has a minimal element.

### Basis for the Induction

Let $T$ have exactly one element $x$.

Since $x \nprec x$ it follows that $x$ is minimal.

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:

Every set with $k$ elements has a minimal element.

from which it is to be shown that:

Every set with $k + 1$ elements has a minimal element.

### Induction Step

Suppose that every subset of $S$ with $k$ elements has a minimal element.

Let $T$ have $k + 1$ elements.

Then:

$T = T_0 \cup \set x$

where $T_0$ has $k$ elements and $x \in T \setminus T_0$.

Then $T_0$ has a minimal element $m_0$.

If $m_0$ is not a minimal element of $T$, then:

$x \prec m_0$

Thus $x$ is a minimal element of $T$.

Thus either $m_0$ or $x$ is a minimal element of $T$.

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

Therefore:

For every finite, non-empty subset $T$ of $S$, $T$ has a maximal element and a minimal element

The result follows by the Principle of Mathematical Induction.

$\blacksquare$