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

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $X$ be a finite, nonempty subset of $S$.

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

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 $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $proposition_n$

Basis for the Induction
Let $X$ have exactly one element $x$.

Since $x \not \prec x$ it follows that $x$ is minimal.

Induction Hypothesis
Now we need to show that, if every set with $n$ elements has a minimal element, where $n \ge 1$, then it logically follows that every set with $n+1$ elements has a minimal element.

So this is the induction hypothesis:


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

Induction Step
Suppose that every subset of $S$ with $n$ elements has a minimal element.

Let $X$ have $n + 1$ elements.

Then:
 * $X = X_0 \cup \left\{{x}\right\}$

where $X_0$ has $n$ elements and $x \in X \setminus X_0$.

Then $X_0$ has a minimal element $m_0$.

If $m_0$ is not a minimal element of $X$, then:
 * $x \prec m_0$

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

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

The result follows by the Principle of Mathematical Induction.