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 maximal elements will follow from duality.

We proceed by induction:

Base case: If $X$ has exactly one element, $x$, then since $x \not\prec x$, $x$ is minimal.

Inductive case: Suppose that every subset of $S$ with $n$ elements has a minimal element.

If $X$ has $n+1$ elements, then $X = X_0 \cup \{x\}$, where $X_0$ has $n$ elements and $x \in X$.

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

If $m_0$ is not a minimal element of $X$, then $x \prec m_0$, so $x$ is a minimal element of $X$.

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