Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements

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

Then every finite $T$ such that $\varnothing \subset T \subseteq S$ has both a minimal and a maximal element.

Proof
Let $A \subseteq \N^*$ such that every $B \subseteq S$ such that $\left|{B}\right| = n$ has a maximal and a minimal element.


 * Any $B \subseteq S$ such that $\left|{B}\right| = 1$ has $1$ element, $b \in B$ say.

Then $b$ is both the maximal and minimal element of $B$.

So $1 \in A$.


 * Let $n \in A$.

Let $B \subseteq S$ such that $\left|{B}\right| = n + 1$.

Then $\exists b \in B$, and $\left|B - \left\{{b}\right\}\right| = n$ elements by Cardinality Less One.

So, by the induction hypothesis, $B - \left\{{b}\right\}$ has a maximal element $c$ and a minimal element $a$, as $n \in A$.

Note that $b \ne c$ as $c \in B - \left\{{b}\right\}$ but $b \notin B - \left\{{b}\right\}$.

So as $\left({S, \preceq}\right)$ is totally ordered, either $b \prec c$ or $c \prec b$ as $b \ne c$.

If $b \prec c$ then $c$ is the maximal element of $B$, otherwise it's $b$.

Similarly, either $b \prec a$ or $a \prec b$, and thus either $a$ or $b$ is the minimal element of $B$.

Either way, $B$ has both a maximal and a minimal element, and therefore $n + 1 \in A$.


 * Therefore, by the Principle of Finite Induction, $A = \N^*$ and the proof is complete.