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 $$A \subseteq \mathbb{N}^*$$ 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, $$A = \mathbb{N}^*$$ and the proof is complete.