Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements/Proof 1

Let $A \subseteq \N_{>0}$ such that every $B \subseteq S$ such that $\left|{B}\right| = n$ has a greatest and a smallest element.

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

Then $b$ is both the greatest and smallest 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 $\left|B \setminus \left\{{b}\right\}\right| = n$ elements by Cardinality Less One.

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

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

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 greatest element of $B$, otherwise it's $b$.

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

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

Therefore, by the Principle of Finite Induction, $A = \N_{>0}$ and the proof is complete.