Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2

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

Then:
 * $$\preceq$$ is a total ordering;
 * $$\left({S, \preceq}\right)$$ has a minimal element.

Proof
By definition of well-ordered, every subset of $$S$$, including $$S$$ itself, has a minimal element under $$\preceq$$.

So that covers the assertion that $$\left({S, \preceq}\right)$$ has a minimal element.

Now consider $$X = \left\{{a, b}\right\}$$ where $$a, b \in S$$.

Because $$S$$ is well-ordered under $$\preceq$$, $$X$$ has a minimal element.

So either $$\inf X = a$$ or $$\inf X = b$$.

But by definition, $$\inf X \le a$$ or $$\inf X \le b$$.

So either $$a \le b$$ or $$b \le a$$.

That is, $$a$$ and $$b$$ are comparable.

We have shown that any arbitrarily selected elements $$a$$ and $$b$$ of $$S$$ are comparable.

So there are no non-comparable pairs in $$\left({S, \preceq}\right)$$.

Hence the ordering $$\preceq$$ is total.