Totally Ordered Set is Well-Ordered iff Subsets Contain Infima

Theorem

Let $\struct {S, \preccurlyeq}$ be a totally ordered set.

Then $\struct {S, \preccurlyeq}$ is a well-ordered set if and only if every non-empty subset of $T \subseteq S$ has an infimum such that $\map \inf T \in T$.

Let $T \subseteq S$ such that $m := \map \inf T \in T$.

By definition, $m$ is a lower bound of $T$ and so:

$\forall x \in T: m \preccurlyeq x$

That is:

$\neg \exists x \in T: x \prec m$

Thus by definition $x$ is a minimal element of $T$.

Thus by definition $S$ is well-founded and so is a well-ordered set.

$\Box$

Let $\struct {S, \preccurlyeq}$ be a well-ordered set.

Let $T \subseteq S$.

Hence, by definition, $\struct {T, \preccurlyeq}$, has a smallest element $m$.

But $m \in T$.

Hence the result.

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings