# Totally Ordered Set is Well-Ordered iff Subsets Contain Infima

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## Theorem

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

Then $\left({S, \preccurlyeq}\right)$ is a well-ordered set iff every non-empty subset of $T \subseteq S$ has an infimum such that $\inf \left({T}\right) \in T$.

## Proof

### Sufficient Condition

Let $T \subseteq S$ such that $m := \inf \left({T}\right) \in T$.

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.

$\Box$

### Necessary Condition

Now suppose $T$ is a well-ordered set.

Then from Well-Ordering is Total Ordering it follows immediately that $\left({S, \preccurlyeq}\right)$ is a totally ordered set.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 14$