Ordered Set may not have Minimal Element

Theorem

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

It may be the case that $S$ has no minimal elements.

Proof

Let $\Q_{>0}$ denote the set of (strictly) positive rational numbers.

From Rational Numbers form Ordered Field, the rational numbers $\Q$ are totally ordered by the usual ordering $\le$.

From Subset of Toset is Toset, $\Q_{>0}$ is also totally ordered by $\le$.

Thus $\struct {\Q_{>0}, \le}$ is an ordered set.

Aiming for a contradiction, suppose there exists a minimal element $m$ of $\struct {\Q_{>0}, \le}$.

From Minimal Element in Toset is Unique and Smallest, $m$ is also the smallest element of $\struct {\Q_{>0}, \le}$

By Smallest Strictly Positive Rational Number does not Exist, $\struct {\Q_{>0}, \le}$ has no smallest element.

Hence $m$ cannot be a minimal element of $\struct {\Q_{>0}, \le}$.

It follows by Proof by Contradiction that $\struct {\Q_{>0}, \le}$ has no minimal elements.

$\blacksquare$

Sources

• 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations