Ordered Set with Multiple Minimal Elements has no Smallest Element

Theorem

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

Let $\struct {S, \preccurlyeq}$ have more than one minimal element.

Then $\struct {S, \preccurlyeq}$ has no smallest element.

Proof

Let $s$ and $t$ both be minimal elements of $\struct {S, \preccurlyeq}$ such that $s \ne t$.

Then by definition:

$\forall x \in S: x \preccurlyeq s \implies s = x$

and:

$\forall x \in S: x \preccurlyeq t \implies t = x$

Aiming for a contradiction, suppose $S$ has a smallest element $m$.

Then by definition:

$\forall y \in S: m \preccurlyeq y$

Hence:

$m \preccurlyeq s$

and also:

$m \preccurlyeq t$

But because $s$ and $t$ are both minimal elements of $\struct {S, \preccurlyeq}$:

$s = m$

and:

$t = m$

This contradicts the fact that $s \ne t$.

Hence by Proof by Contradiction there can be no smallest element of $\struct {S, \preccurlyeq}$.

$\blacksquare$