Number of Minimal Elements is Order Property

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

Let $\map m S$ be the number of minimal elements of $\struct {S, \preccurlyeq}$.

Then $\map m S$ is an order property.

Proof
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be isomorphic ordered sets.

Hence let $\phi: S_1 \to S_2$ be an order isomorphism.

By definition of order property, we need to show that the number of minimal elements of $\struct {S_1, \preccurlyeq_1}$ is equal to the number of minimal elements of $\struct {S_2, \preccurlyeq_2}$.

Let $s \in \struct {S_1, \preccurlyeq_1}$ be a minimal element of $\struct {S_1, \preccurlyeq_1}$.

Then:
 * $\forall x \in S_1: x \preccurlyeq_1 s \implies x = s$

Let $\map \phi y, \map \phi s \in S_2$ such that $\map \phi y \preccurlyeq_2 \map \phi s$.

Then as $\phi$ is an order isomorphism:
 * $y \preccurlyeq_1 s$

and so as $s$ is a minimal element of $\struct {S_1, \preccurlyeq_1}$:
 * $y = s$

From Order Embedding is Injection it follows that $\map \phi s$ is a minimal element of $\struct {S_2, \preccurlyeq_2}$.

Similarly, let $\map \phi s \in \struct {S_2, \preccurlyeq_2}$ be a minimal element of $\struct {S_2, \preccurlyeq_2}$.

Then:
 * $\forall \map \phi y \in S_2: \map \phi y \preccurlyeq_2 \map \phi s \implies \map \phi y = \map \phi s$

Let $x \in S_1: x \preccurlyeq_1 s$.

Then as $\phi$ is an order isomorphism:
 * $\map \phi x \preccurlyeq_2 \map \phi s$

and so as $\map \phi s$ is a minimal element of $\struct {S_2, \preccurlyeq_2}$:
 * $\map \phi x = \map \phi s$

That is:
 * $x = s$

and it follows that $s$ is a minimal element of $\struct {S_1, \preccurlyeq_1}$.

Hence:
 * all minimal elements of $\struct {S_1, \preccurlyeq_1}$ are also minimal elements of $\struct {S_2, \preccurlyeq_2}$

and:
 * all minimal elements of $\struct {S_2, \preccurlyeq_2}$ are also minimal elements of $\struct {S_1, \preccurlyeq_1}$

and the result follows.