Set of Local Minimum is Countable

Theorem
Let $X$ be a subset of $\R$.

The set:
 * $\left\{{x \in X: x}\right.$ is local minimum in $\left.{X}\right\}$

is countable.

Proof
Define:
 * $Y := \left\{{x \in Z: x}\right.$ is local minimum in $\left.{X}\right\}$

By definition of $Y$ and definition of local minimum in set:
 * $\forall x \in Y: \exists y \in \R: y < x \land \left({y \,.\,.\, x}\right) \cap X = \varnothing$

By the Axiom of Choice, define a mapping $f: Y \to \mathcal P \left({\R}\right)$ as:
 * $\forall x \in Y: \exists y \in \R: f \left({x}\right) = \left({y \,.\,.\, x}\right) \land y < x \land f \left({x}\right) \cap X = \varnothing$

We will prove that $f$ is an injection by definition:

Let $x_1, x_2 \in Y$ such that
 * $f \left({x_1}\right) = f \left({x_2}\right)$

By definition of $f$:
 * $\exists y_1 \in \R: f \left({x_1}\right) = \left({y_1 \,.\,.\, x_1}\right) \land y_1 < x_1 \land f \left({x_1}\right) \cap X = \varnothing$

and:
 * $\exists y_2 \in \R: f \left({x_2}\right) = \left({y_2 \,.\,.\, x_2}\right) \land y_2 < x_2 \land f \left({x_2}\right) \cap X = \varnothing$

Then:
 * $\left({y_1 \,.\,.\, x_1}\right) = \left({y_2 \,.\,.\, x_2}\right)$

Thus $x_1 = x_2$.

This ends the proof of injection.

By Cardinality of Image of Injection:
 * $(1): \quad \left\vert{Y}\right\vert = \left\vert{f \left({Y}\right)}\right\vert = \left\vert{\operatorname{Im} \left({f}\right)}\right\vert$

where $\left\vert{Y}\right\vert$ denotes the cardinality of $Y$.

We will prove that $\operatorname{Im} \left({f}\right)$ is mutually disjoint.

Let $A, B \in \operatorname{Im} \left({f}\right)$ such that
 * $A \ne B$.

Then by definition of image:
 * $\exists x_1 \in Y: f \left({x_1}\right) = A$

and
 * $\exists x_2 \in Y: f \left({x_2}\right) = B$.

By difference of $A$ and $B$:
 * $x_1 \ne x_2$

By definition of $f$:
 * $\exists y_1 \in \R: f \left({x_1}\right) = \left({y_1 \,.\,.\, x_1}\right) \land y_1 < x_1 \land f \left({x_1}\right) \cap X = \varnothing$

and:
 * $\exists y_2 \in \R: f \left({x_2}\right) = \left({y_2 \,.\,.\, x_2}\right) \land y_2 < x_2 \land f \left({x_2}\right) \cap X = \varnothing$

Aiming at contradiction suppose that
 * $A \cap B \ne \varnothing$.

$x_1 < x_2$ or $x_1 > x_2$.

In case when $x_1 < x_2$, $x_1 \in f \left({x_2}\right)$ what contradicts with $f \left({x_2}\right) \cap X = \varnothing$.

In case when $x_1 > x_2$, analogically.

This ends the proof that $\operatorname{Im} \left({f}\right)$ is mutually disjoint.

By Set of Mutually Disjoint Intervals is Countable:
 * $\operatorname{Im} \left({f}\right)$ is countable.

Thus by $(1)$ the result:
 * $Y$ is countable.