Set of Local Minimum is Countable
Theorem
Let $X$ be a subset of $\R$.
The set:
- $\leftset {x \in X: x}$ is local minimum in $\rightset X$
is countable.
Proof
Define:
- $Y := \leftset {x \in X: x}$ is local minimum in $\rightset X$
By definition of $Y$ and definition of local minimum in set:
- $\forall x \in Y: \exists y \in \R: y < x \land \openint y x \cap X = \O$
By the Axiom of Choice, define a mapping $f: Y \to \powerset \R$ as:
- $\forall x \in Y: \exists y \in \R: \map f x = \openint y x \land y < x \land \map f x \cap X = \O$
We will prove that $f$ is an injection by definition:
Let $x_1, x_2 \in Y$ such that
- $\map f {x_1} = \map f {x_2}$
By definition of $f$:
- $\exists y_1 \in \R: \map f {x_1} = \openint {y_1} {x_1} \land y_1 < x_1 \land \map f {x_1} \cap X = \O$
and:
- $\exists y_2 \in \R: \map f {x_2} = \openint {y_2} {x_2} \land y_2 < x_2 \land \map f {x_2} \cap X = \O$
Then:
- $\openint {y_1} {x_1} = \openint {y_2} {x_2}$
Thus $x_1 = x_2$.
This ends the proof of injection.
By Cardinality of Image of Injection:
- $(1): \quad \card Y = \card {\map {f^\to} Y} = \card {\Img f}$
where
- $\card Y$ denotes the cardinality of $Y$,
- $\map {f^\to} Y$ denotes the image of $Y$ under $f$,
- $\Img f$ denotes the image of $f$.
We will prove that $\Img f$ is pairwise disjoint by definition.
Let $A, B \in \Img f$ such that
- $A \ne B$.
Then by definition of image:
- $\exists x_1 \in Y: \map f {x_1} = A$
and
- $\exists x_2 \in Y: \map f {x_2} = B$.
By difference of $A$ and $B$:
- $x_1 \ne x_2$
By definition of $f$:
- $\exists y_1 \in \R: \map f {x_1} = \openint {y_1} {x_1} \land y_1 < x_1 \land \map f {x_1} \cap X = \O$
and:
- $\exists y_2 \in \R: \map f {x_2} = \openint {y_2} {x_2} \land y_2 < x_2 \land \map f {x_2} \cap X = \O$
Aiming at contradiction suppose that
- $A \cap B \ne \O$.
$x_1 < x_2$ or $x_1 > x_2$.
In case when $x_1 < x_2$, $x_1 \in \map f {x_2}$ what contradicts with $\map f {x_2} \cap X = \O$.
In case when $x_1 > x_2$, analogically.
This ends the proof that $\Img f$ is pairwise disjoint.
By Set of Pairwise Disjoint Intervals is Countable:
- $\Img f$ is countable.
Thus by $(1)$ and Set is Countable if Cardinality equals Cardinality of Countable Set the result:
- $Y$ is countable.
$\blacksquare$
Sources
- Mizar article TOPGEN_3:19