Set of Subsets of Reals with Cardinality less than Continuum Cardinality of Local Minimums of Union Closure less than Continuum

Theorem
Let $\mathcal B$ be a set of subsets of $\R$.

Let:
 * $\left\vert{\mathcal B}\right\vert < \mathfrak c$

where
 * $\left\vert{\mathcal B}\right\vert$ denotes the cardinality of $\mathcal B$
 * $\mathfrak c = \left\vert{\R}\right\vert$ denotes continuum.

Let
 * $X = \left\{{x \in \R: \exists U \in \left\{{\bigcup \mathcal G: \mathcal G \subseteq \mathcal B}\right\}: x}\right.$ is local minimum in $\left.{U}\right\}$

Then:
 * $\left\vert{X}\right\vert < \mathfrak c$

Proof
We will prove that
 * $(1): \quad \left\vert{\mathcal B}\right\vert \aleph_0 < \mathfrak c$

where $\aleph_0 = \left\vert{\N}\right\vert$.

In the case when $\left\vert{\mathcal B}\right\vert = \mathbf 0$ we have by Zero of Cardinal Product is Zero:
 * $\left\vert{\mathcal B}\right\vert \aleph_0 = \mathbf 0 < \mathfrak c$

In the case when $\mathbf 0 < \left\vert{\mathcal B}\right\vert < \aleph_9$

In the case when $\left\vert{\mathcal B}\right\vert \geq \aleph_0$ we have

Define
 * $Y = \left\{{x \in \R: \exists U \in \mathcal B: x}\right.$ is local minimum in $\left.{U}\right\}$

We will show that $X \subseteq Y$ by definition of subset.

Let $x \in X$.

By definition of $X$
 * $ \exists U \in \left\{{\bigcup \mathcal G: \mathcal G \subseteq \mathcal B}\right\}: x$ is local minimum in $U$


 * $\exists \mathcal G \subseteq \mathcal B: U = \bigcup \mathcal G$

By definition of local minimum
 * $x \in U$

By definition of union
 * $\exists V \in \mathcal G: x \in V$

By definition of subset
 * $V \in \mathcal B$

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

By Set is Subset of Union:
 * $V \subseteq U$

Then
 * $\exists y \in \R: y < x \land \left({y \,.\,.\, x}\right) \cap V = \varnothing$

By definition:
 * $x$ is local minimum in $V$

Thus by definition of $Y$
 * $x \in Y$

So
 * $(2): \quad X \subseteq Y$

Define $\left({Z_A}\right)_{A \in \mathcal B}$
 * $Z_A = \left\{{x \in \R: x}\right.$ is local minimum in $\left.{A}\right\}$

We will prove that
 * $(3): \quad Y \subseteq \displaystyle \bigcup_{A \in \mathcal B} Z_A$

Let $x \in Y$.

By definition of $Y$
 * $\exists U \in \mathcal B: x$ is local minimum in $U$

By definition of $Z_U$
 * $x \in Z_U$

Thus by definition of union
 * $x \in \displaystyle \bigcup_{A \in \mathcal B} Z_A$

This ends the proof of inclusion.

By Set of Local Minimum is Countable
 * $\forall A \in \mathcal B: Z_A$ is countable

By Countable iff Cardinality not greater Aleph Zero
 * $\forall A \in \mathcal B: \left\vert{Z_A}\right\vert \leq \aleph_0$

By Cardinality of Union not greater than Product
 * $(4): \quad \displaystyle \left\vert{\bigcup_{A \in \mathcal B} Z_A}\right\vert \leq \left\vert{\mathcal B}\right\vert \aleph_0$

Thus