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

Let $\BB$ be a set of subsets of $\R$.
Let:
:$\size \BB < \mathfrak c$
where
:$\size \BB$ denotes the cardinality of $\BB$
:$\mathfrak c = \size \R$ denotes the cardinality of the continuum.
Let
:$X = \leftset {x \in \R: \exists U \in \set {\bigcup \GG: \GG \subseteq \BB}: x}$ is a local minimum in $\rightset U$
Then:
:$\size X < \mathfrak c$

Proof:
We will prove that:
:$(1): \quad \size \BB \aleph_0 < \mathfrak c$
where $\aleph_0 = \size \N$ by Aleph Zero equals Cardinality of Naturals.
In the case when $\size \BB = \mathbf 0$ we have by Zero of Cardinal Product is Zero:
:$\size \BB \aleph_0 = \mathbf 0 < \mathfrak c$
In the case when $\mathbf 0 < \size \BB < \aleph_0$:
{{begin-eqn}}
{{eqn | l = \size \BB \aleph_0
      | r = \aleph_0 \size \BB
      | c = Product of Cardinals is Commutative
}}
{{eqn | r = \size {\N \times \BB}
      | c = {{Defof|Product of Cardinals}}
}}
{{eqn | r = \map \max {\size \N, \size \BB}
      | c = Cardinal Product Equal to Maximum
}}
{{eqn | r = \aleph_0
      | c = because $\size \BB < \aleph_0$
}}
{{eqn | o = <
      | r = \mathfrak c
      | c = Aleph Zero is less than Cardinality of Continuum
}}
{{end-eqn}}
In the case when $\size \BB \ge \aleph_0$ we have:
{{begin-eqn}}
{{eqn | l = \size \BB \aleph_0
      | r = \size {\BB \times \N}
      | c = {{Defof|Product of Cardinals}}
}}
{{eqn | r = \map \max {\size \BB, \size \N}
      | c = Cardinal Product Equal to Maximum
}}
{{eqn | r = \size \BB
      | c = because $\size \BB \ge \aleph_0$
}}
{{eqn | o = <
      | r = \mathfrak c
      | c = {{hypothesis}}
}}
{{end-eqn}}
Define:
:$Y = \leftset {x \in \R: \exists U \in \BB: x}$ is a local minimum in $\rightset U$
We will show that $X \subseteq Y$ by definition of subset.
Let $x \in X$.
By definition of $X$:
:$\exists U \in \leftset {\bigcup \GG: \GG \subseteq \BB}: x$ is local minimum in $\rightset U$
:$\exists \GG \subseteq \BB: U = \bigcup \GG$
By definition of local minimum:
:$x \in U$
By definition of union:
:$\exists V \in \GG: x \in V$
By definition of subset:
:$V \in \BB$
By definition of local minimum:
:$\exists y \in \R: y < x \land \openint y x \cap U = \O$
By Set is Subset of Union:
:$V \subseteq U$
Then:
:$\exists y \in \R: y < x \land \openint y x \cap V = \O$
By definition:
:$x$ is local minimum in $V$
Thus by definition of $Y$
:$x \in Y$
So
:$(2): \quad X \subseteq Y$
Define $\family {Z_A}_{A \mathop \in \BB}$ as:
:$Z_A = \leftset {x \in \R: x}$ is local minimum in $\rightset A$
We will prove that:
:$(3): \quad Y \subseteq \ds \bigcup_{A \mathop \in \BB} Z_A$
Let $x \in Y$.
By definition of $Y$:
:$\exists U \in \BB: x$ is local minimum in $U$
By definition of $Z_U$:
:$x \in Z_U$
Thus by definition of union:
:$x \in \ds \bigcup_{A \mathop \in \BB} Z_A$
This ends the proof of inclusion.
{{qed|lemma}}
By Set of Local Minimum is Countable:
:$\forall A \in \BB: Z_A$ is countable
By Countable iff Cardinality not greater Aleph Zero:
:$\forall A \in \BB: \size {Z_A} \le \aleph_0$
By Cardinality of Union not greater than Product:
:$(4): \quad \ds \size {\bigcup_{A \mathop \in \BB} Z_A} \le \size \BB \aleph_0$
Thus:
{{begin-eqn}}
{{eqn | l = \size X
      | o = \le
      | r = \size Y
      | c = $(2)$ and Subset implies Cardinal Inequality
}}
{{eqn | o = \le
      | r = \size {\bigcup_{A \mathop \in \BB} Z_A}
      | c = $(3)$ and Subset implies Cardinal Inequality
}}
{{eqn | o = \le
      | r = \size \BB \aleph_0
      | c = $(4)$
}}
{{eqn | o = <
      | r = \mathfrak c
      | c = $(1)$
}}
{{end-eqn}}
{{qed}}