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

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Theorem

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 $\mathbf 0 < \size \BB < \aleph_0$:

\(\ds \size \BB \aleph_0\) \(=\) \(\ds \aleph_0 \size \BB\) Product of Cardinals is Commutative
\(\ds \) \(=\) \(\ds \size {\N \times \BB}\) Definition of Product of Cardinals
\(\ds \) \(=\) \(\ds \map \max {\size \N, \size \BB}\) Cardinal Product Equal to Maximum
\(\ds \) \(=\) \(\ds \aleph_0\) because $\size \BB < \aleph_0$
\(\ds \) \(<\) \(\ds \mathfrak c\) Aleph Zero is less than Cardinality of Continuum


In the case when $\size \BB \ge \aleph_0$ we have:

\(\ds \size \BB \aleph_0\) \(=\) \(\ds \size {\BB \times \N}\) Definition of Product of Cardinals
\(\ds \) \(=\) \(\ds \map \max {\size \BB, \size \N}\) Cardinal Product Equal to Maximum
\(\ds \) \(=\) \(\ds \size \BB\) because $\size \BB \ge \aleph_0$
\(\ds \) \(<\) \(\ds \mathfrak c\) by hypothesis


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.

$\Box$


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:

\(\ds \size X\) \(\le\) \(\ds \size Y\) $(2)$ and Subset implies Cardinal Inequality
\(\ds \) \(\le\) \(\ds \size {\bigcup_{A \mathop \in \BB} Z_A}\) $(3)$ and Subset implies Cardinal Inequality
\(\ds \) \(\le\) \(\ds \size \BB \aleph_0\) $(4)$
\(\ds \) \(<\) \(\ds \mathfrak c\) $(1)$

$\blacksquare$


Sources

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