# 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 $\aleph_0 = \left\vert{\N}\right\vert$ by Aleph Zero equals Cardinality of Naturals.

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 $\mathbf 0 < \left\vert{\mathcal B}\right\vert < \aleph_9$

 $\displaystyle \left\vert{\mathcal B}\right\vert \aleph_0$ $=$ $\displaystyle \aleph_0 \left\vert{\mathcal B}\right\vert$ Product of Cardinals is Commutative $\displaystyle$ $=$ $\displaystyle \left\vert{\N \times \mathcal B}\right\vert$ definition of Definition:Product of Cardinals $\displaystyle$ $=$ $\displaystyle \max \left({\left\vert{\N}\right\vert, \left\vert{\mathcal B}\right\vert}\right)$ Cardinal Product Equal to Maximum $\displaystyle$ $=$ $\displaystyle \aleph_0$ because $\left\vert{\mathcal B}\right\vert < \aleph_0$ $\displaystyle$ $<$ $\displaystyle \mathfrak c$ Aleph Zero is less than Continuum

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

 $\displaystyle \left\vert{\mathcal B}\right\vert \aleph_0$ $=$ $\displaystyle \left\vert{\mathcal B \times \N}\right\vert$ definition of Definition:Product of Cardinals $\displaystyle$ $=$ $\displaystyle \max \left({\left\vert{\mathcal B}\right\vert, \left\vert{\N}\right\vert}\right)$ Cardinal Product Equal to Maximum $\displaystyle$ $=$ $\displaystyle \left\vert{\mathcal B}\right\vert$ because $\left\vert{\mathcal B}\right\vert \geq \aleph_0$ $\displaystyle$ $<$ $\displaystyle \mathfrak c$ assumption

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

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

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.

$\forall A \in \mathcal B: Z_A$ is countable
$\forall A \in \mathcal B: \left\vert{Z_A}\right\vert \leq \aleph_0$
$(4): \quad \displaystyle \left\vert{\bigcup_{A \in \mathcal B} Z_A}\right\vert \leq \left\vert{\mathcal B}\right\vert \aleph_0$

Thus

 $\displaystyle \left\vert{X}\right\vert$ $\leq$ $\displaystyle \left\vert{Y}\right\vert$ $(2)$ and Subset implies Cardinal Inequality $\displaystyle$ $\leq$ $\displaystyle \left\vert{\bigcup_{A \in \mathcal B} Z_A}\right\vert$ $(3)$ and Subset implies Cardinal Inequality $\displaystyle$ $\leq$ $\displaystyle \left\vert{\mathcal B}\right\vert \aleph_0$ $(4)$ $\displaystyle$ $<$ $\displaystyle \mathfrak c$ $(1)$

$\blacksquare$