Set of Subset of Reals with Cardinality less than Continuum has not Interval in Union Closure
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Theorem
Let $\BB$ be a set of subsets of $\R$, the set of all real numbers.
Let:
- $\card \BB < \mathfrak c$
where
- $\card \BB$ denotes the cardinality of $\BB$
- $\mathfrak c = \card \R$ denotes continuum.
Let $\FF = \set {\bigcup \GG: \GG \subseteq \BB}$.
Then:
- $\exists x, y \in \R: x < y \land \hointr x y \notin \FF$
Proof
Define:
- $ Z = \leftset {x \in \R: \exists U \in \FF: x}$ is local minimum in $\rightset U$
- $\card Z < \mathfrak c$
Then by Cardinalities form Inequality implies Difference is Nonempty:
- $\R \setminus Z \ne \O$
Hence by definition of empty set:
- $\exists z: z \in \R \setminus Z$
By definition of difference:
- $z \in \R \land z \notin Z$
Thus $z < z + 1$.
We will show that $z$ is local minimum in $\hointr z {z + 1}$.
Thus:
- $z \in \hointr z {z + 1}$
Hence:
- $z - 1 < z$
Thus:
- $\openint {z - 1} z \cap \hointr z {z + 1} = \O$
Then by definition $z$ is a local minimum in $\hointr z {z + 1}$.
Because $z \notin Z$:
- $\hointr z {z + 1} \notin \FF$
$\blacksquare$
Sources
- Mizar article TOPGEN_3:32