Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions/Proof
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Theorem
Let $S$ be a set of sets.
Let $S$ be such that:
- $\forall x \in S: x$ is closed under chain unions.
Then the intersection $\ds \bigcap S$ of $S$ is also closed under chain unions.
Proof
First we note that by definition of intersection of $S$:
- $\ds \bigcap S := \set {y: \forall x \in S: y \in x}$
Recall the definition of closed under chain unions:
$S$ is closed under chain unions if and only if:
Let $C_\cap$ be a chain in $\ds \bigcap S$.
Then by Definition of Intersection of Set of Sets:
- $\forall x \in S: C_\cap$ is a chain in $x$.
Hence as $x$ is closed under chain unions for all $x \in S$:
- $\forall x \in S: \ds \bigcup C_\cap$ is in $x$
Hence by Definition of Intersection of Set of Sets:
- $\ds \bigcup C_\cap$ is in $S$
Hence the result by definition of closed under chain unions.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Exercise $2.1$