Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping/Proof

Theorem

Let $S$ be a set of sets.

Let $g$ be a mapping such that:

for every $x \in S$, $x$ is closed under $g$.

Then the intersection $\bigcap S$ of $S$ is also closed under $g$.

The domain of $g$ is not made clear, but the assumption is that:

$\forall x \in S: \forall y \in x: y \in \Dom g$

First we note that by definition of intersection of $S$:

$\bigcap S := \set {y: \forall x \in S: y \in x}$

Then:

$\ds y$

$\in$

$\ds \bigcap S$

$\ds \leadsto \ \ $

$\ds \forall x \in S: \, $

$\ds y$

$\in$

$\ds x$

$\ds \leadsto \ \ $

$\ds \forall x \in S: \, $

$\ds \map g y$

$\in$

$\ds x$

$\ds \leadsto \ \ $

$\ds \map g y$

$\in$

$\ds \bigcap S$

Hence the result by definition of closed under $g$.

$\blacksquare$

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 $2.1$