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

Proof
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:

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