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

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 :


 * for every chain $C$ of elements of $S$, $\ds \bigcup C$ is also in $S$.

Let $C_\cap$ be a chain in $\ds \bigcap S$.

Then by :
 * $\forall x \in S: C_\cap$ 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 :
 * $\ds \bigcup C_\cap$ is in $S$

Hence the result by definition of closed under chain unions.