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

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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:

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 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.

$\blacksquare$


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