Characterizing elements of a collection closed under countable intersections

Question

Suppose we have a collection $\mathcal{E}$ containing some subsets of a nonempty set $E$. Assume $E \in \mathcal{E}$. Define $\hat{\mathcal{E}}$ to be the smallest collection containing elements of $\mathcal{E}$ and closed under countable intersections. It is clear that $\hat{\mathcal{E}}$ is simply the intersection of all collections containing elements of $\mathcal{E}$ and closed under countable intersections.

What can we say about the elements of $\hat{\mathcal{E}}$?

My hunch is that if $A \in \hat{\mathcal{E}}$, then $A$ is a countable intersection of elements of $\mathcal{E}$. I am unable to prove or disprove this claim though.

If the claim is false, then when can we say the above property holds? If $\mathcal{E}$ is countable, then the above property can be shown to be true.

Answer 1

Let $\mathcal{E}'$ be the collections of all countable intersections of elements of $\mathcal{E}$. We would like to show that $\mathcal{E}' = \hat{\mathcal{E}}$.

By definition $\mathcal{E}'$ contains all elements on $\mathcal{E}$ and is closed under countable intersections because a countable intersection of countable intersections is still a countable intersection. Since $\hat{\mathcal{E}}$ is the smallest collection with these properties, we have $\hat{\mathcal{E}} \subseteq \mathcal{E}'$. But $\hat{\mathcal{E}}$ contains all elements of $\mathcal{E}$, and hence must contain all countable intersections of elements in $\mathcal{E}$, so $\mathcal{E}' \subseteq \hat{\mathcal{E}}$, thus we have shown $\mathcal{E}' = \hat{\mathcal{E}}$.

Answer 2

Your conjecture is correct.

Let $\mathscr{C}$ be the family of all countable intersections of members of $\mathscr{E}$, and let $\mathscr{A}$ be a countable subset of $\mathscr{C}$. Then for each $A\in\mathscr{A}$ there is a countable $\mathscr{E}_A\subseteq\mathscr{E}$ such that $A=\bigcap\mathscr{E}_A$. Let $\mathscr{B}=\bigcup_{A\in\mathscr{A}}\mathscr{E}_A$; then $\mathscr{B}$ is the union of countably many countable subsets of $\mathscr{E}$, so $\mathscr{B}$ is a countable subset of $\mathscr{E}$, and $\bigcap\mathscr{B}\in\mathscr{C}$. Finally,

$$\bigcap\mathscr{B}=\bigcap_{A\in\mathscr{A}}\bigcap\mathscr{E}_A=\bigcap_{A\in\mathscr{A}}A=\bigcap\mathscr{A}\,,$$

so $\bigcap\mathscr{A}\in\mathscr{C}$, and $\mathscr{C}$ is closed under countable intersections.

Now let $\mathfrak{C}$ be the collection of all families of subsets of $E$ that are closed under countable intersection and contain $\mathscr{E}$; we’ve just shown that $\mathscr{C}\in\mathfrak{C}$. If $\mathscr{F}\in\mathfrak{C}$, then $\mathscr{E}\subseteq\mathscr{F}$, and $\mathscr{F}$ is closed under countable intersection, so $\mathscr{C}\subseteq\mathscr{F}$. Thus, $\mathscr{C}\subseteq\bigcap\mathfrak{C}\subseteq\mathscr{C}$, so $\mathscr{C}=\bigcap\mathfrak{C}=\hat{\mathscr{E}}$.

Answer 3

Let $\mathcal{E}'$ be the set of countable intersections of members of $\mathcal{E}$.

Facts: $\mathcal{E} \subseteq \mathcal{E}'$ as $\{E\}$ is a countable (even finite) subcollection of $\mathcal{E}$ and its intersection is $E$; $\mathcal{E}'$ is closed under countable intersections. (uses that a countable union of countable subfamilies is still countable).

As $\hat{\mathcal{E}}$ is the minimal collection with those properties, $\hat{\mathcal{E}} \subseteq \mathcal{E}'$, so your property holds. The other inclusion is also true: if $\mathcal{E}' \ni E' = \bigcap_n E_n$ is a countable intersection of members of $\mathcal{E}$, then as $E_n \in \hat{\mathcal{E}}$ and the latter collection is closed under countable intersections, $\mathcal{E}' \subseteq \hat{\mathcal{E}}$ and we have equality.