If a set is closed under countable unions, is it closed under countable intersections?

Question

I am trying to think through the intuition of DeMorgan's laws. If I have a set $\Omega$ and a sequence of subsets ($A_n$)={$X_1$, $X_2$, ...} how can I know that given the fact $\Omega$ is closed under countable unions, $\Omega$ is closed under countable intersections, or vice-versa.

Moreover - given the answer to the above, is it enough to show that $\Omega$ is closed under countable intersections OR countable unions as part of proving that $\Omega$ is a $\sigma$-algebra?

Thanks in advance!

Answer 1

It is false. Consider as $\Omega$ the family of subsets of $\mathbb{N}$ whose complement is finite plus the empty set.

Then $\Omega$ is closed under arbitrary unions (in particular, countable unions), but it is easy to find a countable family whose intersection doesn't belong to $\Omega$: take $A_n=\mathbb{N}\setminus\{1,2,\dots,n\}$; then $\bigcap_n A_n=\{0\}\notin\Omega$.

Closure under countable unions implies closure under countable intersections if the family is closed under complements (this follows from De Morgan’s laws).

Answer 2

I believe it is if it is also closed under complements.

Suppose $A \in \Omega$ and $B \in \Omega$.

Then $A^c \in \Omega$ and $B^c \in \Omega$.

Then $(A^c \cup B^c)^c = A \cap B \in \Omega$

Answer 3

Consider the family of sets of natural numbers with at least seventeen elements.

The union of any number of such sets still has at least seventeen elements. But the intersection ...

Answer 4

For a small example:

$\mathcal{F}=\{\{1\},\{2\},\{1,2\}\}$ is closed under unions... (Each of $\{1\}\cup\{1\},\{1\}\cup\{2\},\{1\}\cup\{1,2\},\dots$ etc... are elements of $\mathcal{F}$)

...however it is not closed under intersections. ($\{1\}\cap\{2\}=\emptyset\notin\mathcal{F}$)