Why isn't a ring a delta-ring? A ring is closed under intersection. And delta-ring is closed under countable intersection. According to my understanding, if a family of sets is closed under intersection, than it should also be closed under countable intersection.
Let's say $R$ is a ring. To $R$ be a delta-ring, $A_n\cap A_{n+1}\cap A_{n+2}\cap ...$ should also be in $R$. And, as a ring is closed under intersection, $A_n\cap A_{n+1}$ is in $R$, then $A_n\cap A_{n+1}\cap A_{n+2}$ is in R, and so on. So, any ring is a delta-ring.
What am I missing? I guess I don't properly understand what countable intersection is. Thank you for reading!

Closure under intersection (i.e. of two elements) implies closure under finite intersection by induction. It does not imply closure under countable (i.e. a type of infinite) intersection.
Take a simple example from topology. The intervals $$\Big( -\frac{1}{i}, \frac{1}{i} \Big)$$ for $i = 1, 2, \dots$ are open, and it's known in topology (and reasonably intuitive, I think) that the intersection of finitely many, i.e. $$\bigcap_{i = 1}^n \Big( -\frac{1}{i}, \frac{1}{i} \Big)$$ is still an open set. But I think it's also quite clear that the countable intersection $$\bigcap_{i = 1}^{\infty} \Big( -\frac{1}{i}, \frac{1}{i} \Big)$$ is not an open set.
Note: I should mention that "countable" is sometimes used to mean "countably infinite", but in certain places it's used as an umbrella term for "finite or denumerable", where denumerable means "can be put into bijection with the natural numbers".
