What is the difference between closed under intersections of finitely many subsets and closed under countable intersections? It is given in definition, that $\sigma$ algebra is a type of algebra of sets. So if algebra of sets is defined as being closed under intersections with finitely many subsets, does that mean that being closed under countable intersections implies being closed under intersections with finitely many subsets? Countable set is a set which is bijective to a subset of natural numbers? If so, does finitely many mean that you can also have a bijective function with natural number set? And then, why is that important in probability? 
 A: $$
\bigcap_{n=1}^\infty \left( \frac{-1} n , \frac 1 n \right) = \{0\}.
$$
This is an intersection of countably many open sets, but it is not an intersection of finitely many open sets. An intersection of finitely many open sets would be open, but this set is not open. “Open” in this context would mean containing an open interval about each of its points.
A: Being closed under arbitrary finite intersections does not imply being closed under countable intersections. For example, take the family of co-finite subsets of the natural numbers, that is, the family $F$ of sets whose complement is finite. So, e.g., $\{3,4,5,6,\ldots\}$, which omits the finite set {$0,1,2$}, is in $F$, but the set of even numbers is not (because it’s complement is infinite).
Now, any intersection of finitely many sets from $F$ is in $F$ because the complement of that intersection is the union of the complements, i.e. a finite union of finite sets, which is finite. 
But if you look at the countably many sets $A_n = N - \{n\}$ for $n = 0, 1,2,3,\ldots$, each one is in $F$, but their intersection is empty, and the empty set is not in $F$. So $F$ is not closed under countable intersections.
