Prove that only principal filters will have the infinite intersection property. Let, $X$ be an infinite set. And also let $X=\{x_i:i\in\mathbb{N}\}$. I have considered a filter defined on $X$ but it doesn't have the infinite intersection property. The filter that I considered is
$$\mathscr{F}(X)=\{S\subseteq X: X\setminus S<\infty\}$$
Where by $X\setminus S<\infty$, I mean that $X\setminus S$ is finite. Using the axioms of filters I showed that the following set is a filter and doesn't have the infinite intersection property.
My question:-
$(1)$- I was wondering if all filters with infinite collection of subsets from $X$, which doesn't have the infinite intersection property have something in common or they are some kind of filters. I am willing to know why and why not?
$(2)$- Secondly are there any filters on this infinite set $X$ which does have the infinite intersection property?
 A: I agree with Eric Wofsey about the definition of "principal filter", but the first sentence of Robert Shore's answer is correct. To prove it, suppose $\mathcal F$ is a filter closed under arbitrary intersections, and let $A$ be the intersection of all the sets in $\mathcal F$. So the assumption implies that $A\in\mathcal F$, and from this (and the fact that $\mathcal F$ is a filter) it follows that $\mathcal F$ consists of exactly the supersets of $A$. So $\mathcal F$ is the principlal filter generated by $A$.
A: No filter on the space of subsets other than a principal filter can possibly have the "infinite intersection property," if by that you mean that that the filter contains arbitrarily large intersections of filter elements.  To see this, for each $x \in X$ choose $A_x \in \mathscr F$ such that $x \notin A_x$.  Since by assumption $\mathscr F$ is not principal, it's always possible to find such an $A_x$ for any $x \in X$.
But then $$\bigcap_{x \in X} A_x = \varnothing.$$
This proves that no filter on the subsets of a countable set can be closed under the intersection of countably many filter elements.
The more interesting question is whether a filter on the subsets of $X$, where $\vert X \vert = \kappa \gt \omega$, can satisfy the $\lambda$-intersection property for any (or all) $\lambda \lt \kappa$.  The answer to that question turns out to be yes.
