@TedShifrin yeah I'm looking for a club with kappa being a regular Cardinal but by taking the intersection of all closed sets, the end result is an empty set so either there's nothing in common (disjoint) or just empty sequence
Club filter is the intersections are also club which is closed and unbounded but someone suggested looking at a URL which I saw already so I deleted my question and would just attempt and list as a reference being used
Fix $\kappa > \omega_{1}$. The club filter on $\kappa$ is not an ultrafilter . So do I prove that $\kappa$ is not an ultrafilter using the definitions of filter ? :/
How to show C being a club if C is defined as a collection of limit ordinals $< \kappa$?? I know that definition of club means closed and unbounded so I have to use proof by cases to show that C is closed and C is unbounded. What I got for unbounded is that since $\kappa \in C$ so that implies sup($C \cap \kappa) = \kappa$ but that's just a one liner.
Maybe I can prove it directly since $\omega$ is countably closed (which by the way $\omega$ is the set of all natural numbers)...it can't be a filter because filters are usually large and ideals are small. Hmm
Maybe $\omega$ is countably closed if it's an ideal... Since ideals are small subsets of X and countably closed may possibly mean that there aren't that many elements
So what I'm thinking is that there exists an uncountable sequence and if there can't be a filter then it's an ideal but that also doesn't make sense either because ideals are small subsets of X. Bddkmabs
