For 1, an obvious idea is this: if the intersection of a finite subcollection is not in M, add it. If the result has the FIP, this would contradict the maximality of M. So, it is left to prove that the result of addition does, in fact, have the FIP.
I'm working on a problem, but I just can't figure it out.
Let be any set and suppose that is a collection of subsets of that is maximal with respect to the finite intersection property. Prove the following statements are true.
1. The intersection of any finite nonempty subcollection of is a member of .
2. Any subset of that is not disjoint with every member of is contained in .
I'm working on the first one, but I have found nothing that indicates that is closed under finite intersection. Any help is greatly appreciated.
For 1, an obvious idea is this: if the intersection of a finite subcollection is not in M, add it. If the result has the FIP, this would contradict the maximality of M. So, it is left to prove that the result of addition does, in fact, have the FIP.
Let's say we take a finite subcollection, say , of and let where are the sets from . Can we say that, since and has nonempty intersection with every member of , must also have nonempty intersection with members from ?
The idea is that for the purposes of intersection, can replace . If you have any collection that contains , you can remove , add , and the intersection won't change. This is because . Moreover, is finite, so using it to replace in any finite collection also produces a finite collection. All this is to say that, in the context of post #4, is finite and its intersection is empty, just like for . But , which contradicts the FIP of .