Say there's a family of set M with infinitely many elements with the property that whenever X and Y belong to M, so does their intersection. How to justify that the intersection of all elements in in M, N, (interpreted here as the largest subset common to every element in M) is also in M? Since N can't be constructed in finite number of steps, I'm having trouble seeing what justifies the conclusion. Maybe there's a way to establish the existence of two elements in M whose intersection is exactly N?
tiny-tim
Aug10-08, 04:07 AM
Hi Werg22! :smile:
If M is all the subsets [0,x) of the set [0,1],
then N is {0}, which is not in M.
HallsofIvy
Aug10-08, 06:29 AM
Hi Werg22! :smile:
If M is all the subsets [0,x) of the set [0,1],
For x> 0. Without that [0, 0]= {0} is in M.
then N is {0}, which is not in M.
CRGreathouse
Aug10-08, 11:00 AM
For x> 0. Without that [0, 0]= {0} is in M.
I agree that the example requires x > 0, but wouldn't [0, 0) = {}?
Werg22
Aug10-08, 04:51 PM
I see, thanks for the counter-example tiny-tim. What if all the elements of M are finite?
Werg22
Aug10-08, 10:48 PM
Oh, silly me. If the elements of M are finite, it implies that there is a smallest element, making it the intersection of all elements in M (keeping in mind that the intersection of two sets in M is also in M).
HallsofIvy
Aug22-08, 06:54 AM
It looks to me like you are talking about the "finite intersection property" which requires compactness: If every finite collection of a family of compact sets is non-empty, then the intersection of all sets in the family is non-empty".
Generally, "compact" requires the specification of a topology but it is true that any finite set is compact in any topology.