# Intersection of all sets in a family of sets

1. Aug 9, 2008

### Werg22

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?

2. Aug 10, 2008

### tiny-tim

Hi Werg22!

If M is all the subsets [0,x) of the set [0,1],

then N is {0}, which is not in M.

3. Aug 10, 2008

### HallsofIvy

Staff Emeritus
For x> 0. Without that [0, 0]= {0} is in M.

4. Aug 10, 2008

### CRGreathouse

I agree that the example requires x > 0, but wouldn't [0, 0) = {}?

5. Aug 10, 2008

### Werg22

I see, thanks for the counter-example tiny-tim. What if all the elements of M are finite?

6. Aug 10, 2008

### Werg22

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).

7. Aug 22, 2008

### HallsofIvy

Staff Emeritus
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.