Let . Suppose we have a family of subsets of such that the intersection of any three members of the family is nonempty. Show that there exists which is common to all the elements of .
Let . Suppose we have a family of subsets of such that the intersection of any three members of the family is nonempty. Show that there exists which is common to all the elements of .
Let be the power set of .
We show that given , either or . Indeed, both and cannot be in because their intersection is empty. Thus the map which maps is well-defined, and a bijection because contains elements. Since the domain and the image of this map partition , one of or must lie in .
Now suppose is not closed under intersection. Then there exists such that . By the above, we must have that is in . But , which contradicts the hypothesis that any three elements of have a nonempty intersection. Therefore is closed under intersection, and ; since does not contain we are done.
Last edited by Bruno J.; November 9th, 2009 at 10:48 PM.
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