proof of delta system lemma
Since there are only possible cardinalities for any element of , there must be some such that there are an uncountable number of elements of with cardinality . Let for this . By induction, the lemma holds:
If then there each element of is distinct, and has no intersection with the others, so and .
Suppose . If there is some which is in an uncountable number of elements of then take . Obviously this is uncountable and every element has elements, so by the induction hypothesis there is some of uncountable cardinality such that the intersection of any two elements is . Obviously satisfies the lemma, since the intersection of any two elements is .
On the other hand, if there is no such then we can construct a sequence such that each and for any , by induction. Take any element for , and given , since is countable, is countable. Obviously each element of is in only a countable number of elements of , so there are an uncountable number of elements of which are candidates for . Then this sequence satisfies the lemma, since the intersection of any two elements is .
|Title||proof of delta system lemma|
|Date of creation||2013-03-22 12:55:03|
|Last modified on||2013-03-22 12:55:03|
|Last modified by||Henry (455)|