one more thing you need to do is show that U is contained in the ring. very easy!
Hello friends. I am having trouble interpreting what the following question is asking.
I can show that if a ring of sets contains then it is a Boolean algebra of sets, and it is trivially true that every Boolean algebra of sets is a ring of sets. So is that all I have to do?Assuming that the universal set is non-empty, show that a Boolean algebra of sets can be described as a ring of sets which contain .
P.S. A Boolean algebra of sets is a class of sets that is closed under intersection, union, and complementation.
A ring of sets is a set that is closed under intersection and symmetric difference.
There is a question whether the last part reads (a ring of) (sets that contain U), as the grammar suggest, or (a ring of sets) (which contain U). Since U is the universal set, I don't see much sense in talking about sets that contain U. So, I think, the last part should read "a ring of sets which contains U".Assuming that the universal set U is non-empty, show that a Boolean algebra of sets can be described as a ring of sets which contain U.
I think you need the other direction. "A Boolean algebra of sets can be described as a ring of sets" to me means that every Boolean algebra (1) is a ring of sets and (2) contains U."Show that if a ring of sets contains the universal set then it is a Boolean algebra."
Thank you emakarov I had a feeling you would answer this, this area seems to be your forte. Tell me what you think of the following (I am going to write this as if I was doing it on my worksheet)
Problem: Prove that A) a ring of sets that contains the universal set is a Boolean algebra of sets and that B) a Boolean algebra of sets is a ring of sets that contains the universal set
Proof:
A) We must show that given a ring of sets with that is also a Boolean algebra of sets. We know that is closed under intersection and union (by previous exercises) so it just remains to show that is closed under complementation. To see this merely note that by definition given any we may conclude that , and the conclusion follows.
B) We must show that given a Boolean algebra of sets that and that it is a ring of sets. To see the first part note that since is closed under complementation and union that . Now since is closed under intersection by definition we must merely show that it is closed under symmetric difference, but note that and since is closed under complementation, union, and intersection it is clear that it is closed under symmetric difference.
Did that seem correct? Do you think I fully answered the question?
There is a nice discussion of this problem in the text MEASURE AND INTEGRATION by Sterling K Berberian.
SK uses a slightly different definition for ring of sets. He says a collection of subsets of is a ring if it is closed under difference and finite unions. That is the way I learned the concept.