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Math Help - Ring of sets

  1. #1
    MHF Contributor Drexel28's Avatar
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    Ring of sets

    Hello friends. I am having trouble interpreting what the following question is asking.

    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 can show that if a ring of sets \mathbb{T} contains U 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?

    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.
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  2. #2
    Senior Member Shanks's Avatar
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    one more thing you need to do is show that U is contained in the ring. very easy!
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Shanks View Post
    one more thing you need to do is show that U is contained in the ring. very easy!
    What? A Boolean algebra by definition contains U so a ring of sets that is describing it must contain U. Maybe I am misunerstanding the question. Here is how I interpret it:

    "Show that if a ring of sets contains the universal set then it is a Boolean algebra."
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    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.
    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".

    "Show that if a ring of sets contains the universal set then it is a Boolean algebra."
    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.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by emakarov View Post
    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".

    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.
    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 U is a Boolean algebra of sets and that B) a Boolean algebra of sets is a ring of sets that contains the universal set U

    Proof:

    A) We must show that given a ring of sets \mathbb{T} with U\in\mathbb{T} that \mathbb{T} is also a Boolean algebra of sets. We know that \mathbb{T} is closed under intersection and union (by previous exercises) so it just remains to show that \mathbb{T} is closed under complementation. To see this merely note that by definition given any A\in\mathbb{T} we may conclude that A\text{ }\Delta\text{ }U=\left(A\cap U'\right)\cup\left(U\cap A'\right)=\left(A\cap\varnothing\right)\cup A'=\varnothing\cup A'=A'\in\mathbb{T}, and the conclusion follows.

    B) We must show that given a Boolean algebra of sets \mathbb{B} that U\in\mathbb{B} and that it is a ring of sets. To see the first part note that since \mathbb{B} is closed under complementation and union that A\in\mathbb{B}\implies A\cup A'=U\in\mathbb{B}. Now since \mathbb{B} is closed under intersection by definition we must merely show that it is closed under symmetric difference, but note that A\text{ }\Delta\text{ }B=\left(A\cap B'\right)\cup\left(B\cap A'\right) and since \mathbb{B} is closed under complementation, union, and intersection it is clear that it is closed under symmetric difference. \blacksquare


    Did that seem correct? Do you think I fully answered the question?
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  6. #6
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    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 X is a ring if it is closed under difference and finite unions. That is the way I learned the concept.
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  7. #7
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    Yes, I think it's correct.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Plato View Post
    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 X is a ring if it is closed under difference and finite unions. That is the way I learned the concept.
    That is strange. I have also, on planetmath.com, seen rings of sets defined as being a class closed under union and intersection. Ring of sets is quite a varied term apparently.
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