# Ring of sets

• Dec 15th 2009, 09:48 PM
Drexel28
Ring of sets
Hello friends. I am having trouble interpreting what the following question is asking.

Quote:

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.
• Dec 16th 2009, 12:47 AM
Shanks
one more thing you need to do is show that U is contained in the ring. very easy!
• Dec 16th 2009, 12:51 AM
Drexel28
Quote:

Originally Posted by Shanks
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."
• Dec 16th 2009, 04:54 AM
emakarov
Quote:

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".

Quote:

"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.
• Dec 16th 2009, 01:40 PM
Drexel28
Quote:

Originally Posted by emakarov
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?
• Dec 16th 2009, 03:39 PM
Plato
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.
• Dec 16th 2009, 03:42 PM
emakarov
Yes, I think it's correct.
• Dec 16th 2009, 08:35 PM
Drexel28
Quote:

Originally Posted by Plato
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.