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 Posted: Wed Sep 09, 2009 7:01 pm
So, I was working my way through the first assignment of my probability course, when I came across the following problem: Quote: Give a set S and an algebra A on S, such that A isn't also a sigma-algebra on S. Now perhaps it's just my limited understanding, or maybe it's just how the prof gave the lecture, (No textbook attached to this class, all I've got is the notes) but I've been puzzling over this for the last hour now, and I can't even begin to fathom what such an algebra/set pair might be. The only variation I was given on the two is that: arrow In an algebra, if E1, E2, E3, ..., En are in A, then (No idea how to write this, but...) [union of E1->En] is in A. arrow In a sigma-algebra if E1, E2, E3,... are in A then [union E1->Einf] are in A. It's a start, but I can't figure out how to work the two to come up with a solution. I don't want to be given a solution, (Unless it's vital to the explanation) but any help you can give would be much appreciated. Thanks! whee
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Posted: Wed Sep 09, 2009 10:52 pm
So you want a bunch of subsets of S that is closed under finite union and complements, but not countable union.
Um, I suppose you could take the finite subsets of N and their complements. That would be closed under finite union and complementation, but it wouldn't have, for instance, the union of all the sets each containing a single odd number, i.e. it wouldn't have {1}u{3}u{5}u... = {1, 3, 5,...} since both that set and its complement are infinite.
Is that the kind of thing you're looking for?
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Posted: Wed Sep 09, 2009 11:46 pm
Maybe. What I really can't get my head around is how to make the countable union one fail.
You probably already know, but the other axiom that apply to both algebra types are:
-> A is a subset of the power set of S
-> if E is in A, then (S-E) is in A
-> S is in A.
So, since every element and it's complement are in A, the union of its elements would be S. So then A is an algebra on S.
Now for the part I don't get.
Can't one just add infinitely many empty sets onto A and thus it would also be a sigma-algebra on S? Or does that inherently change the nature of A? Because if that can't be done without changing the value/meaning/nature of A, I suppose any finite A would solve the problem.
But if it doesn't change anything, I'm stuck.
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Posted: Thu Sep 10, 2009 3:32 pm
Unless otherwise specified, you can only union two things at a time and you can only perform union finite number of times; this finite number can be as high as you want (i.e. you can union any finite number of elements) but you can't perform union an infinite number of times.
In a sigma-algebra, you can union an infinite number of elements all at once, which makes it different from the situation above.
It's like how the natural numbers are closed under addition of any two natural numbers, but you can't add an infinite number of natural numbers together and still get a natural number.
Also a finite A wouldn't work, because any finite algebra is a sigma-algebra automatically. The difference between sigma-algebras and non-sigma-algebras is in how they treat infinite unions, and if there are only a finite number of elements in A then the notion of infinite union doesn't apply.
So in order to have an example, you'll need an infinite A, and an infinite subset of A such that the union of all the elements in that subset is not in A.
So let's look at the example I gave: the finite and cofinite subsets of N. The union of any two finite sets is finite, and the union of any finite number of finite sets is finite. The union of any two cofinite sets is cofinite, since if cofinite sets A and B are missing |N-A| and |N-B| elements respectively, their union certainly isn't going to be missing more elements. Hence the union of any finite number of cofinite sets is cofinite.
Hence the set of finite and cofinite subsets of N is closed under complementation and finite union, and hence is an algebra.
Now for the fun part: the set of all odd numbers is neither finite nor cofinite. Hence you can't get to it via a finite number of complementation and union operations on finite and cofinite sets. But you can get the set of all odd numbers by an infinite union of finite sets.
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Posted: Thu Sep 10, 2009 11:45 pm
Smartassery: the algebra of matrices S = M(n,ℝ) is not a σ-algebra, and hence fulfills the stated requirements of the problem.
More seriously, you're looking a Boolean algebra that's not complete over infinite joins (or meets, since they're dual), since they're dual to each other. A great example has already been provided by Layra-chan, and many more can be exhibited. And another, take real intervals of the form [a,b), where b can be infinite, and generate the algebra by taking all possible finite unions, intersections, and complements. This forms a Boolean algebra, but not a σ-algebra (exercise: prove this).
Another one is to take from rationals ℚ the subsets in the form {q in ℚ: a < q < b} for irrational a,b to generate the algebra. Once again, this forms a Boolean algebra that is not a σ-algebra (exercise: prove this too).
Side note: All of the above examples, including Layra-chan's, have a common pattern. Unfortunately, it's a bit hard to explain without topology--what we're actually looking at is a totally disconnected but not extremally disconnected spaces. Any such space naturally generates an incomplete Boolean algebra through its clopen subsets.
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Posted: Tue Sep 22, 2009 9:16 am
The solution presented in class, for anyone interested was as follows:
S=Z
A={{All EcS : |E|
Thus it obeys the rules of an algebra, but if you take something like the union of all the singleton sets containing even numbers, which are in the first part, that union won't be in the final set, as it is infinite, and our two groups contain only the finite and cofinite subsets of S. And thus, it isn't a sigma-algebra.
Similarly, All the odd numbers, all the negative numbers, all the positive numbers... pretty much any infinite collection of integers so long as it !=S.
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Posted: Wed Sep 23, 2009 12:54 pm
So that's essentially the algebra I gave, only over Z as opposed to N. But you see why it works, right?
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Posted: Wed Sep 23, 2009 4:20 pm
Yes, I get why it works. In class he actually just gave the example and didn't have time to explain why. I worked that part out on my own. (With a minor hint that I was looking for something neither finite nor cofinite.)
And thanks for your example, I just didn't understand it at the time. Mathematics is hard to convey purely through the written word. Seeing the complete set written out on the board helped it click.
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Posted: Sat Sep 26, 2009 3:40 pm
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Posted: Sun Sep 27, 2009 6:08 am
In my analysis class we had to deal with the Borel algebra over R (the sigma algebra generated by open intervals) and give an example of an element of the Borel algebra that isn't a countable intersection of open sets or a countable union of closed sets. That was a decently interesting question.
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