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Quasi-sigma algebra question

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  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Let ε = { (-∞,a] : a∈ℝ } be the collection of all intervals of the form (-∞,a] = {x∈ℝ : x≤a} for some a∈ℝ.
    Is ε closed under countable unions?

    2. Relevant equations
    Potentially De Morgan's laws?

    3. The attempt at a solution

    Hi everyone,

    Thanks in advance for looking at my question. I know intuitively this IS closed under countable unions; the problem is showing this rigorously.

    Here is what I have done so far:

    ε is closed under finite unions.

    Proof:

    Use induction:

    n=2 case:
    A1 = (-∞,a1], A2 = (-∞,a2], a1<a2 (trivial if equal).

    Then A1 ∪ A2 = A2 ∈ ε

    Assume true for n=k.

    i.e. A1 ∪...∪ Ak = (-∞,am], where am = {max(ai : i = [1,k]}

    For n=k+1:

    A1 ∪...∪ Ak ∪ Ak+1 = (-∞,am] ∪ (-∞,ak+1], which is in ε by the k=2 case.

    How do I then extend this argument to the infinite case?

    Also, using a similar induction argument, ε is closed under finite intersections.
    ε is NOT closed under complements e.g. if A = (-∞,a], then Ac = (a,∞) ∉ ε. As it is not closed under complements, I don't think the answer can involve De Morgan's laws.

    Thanks for any help!
     
  2. jcsd
  3. Oct 25, 2015 #2

    andrewkirk

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    Why would you think it should be closed under countable unions just because it is closed under finite unions? There are very many rules that hold true for finite collections but not for infinite ones.

    In cases like this, it's often best to start by looking for a counterexample - in this case a countable union that is not in the set. If after some diligent searching you can't find one, you'll start to get an idea of why and be able to construct a proof. Think about what, if any, sort of sets might be able to be made out of infinite (but countable) unions of closed sets that can't be made out of finite ones.
     
  4. Oct 25, 2015 #3

    Krylov

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    Are you sure that ##\varepsilon## is actually closed under countable unions?
    I think that ##\varepsilon## is even closed under countable intersections.
     
  5. Oct 25, 2015 #4
    Thanks for your response, Andrew!

    a counterexample - in this case a countable union that is not in the set: To get this, as the outcome we would have to have some set that was in the form (-∞,a) (i.e. right interval is open) but that would mean there is some set in our union of this form, which is not allowed.
    If we say we are having a countably infinite union of sets of ε though, I still don't see how this creates a problem? Because we can always go to -∞ in the negative direction, so have infinitely many (and thus a subset of countably-infinite) subsets from which to choose for our union.
     
  6. Oct 25, 2015 #5

    Krylov

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    This is not true.
    This is mysterious.
     
  7. Oct 25, 2015 #6

    andrewkirk

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    Why? Are you sure you're not leaping to conclusions here?
     
  8. Oct 25, 2015 #7
    Hi Krylov/Andrew,

    If I take the (finite) union of n number of intervals on the real line in the form (-∞,a], then the resultant union is going to be in the form (-∞,a], yes? Because there will always be some a' that is bigger than all the other a's, hence it will be the right-most interval for the union.

    I assume there is something that happens when we go to the countably infinite case that changes this, which I am not really understanding. By my understanding, the fact we can have countably infinite many subsets in our union would seem to suggest we need to push past a' to head towards +∞... but the fact that we can go to -∞ in the left direction means that there can still be countably infinitely many intervals of the form (-∞,a] between -∞ and a', because both the integers and rational numbers are countably infinite sets that extend in the negative direction...
     
  9. Oct 25, 2015 #8

    andrewkirk

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    Why? What is the biggest number in (0,1)?
     
  10. Oct 25, 2015 #9
    Ah, because an open interval can be written as the countably infinite intersection of closed intervals...

    Hmm, ok, let me think about it some more :) Thanks for your help!
     
  11. Oct 25, 2015 #10

    Krylov

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    Careful, any arbitrary intersection of closed sets is closed. However, it seems now you are on to something.
     
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