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Math Help - Help with proof

  1. #1
    Junior Member TheProphet's Avatar
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    Help with proof

    I need help understanding a certain part of the following proof.

    Theorem: (Monotone Class Theorem)
    Let  \mathcal{C} be a class of subsets of \Omega, closed under finite intersections and containing \Omega. Let \mathcal{B} be the smallest class containing \mathcal{C} which is closed under increasing limits and by difference. Then \mathcal{B} = \sigma(\mathcal{C}).

    Proof:

    The intersection of classes of sets closed under increasing limits and differences is a class of that type. So by taking the intersection of all such classes, there always exists a smallest class containing  \mathcal{C} which is closed under increasing limits and by differences. For each set  B , denote
    \mathcal{B}_{B} to be the collection of sets A such that A \in \mathcal{B} and A \cap B \in \mathcal{B}.

    Clearly \mathcal{B}_{B} is closed under increasing limits and by difference.

    Let B \in \mathcal{C}; for each C \in \mathcal{C} one has
    B \cap C \in \mathcal{C} \subset \mathcal{B} and C \in \mathcal{B}, so C \in \mathcal{B}_{B}.

    HERE THE PROBLEM STARTS
    Now let B \in \mathcal{B}. For each C \in \mathcal{C} we have B \in \mathcal{B}_{C}, and because of the preceeding
    B \cap C \in \mathcal{B}, hence C \in \mathcal{B}_{B}, whence \mathcal{C} \subset \mathcal{B}_{B} \subset \mathcal{B}, hence \mathcal{B} = \mathcal{B}_{B}.

    The rest of the proof is just concluding that \mathcal{B} is a \sigma-algebra.

    Back to the problem part. Why must B \in \mathcal{B}_{C}?
    It's easy to see when B \in \mathcal{C}, because \mathcal{C} is closed under finite intersections.
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  2. #2
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    Re: Help with proof

    Hi,

    B \in \mathcal{B}_C iff B \cap C \in \mathcal{B}.
    But both B and C belong to \mathcal{B}, since \mathcal{C} \subset \mathcal{B}, and \mathcal{B} is closed under finite intersection, so B \cap C \in \mathcal{B}.
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  3. #3
    Junior Member TheProphet's Avatar
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    Re: Help with proof

    \mathcal{B} is not closed under finite intersection, \mathcal{C} is.
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  4. #4
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    Re: Help with proof

    You have proved that \mathcal{B}_B is closed under increasing limits and difference, and we know that \mathcal{C} \subset \mathcal{B}_B when B \in \mathcal{C} : hence we have \mathcal{B} \subset \mathcal{B}_B, since \mathcal{B} is the smallest class of subsets closed under increasing limits and difference containing \mathcal{C}.

    We can apply now the same trick when B \in \mathcal{B} and conclude the proof.
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  5. #5
    Junior Member TheProphet's Avatar
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    Re: Help with proof

    Quote Originally Posted by Ahriman View Post
    We can apply now the same trick when B \in \mathcal{B} and conclude the proof.
    Could you please do this? I don't really see how we can conclude
    B \cap C \in \mathcal{B}_{C}, for every C \in \mathcal{C}.

    When you assume B \in \mathcal{B} you can't assume
    \mathcal{B} = \mathcal{B}_{B} ?
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  6. #6
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    Re: Help with proof

    This is absolutely the same scheme of proof than before.

    We have just showed that for all B \in \mathcal{B} and all C \in \mathcal{C}, we have B \cap C \in \mathcal{B}.
    This means that for every B \in \mathcal{B}, we have \mathcal{C} \subset \mathcal{B}_B, and so we get \mathcal{B}_B = \mathcal{B} since \mathcal{B} is the smallest bla bla bla ...
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  7. #7
    Junior Member TheProphet's Avatar
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    Re: Help with proof

    I think you missed the part where I have trouble. I have trouble understanding why
    B \cap C \in \mathcal{B} when B \in \mathcal{B}. I see how you conclude this when you say B \in \mathcal{C}.

    If B \in \mathcal{C}, then obviously B \cap C \in \mathcal{C} for each C because \mathcal{C} is closed under finite intersections. This of course implies C \subset \mathcal{B}_{B} \subset \mathcal{B}, whence \mathcal{B} = \mathcal{B}_{B}, because
    \mathcal{B} is the smallest collection of subsets containing \mathcal{C} that is also closed under increasing limits and differences.

    So, if B \in \mathcal{C}, we could conclude \mathcal{B} = \mathcal{B}_{B}.

    If we now assume B \in \mathcal{B}, then B need not be in \mathcal{C}. And only \mathcal{C} is closed under finite intersections. The whole \mathcal{B} is not. How can you then know that B \cap C \in \mathcal{B} for each C \in \mathcal{C}?
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  8. #8
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    Re: Help with proof

    My second message shows that for B \in \mathcal{C}, we have \mathcal{B}_B = \mathcal{B}. This exactly says that for every B \in \mathcal{C} and C \in \mathcal{B} we have B \cap C \in \mathcal{B} ! (rename B by C and C by B, and you find what you are expecting)

    Next, if B \in \mathcal{B}, we have that \mathcal{C} \subset \mathcal{B}_B by the preceding, and so we conclude the proof ...
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  9. #9
    Junior Member TheProphet's Avatar
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    Re: Help with proof

    Ah, so by the first part (when B \in \mathcal{C} ) we conclude that
    \mathcal{B}_{B} = \mathcal{B}, so \mathcal{B} actually contains all intersections B \cap C, for each C \in \mathcal{C} and B \in \mathcal{B}?
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  10. #10
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    Re: Help with proof

    Yes, by definition of \mathcal{B}_B.
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