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Math Help - sigma-algebra in separable space

  1. #1
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    sigma-algebra in separable space

    Hi,
    I don't know how to prove this exercise, actually I don't even know how to start :

    Let (E,d) is separable metric space. Prove that its borel \sigma-algebra B(E) is generated by system U(x,\epsilon) = \left\{y \in E: d(y,x)<\epsilon\right\}.

    I would appreciate any hint.
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  2. #2
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    Re: sigma-algebra in separable space

    Quote Originally Posted by drabi View Post
    Let (E,d) is separable metric space. Prove that its borel \sigma-algebra B(E) is generated by system U(x,\epsilon) = \left\{y \in E: d(y,x)<\epsilon\right\}.
    I would appreciate any hint.
    This notation uses vocabulary that is hard for me to follow. I think it means that you are to show that the set of balls. \mathcal{U}(x;\epsilon) forms a basis for the metric topology.

    The countable part comes from the given separable. So perhaps you need to post some more about the terms here. Does "its borel \sigma-algebra B(E)" simply mean the metric generated topology?

    Do you have a definitions of basis?
    Last edited by Plato; October 3rd 2013 at 03:47 AM.
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  3. #3
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    Re: sigma-algebra in separable space

    Hi,
    thanks for answer.

    I'm sorry for the vocabulary, I'll try to explain:
    borel \sigma-algebra \mathcal{B}(E) means \sigma-algebra generated by open subsets of E.

    So I'm supposed to show, that it's ok to take only the sets of open balls \mathcal{U}(x,\epsilon) instead of all open subsets.

    It is clear now?
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  4. #4
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    Re: sigma-algebra in separable space

    Quote Originally Posted by drabi View Post
    borel \sigma-algebra \mathcal{B}(E) means \sigma-algebra generated by open subsets of E.
    So I'm supposed to show, that it's ok to take only the sets of open balls \mathcal{U}(x,\epsilon) instead of all open subsets.
    The fact that E is separable means that it contains a countable dense subset \mathfrak{D}.

    If \mathcal{O} is an open set in E then for each x\in\mathcal{O} then \exists\delta_x such that \mathcal{U}(x,\delta_x)\subset\mathcal{O}.

    Now let \epsilon_x=\delta_x/2. Then (\exists t_x\in\mathfrak{D})[d(x,t_x)<\epsilon_x].
    Thus we see that x\in\mathcal{U}(t_x,\epsilon_x)]\subset\mathcal{O}.

    \mathcal{O}=\bigcup\limits_{x \in \mathcal{O}} \mathcal{U}(t_x,\epsilon_x)

    I have done far too much for you. So you fill in details.
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  5. #5
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    Re: sigma-algebra in separable space

    Let \mathcal{O}(E) be the set of all open sets of E. Let \mathcal{P}(E) be the power set of E.
    Let \mathcal{B}(E) = \left\{B \subseteq \mathcal{P}(E) \mid B \mbox{ is a }\sigma\mbox{-algebra and }\mathcal{O}(E) \subseteq B \right\}

    Then by definition, we know the Borel \sigma-algebra B(E) = \bigcap_{B \in \mathcal{B}(E)} B.

    Next, let U(E) be the \sigma-algebra generated by the sets \mathcal{U}(x;\epsilon). In other words, it is the intersection of all \sigma-algebras containing the sets \mathcal{U}(x;\epsilon).

    Now, we want to show that U(E) = B(E). How do you show that two sets are equal? You show that U(E) \subseteq B(E) and B(E) \subseteq U(E).

    For any \epsilon>0 and any x\in E, \mathcal{U}(x;\epsilon) is an open set of E, so it is an element of \mathcal{O}(E). Since \mathcal{U}(x;\epsilon) \in \mathcal{O}(E) \subseteq B(E), and U(E) is the smallest \sigma-algebra containing all sets \mathcal{U}(x;\epsilon), it must be that U(E) \subseteq B(E).

    Next, how do we show that B(E) \subseteq U(E)? We take an arbitrary open set O \in \mathcal{O}(E). Then we need to show that O \in U(E). If this is true for any arbitrary O, then since B(E) is the smallest \sigma-algebra containing \mathcal{O}(E), it must be that B(E) \subseteq U(E).

    So, question: What exactly is the system \mathcal{U}(x;\epsilon)? Is x\in E fixed? Is \epsilon>0 fixed? Are neither fixed? If neither is fixed, then it is easy to find a union of sets of the form \mathcal{U}(x;\epsilon) that will contain every point of O\in \mathcal{O}(E) and you don't even need complements. If one is fixed, but not another, this becomes a little more difficult. If both are fixed, then you most definitely do not get U(E)=B(E).

    Edit: If x\in E is fixed, but \epsilon>0 is arbitrary, then you do not get U(E) = B(E) either. Alternately, if x\in E is arbitrary, but \epsilon>0 is fixed, you might still be able to show U(E) = B(E)... I would have to think about that.

    Edit 2: If \epsilon>0 is fixed, but x\in E is arbitrary, then if for all y \in E, \{y\} = \left( \left\{ x \in \mathcal{U}(z;\epsilon) \mid z\in E \mbox{ and }y \notin \mathcal{U}(z;\epsilon) \right\} \right)^c, then you can show U(E) = B(E).
    Last edited by SlipEternal; October 5th 2013 at 10:32 AM.
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