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

    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 $\displaystyle (E,d)$ is separable metric space. Prove that its borel $\displaystyle \sigma$-algebra $\displaystyle B(E)$ is generated by system $\displaystyle U(x,\epsilon) = \left\{y \in E: d(y,x)<\epsilon\right\}$.
    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. $\displaystyle \mathcal{U}(x;\epsilon)$ 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 $\displaystyle \sigma$-algebra $\displaystyle B(E)$" simply mean the metric generated topology?

    Do you have a definitions of basis?
    Last edited by Plato; Oct 3rd 2013 at 02:47 AM.
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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 $\displaystyle \sigma$-algebra $\displaystyle \mathcal{B}(E)$ means $\displaystyle \sigma$-algebra generated by open subsets of $\displaystyle E$.

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

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

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

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

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

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

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

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

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

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

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

    So, question: What exactly is the system $\displaystyle \mathcal{U}(x;\epsilon)$? Is $\displaystyle x\in E$ fixed? Is $\displaystyle \epsilon>0$ fixed? Are neither fixed? If neither is fixed, then it is easy to find a union of sets of the form $\displaystyle \mathcal{U}(x;\epsilon)$ that will contain every point of $\displaystyle O\in \mathcal{O}(E)$ 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 $\displaystyle U(E)=B(E)$.

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

    Edit 2: If $\displaystyle \epsilon>0$ is fixed, but $\displaystyle x\in E$ is arbitrary, then if for all $\displaystyle y \in E$, $\displaystyle \{y\} = \left( \left\{ x \in \mathcal{U}(z;\epsilon) \mid z\in E \mbox{ and }y \notin \mathcal{U}(z;\epsilon) \right\} \right)^c$, then you can show $\displaystyle U(E) = B(E)$.
    Last edited by SlipEternal; Oct 5th 2013 at 09:32 AM.
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