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Thread: Borel-sets

  1. #1
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    Borel-sets

    Hello,

    I wish to show the following:

    The Borel-sets of the real line $\displaystyle \mathbb{R}$ are also generated by the system $\displaystyle \left\{(-\infty,a):a\in \mathbb{Q}\right\}$, i.e. that $\displaystyle \mathcal{B}=\mathcal{B(\mathbb{R})}=\sigma(\left\{ (-\infty,a):a\in \mathbb{Q}\right\})$.

    This is what I have done:

    1)

    I will use the following information:

    (a) the Borel $\displaystyle \sigma$-algebra is also generated by the system $\displaystyle \left\{[a,b):a,b\in \mathbb{Q}\right\}$

    (b) if $\displaystyle \mathcal{F} \subset \mathcal{G} \subset \mathcal{A}$ then $\displaystyle \sigma(\mathcal{F}) \subset \sigma(\mathcal{G}) \subset \sigma(\mathcal{A})$

    2)

    Notice that $\displaystyle (\infty,a) \in \mathcal{O}$, where $\displaystyle \mathcal{O}$ is the system of open sets.

    Notice also that $\displaystyle \mathcal{O} \subseteq \mathcal{B}=\mathcal{B(\mathbb{R})}$.

    3)

    Using (b) we have that $\displaystyle \mathcal{O} \subseteq \mathcal{B(\mathbb{R})} \Leftrightarrow \sigma(\mathcal{O}) \subseteq \sigma(\mathcal{B(\mathbb{R})})=\mathcal{B(\mathbb {R})} $

    Since $\displaystyle (\infty,a)$ is only an element of $\displaystyle \mathcal{O}$ we have that $\displaystyle \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\}) \subseteq \sigma(\mathcal{O})$.

    Therefore we have that $\displaystyle \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\}) \subseteq \mathcal{B(\mathbb{R})}$

    4)

    Using (a) we have for $\displaystyle a,b \in \mathbb{Q}$ and $\displaystyle a<b$ that:

    $\displaystyle \mathcal{B}=\left\{[a,b):a,b\in \mathbb{Q}\right\}= \left\{ (-\infty,b) \setminus (-\infty,a):a,b\in \mathbb{Q}\right\} = \left\{(-\infty,b) \cap (-\infty,a)^{c}:a,b\in \mathbb{Q}\right\} $

    By the properties of a $\displaystyle \sigma$-algebra we have that $\displaystyle (-\infty,b) \cap (-\infty,a)^{c} \in \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\})$.

    We therefore have that $\displaystyle \mathcal{B}= \sigma(\left\{[a,b):a,b\in \mathbb{Q}\right\}) \subseteq \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\})$

    5)

    Since we have shown that $\displaystyle \mathcal{B} \subseteq \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\})$ and $\displaystyle \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\}) \subseteq \mathcal{B}$ we conclude that $\displaystyle \mathcal{B}= \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\})$

    Could someone comment on the above proof? Is it fine? Is there something missing?

    Thanks.
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  2. #2
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    Re: Borel-sets

    How should I interpret a missing response?
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  3. #3
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    Re: Borel-sets

    Quote Originally Posted by detalosi View Post
    How should I interpret a missing response?
    That simply means that a person who can give you an answer hasn't gotten to it yet. We've got pretty good coverage but sometimes the only person who can answer hasn't been on.

    The thread is up so you may get a response but if you need an answer ASAP I'd recommend for this specific instance you might want to check in with another site as well.

    -Dan
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