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Thread: Jordan Regions

  1. #1
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    Jordan Regions

    Prove that all balls in R^n are Jordan regions.

    For rectangles it is easy enough, but I can't seem to find a way of proving it for balls. Thanks for the help in advance!
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  2. #2
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    Quote Originally Posted by h2osprey View Post
    Prove that all balls in R^n are Jordan regions.

    For rectangles it is easy enough, but I can't seem to find a way of proving it for balls. Thanks for the help in advance!
    Do you mean that they're Jordan measurable sets?
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    Hmm I'm not sure about the terminology, but basically a Jordan set is one that's bounded, and its boundary has volume zero.
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    Quote Originally Posted by h2osprey View Post
    Hmm I'm not sure about the terminology, but basically a Jordan set is one that's bounded, and its boundary has volume zero.
    Yes they're the same, and you actually have the definition that's useful in this case!

    Remember that $\displaystyle \mathbb{Q} ^n$ is countable and ennumerate it by $\displaystyle (x_n)_{n\in \mathbb{N} }$. Given $\displaystyle a>0$ take cover the sphere (boundary of the ball) with sets of the form $\displaystyle \{ B_{r_{x_n}}(x_n) : r_{x_n} < \frac{a}{2^n} \}=A$ then $\displaystyle A$ certainly covers your sphere and since this last one is compact it can be covered by a finite number of these, say (after possibly renaming them) $\displaystyle B=(B_{r_{x_i}}(x_i))_{i=1}^k$ then $\displaystyle vol(sphere) \leq vol(\cup B ) \leq \sum_{i=1}^{\infty } \frac{a}{2^i} =a$ so the volume of the sphere is bounded by an arbitrarily small number.
    Last edited by Jose27; Jan 8th 2010 at 04:49 PM.
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    Quote Originally Posted by Jose27 View Post
    Yes they're the same, and you actually have the definition that's useful in this case!

    Remember that $\displaystyle \mathbb{Q} ^n$ is countable and ennumerate it by $\displaystyle (x_n)_{n\in \mathbb{N} }$. Given $\displaystyle a>0$ take cover the sphere (boundary of the ball) with sets of the form $\displaystyle \{ B_{r_{x_n}}(x_n) : r_{x_n} < \frac{a}{2^n} \}=A$ then $\displaystyle A$ certainly covers your sphere and since this last one is compact it can be covered by a finite number of these, say (after possibly renaming them) $\displaystyle B=(B_{r_{x_i}}(x_i))_{i=1}^k$ then $\displaystyle vol(sphere) \leq vol(\cup B ) \leq \sum_{i=1}^{\infty } \frac{a}{2^i} =a$ so the volume of the sphere is bounded by an arbitrarily small number.
    Thanks, I understand almost all of it, except the last part -

    $\displaystyle vol(\cup B ) \leq \sum_{i=1}^{\infty } \frac{a}{2^i} =a$

    Why is this inequality true? If we were trying to contain the spheres in cubes, should it not be $\displaystyle \sum_{i=1}^{\infty } (\frac{a}{2^i})^n$ instead?
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