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Thread: Convex bodies Shadow.

  1. #1
    Member kezman's Avatar
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    Convex bodies Shadow.

    For $\displaystyle X  \subseteq E^3 $
    and $\displaystyle u \in E^3$

    we set
    $\displaystyle S(u;X) = \left \{ tu + x: t > 0,x \in X \right \} $
    $\displaystyle S(u;X)$ is the shadow of X when the light is coming from the direction u.
    Let $\displaystyle u \in E^3$

    .Prove .

    (a) If X is a convex subset of $\displaystyle E^3$
    , then $\displaystyle S(u;X)$ is also convex.

    (b) If X is open in $\displaystyle E^3$

    , then $\displaystyle S(u;X)$ is open and $\displaystyle X \subseteq S(u;X)$.



    Im not sure how to aproach both of them.

    Thanks.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Convex bodies Shadow.

    Well for
    For
    a) Any element of S(u;X) has the form $\displaystyle ut + x \text{ where } t >0 \text{ and } x \in X $ Well, to show convexity you have to show that for any $\displaystyle p \in [0,1] $ $\displaystyle p(tu + x) + (1-p)(tu + x) \in S(u;X) $ So multiplying the p through we get $\displaystyle ptu + px + (1-p)tu + (1-p)x $ since X is convex, $\displaystyle px + (1-p)x = j \in X $. So we got $\displaystyle ptu + (1-p)tu + j $, then factor out the t to get $\displaystyle (pu+(1-p)u)t + j $, which is $\displaystyle (pu + u - pu)t + j $ which is $\displaystyle ut + j \in S(u; X) $ since $\displaystyle j \in X $

    b)
    Assume X is open.
    To show that S(u;X) is open, for any element $\displaystyle j \in S(u;X) $ i need to show that there exists a positive radius r such that this implication holds, if $\displaystyle p \in B_r(j)$ then p must be in S(u;X).
    Now, since $\displaystyle j \in S(u;X) $ j is of the form $\displaystyle j = tu + x $ where $\displaystyle x \in X $. Since X is open, there exists a radius $\displaystyle r_1 > 0 $ such that this implication holds, if $\displaystyle \bar{x} \in B_{r_1}(x)$ then $\displaystyle \bar{x} \in X $.

    Now i will show that for the radius $\displaystyle r_1 > 0 $ derived from j = tu + x (it is the radius for x, an element of X ), if $\displaystyle p \in B_{r_1}(j) $ then $\displaystyle p \in S(u;X) $.
    Now, assume $\displaystyle p \in B_{r_1}(j) $ So $\displaystyle ||p -(ut + x) || = ||p - ut - x || < r_1 $. So this means that $\displaystyle ||(p-ut) - x || < r_1 $ which means $\displaystyle p - ut \in X $ Thus $\displaystyle ut + (p-ut) \in S(u;X) $, thus $\displaystyle p \in S(u; X) $ which means that S(u;X) is open.
    Last edited by jakncoke; Dec 29th 2012 at 04:57 PM.
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  3. #3
    Member kezman's Avatar
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    Re: Convex bodies Shadow.

    thanks for the help jakncoke !!!!!!
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  4. #4
    Member kezman's Avatar
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    Re: Convex bodies Shadow.

    I re open the thread, cause im missing the inclusion X in S(u,X) knowing that X is open. I have no clue.
    Thanks!
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