Convex bodies Shadow.

**kezman** · December 29th 2012, 10:59 AM

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

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

.Prove .

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

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

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

Im not sure how to aproach both of them.

Thanks.

**jakncoke** · December 29th 2012, 11:35 AM

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

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

Now i will show that for the radius $r_1 > 0$ derived from j = tu + x (it is the radius for x, an element of X ), if $p \in B_{r_1}(j)$ then $p \in S(u;X)$ .
Now, assume $p \in B_{r_1}(j)$ So $||p -(ut + x) || = ||p - ut - x || < r_1$ . So this means that $||(p-ut) - x || < r_1$ which means $p - ut \in X$ Thus $ut + (p-ut) \in S(u;X)$ , thus $p \in S(u; X)$ which means that S(u;X) is open.

**kezman** · December 30th 2012, 07:23 AM

thanks for the help jakncoke !!!!!!

**kezman** · January 25th 2013, 06:38 PM

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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