• Dec 29th 2012, 10:59 AM
kezman
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.
• Dec 29th 2012, 11:35 AM
jakncoke
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.
• Dec 30th 2012, 07:23 AM
kezman