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More specifically in, the set notation for the original set is: $$ S = \begin{Bmatrix} \begin{bmatrix} x\\ y\\ \end{bmatrix} : y = \sin x \end{Bmatrix} $$

So, how would you find the convex hull for this set and write it the same way as the original set?

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Have you drawn a diagram of $S$ and tried to sketch what the complex hull looks like it ought to be? – Henning Makholm Nov 14 '12 at 21:00
Yes, that was a typo. It has been removed. – Tejen Shrestha Nov 14 '12 at 21:03
@HenningMakholm Well, I know that it is a sine curve with $ x \in \left(-\infty, \infty\right) $ and $ y \in \left(-1, 1\right) $ – Tejen Shrestha Nov 14 '12 at 21:08
Then what about $\mathbb{R}\times[-1,1]$ for the interior of the hull and $\mathbb{R}\times\{-1,1\}$ for its boundary ? – Vincent Nivoliers Nov 14 '12 at 21:13
You should prove that $\mathbb{R}\times[-1,1]$ is convex, and that any convex set containing $S$ contains this set. For the first one, prove that any point on a segment between two points of $\mathbb{R}\times[-1,1]$ is in $\mathbb{R}\times[-1,1]$, for the second one, prove that any convex set containing $S$ contains the lines $\mathbb{R}\times\{1\}$ and $\mathbb{R}\times\{-1\}$, and therefore all the points in between. – Vincent Nivoliers Nov 14 '12 at 21:38
up vote 1 down vote accepted

Consider the points where $x\in X_1=\{(4k+1)\pi/2:\ k\in\mathbb{Z}\}$. For any $x\in X_1$, $\sin x=1$. Now consider any $t\in \mathbb R$; there exists $k\in\mathbb Z$ such that $(4k+1)\pi/2\leq t<(4(k+1)+1)\pi/2$. Then there exist $\alpha\in[0,1]$ with $t=\alpha(4k+1)\pi/2+(1-\alpha)(4(k+1)+1)\pi/2$. Thus $$ \begin{bmatrix}t\\1\end{bmatrix}=\alpha\begin{bmatrix}(4k+1)\pi/2\\1\end{bmatrix}+(1-\alpha)\begin{bmatrix}(4(k+1)+1)\pi/2\\1\end{bmatrix}. $$ This shows that $\begin{bmatrix}t\\1\end{bmatrix}$ is in the convex hull of $S$ for all $t\in \mathbb R$ (i.e. the whole line $y=1$ is).

Similarly, working with the numbers $(4k+3)\pi/2$ we get that $\begin{bmatrix}t\\-1\end{bmatrix}$ is in the convex hull for all $t$.

Now take $x\in\mathbb R$, $y\in[-1,1]$. Then there exists $\alpha\in[0,1]$ such that $y=\alpha\,1+(1-\alpha)(-1)$. Then $$ \begin{bmatrix}x\\ y\end{bmatrix}=\alpha\begin{bmatrix}x\\1\end{bmatrix}+(1-\alpha)\begin{bmatrix}x\\-1\end{bmatrix}. $$

In other words, $$ \text{conv}\,S=\left\{\begin{bmatrix}x\\ y\end{bmatrix}:\ x\in\mathbb R,\ y\in[-1,1]\right\}. $$

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