Prove the existence of minimal height of a convex polygon Suppose we have a polygon in $\mathbb{R}^n$. Obviously, we can always trap the polygon into two parallel hyperplanes perpendicular to any given direction. Now the question is, how to prove the existence of the two hyperplanes whose distance is minimum among all parallel hyperplanes sandwiching the polygon?
I think this should be a silly question and basic calculus knowledge should be enough, but I tried a lot and just couldn't make the question mathematically rigorous. There is the same question asked here https://stackoverflow.com/questions/23644372/rotating-a-polygon-to-its-minimal-height, but it is about the algorithm solution, which is not what I want. 
Anyone can give me a rigorous proof? Thanks in advance!
 A: The width of a set $K$ in direction $u$ (where $u$ is a unit vector) is
$$ \max_{x\in K}\ \langle u,x\rangle - \min_{x\in K}\ \langle u,x\rangle $$
Your goal should be to prove that this is a continuous function of $u$, then use the compactness of the unit sphere.  To prove that it is a continuous function of $u$, start with the fact that $\langle u,x\rangle$ is jointly continuous in $u$ and $x$; what does it take to conclude that $u\mapsto\max_{x\in K}\ \langle u,x\rangle$ is continuous?  (For that matter, what does it take to show that it's max and not sup?)
A: Let $P$ be the polygon, which is compact.
Let $\phi(h) = \max_{x,y \in P} \langle h , x-y \rangle$. It is straightforward to show that $\phi$ is continuous.
The set $S=\partial B(0,1)$ is compact, hence $\phi$ is minimised on $S$ at some $\hat{h} \in S$.
Now choose $x,y \in P$ such that $\phi(\hat{h}) = \langle \hat{h} , x-y \rangle$. Then the requisite hyperplanes are
$H_1 = \{ z | \langle \hat{h} , z-x \rangle = 0 \}$ and
$H_2 = \{ z | \langle \hat{h} , z-y \rangle = 0 \}$.
Addendum: To see that $\phi$ is continuous, suppose $h_n \to h$, and let $K$ be any subsequence (that is, $K$ is an infinite subset of $\mathbb{N}$). Let
$x_n,y_n \in P$ such that $\phi(h_n) = \langle h_n , x_n-y_n \rangle$. Since
$P$ is compact, there is some subsequence $K' \subset K$ such that $x_n \overset{K'} {\rightarrow} x$,  $y_n \overset{K'}{\rightarrow} y$, and so 
$\langle h_n , x_n-y_n \rangle \overset{K'} {\rightarrow} \langle h , x-y \rangle \le \phi(h)$. Since $\langle h_n , x_n-y_n \rangle \ge \langle h_n , w-z \rangle$ for any $w,z \in P$, we see that
$\langle h , x-y \rangle \ge \langle h , w-z \rangle$ for any $w,z \in P$,
and so $\phi(h) = \langle h , x-y \rangle$, that is
$\phi(h_n) \overset{K'}{\rightarrow} \phi(h)$. Since $K$ was an arbitrary subsequence, we have $\phi(h_n) {\rightarrow} \phi(h)$.
