2
$\begingroup$

After doing some reading on the V-representation of a convex polytope (finite set of extreme points, also the convex hull?), it's often simply stated that the convex hull is compact. Can anyone show or explain how the proof would go?

Can you write the finite number of points as a convex combination and go from there?

$\endgroup$
4
$\begingroup$

Let $V = \{ v_k \}_{k=1}^p$ and suppose $C = \operatorname{co} V$.

Let $\Sigma = \{ \lambda \in [0,1]^p | \sum_l \lambda_k = 1 \}$. We see that $\Sigma$ is compact (closed & bounded). Define $f: \mathbb{R}^p \to \mathbb{R}^p$ by $f(\lambda) = \sum_k \lambda_k v_k$. It is easy to see that $f$ is continuous, and so $f(\Sigma) = C$ is compact.

Another approach is to use Carathéodory's theorem, and let $\Sigma = \{ \lambda \in [0,1]^{n+1} | \sum_l \lambda_k = 1 \}$, where $n$ is the dimension of the ambient space. Then define $\phi:\Sigma \times (\mathbb{R}^n)^{n+1} \to \mathbb{R}^n$ by $\phi(\lambda, v_1,...,v_{n+1}) = \sum_{k=1}^{n+1} \lambda_k v_k$ and note that $\phi$ is continuous. Then note that $\Sigma \times (V)^{n+1}$ is compact, and so $C = \phi(\Sigma \times (V)^{n+1})$ is compact too.

The advantage of the latter approach is that it shows that for any compact $V$ (not just a finite set of points), the set $\operatorname{co} V$ is compact.

$\endgroup$
  • $\begingroup$ Can you explain what $\in [0,1]^p$ stands for? It refers to the $p^{th}$ $\lambda$ right? $\endgroup$ – alfien Feb 23 '14 at 8:30
  • $\begingroup$ $[0,1]$ is the interval $0 \le x \le 1$, $[0,1]^p$ is the product of $p$ of these intervals. $\lambda \in [0,1]^p$ means $\lambda = (\lambda_1,...,\lambda_p)$ where each $\lambda_k$ satisfies $0 \le \lambda_k \le 1$. $\endgroup$ – copper.hat Feb 23 '14 at 8:35
  • $\begingroup$ Thanks. I thought that was what it was suppose to mean, but didn't understand the notation. $\endgroup$ – alfien Feb 23 '14 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.