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Looking for the proof of the lemma asserting that the conical surface (envelope) is a closed space. Thank you.

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Please give more details, so that we know what you have in mind. Also, writing complete words is usually a great idea! :) – Mariano Suárez-Alvarez Jan 8 '11 at 1:10
If by "conical surface" you mean this: and if by "closed" you mean this: then the lemma is false and there is no proof. You need some additional condition like having a closed directrix. – Ryan Budney Jan 8 '11 at 1:21
Please see my answer below. – mira Jan 8 '11 at 11:36

1 Answer 1

We are talking here about the convex cone $C:={\rm co}(A)$ spanned by a finite set $A$ of vectors $a\in {\mathbb R}^n$. Consider an arbitrary point $x\in C$. It has a representation $x=\sum_{i=1}^m \lambda_i a_i$ with $a_i\in A$, $\lambda_i>0$ and minimal $m$. I claim that the $a_i$ occurring in this representation are linearly independent. (This is essentially Caratheodory's theorem.)

Proof: If one had $\sum_{i=1}^m \mu_i a_i=0$, where not all $\mu_i$ are $0$, then after rearranging one may assume that for some $s\leq m$ one has $$\mu_i\leq 0 \quad (1\leq i\leq s-1), \qquad \mu_i > 0 \quad (s\leq i\leq m), \qquad {\lambda_i\over\mu_i}\geq {\lambda_m\over\mu_m} \quad (s\leq i\leq m).$$ It follows that $$x=\sum_{i=1}^{m-1} (\lambda_i-{\lambda_m\over\mu_m} \mu_i) a_i$$ were a representation of $x$ with nonnegative coefficients using at most $m-1$ of the $a_i$.

Now the cone spanned by a linearly independent set of $m$ vectors $a_i$ is homeomorphic to the set $$\{x\in {\mathbb R}^n\ |\ x_i\geq 0 \ (1\leq i\leq m), \ x_i=0 \ (m+1\leq i\leq n)\}$$ which is clearly a closed set. As $C$ is a union of finitely many such sets it is closed as well.

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