# Math Help - Prove C is a cone...

1. ## Prove C is a cone...

Let $S\subset{R^n} \neq{\emptyset}, x'\in{S}$. Let $C = \left\{{y: y = \lambda (x-x'), \lambda \geq{0}, x\in{S}}\right\}$.

a) Prove that C is a cone and interpretate C geometrically (C is a cone if $\forall{} x\in{C}, \lamba x\in{C}, \lambda \geq{0}$).

b) Suposse S is a close set. ¿Is C necessarely close? If not, ¿what are the conditions for him to be close?

Thank you for every aportation.

2. ## Re: Prove C is a cone...

Well to prove that C is a cone, you must have your set containing the vectors $\lambda $$x-x'$$$ for every $\lambda\geq 0$. If so, then if u take any positive scalar $k$, then $k\lambda $$x-x'$$$ is also in the set and hence your set is a cone by the definition of a cone.