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Math Help - Prove C is a cone...

  1. #1
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    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.
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  2. #2
    Member mohammadfawaz's Avatar
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    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.
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