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Are the following sets convex cones:

1.$\{x\in\Bbb R^n:\langle a,x\rangle\leq 0, a\neq 0 \}$
2.$\{x\in\Bbb R^n:\langle a,x\rangle\lt 0, a\neq 0 \}$ ?

We say $C\subset \Bbb R^n$ is a cone if $\forall x \in C$ and $\lambda>0$, $\lambda x \in C$. A convex cone is a cone that is a convex set.
I drew the sets in $\Bbb R^2$ and from that we can see that both sets are convex cones. Is there a more formal proof?

elsadd
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  • Yes, there are more formal proofs. Namely, the ones where you check if the sets satisfy the conditions you are asking. –  Oct 15 '18 at 14:26
  • I know how to check if it is convex, but how should I go about checking if it is a cone? – elsadd Oct 15 '18 at 14:29
  • It appears you've written it, have you not? –  Oct 15 '18 at 14:30
  • Yes. Let $z, y \in $ ${x\in\Bbb R^n| \langle a, x \rangle \leq0, a\neq 0$}. Then $\langle a, z \rangle \leq0$ and $\langle a, y \rangle \leq0$. Let $\lambda \in [0,1]$. Then $\langle a, \lambda z+ (1-\lambda) y\rangle = \langle a, \lambda z\rangle + \langle a, (1-\lambda) y\rangle = \lambda\langle a, z\rangle + (1-\lambda)\langle a, y \rangle \leq \lambda0 + (1-\lambda)0= 0$. Hence $\lambda z+ (1-\lambda) y \in {x\in\Bbb R^n| \langle a, x \rangle \leq0, a\neq 0$}. – elsadd Oct 15 '18 at 14:41
  • Let $z \in $ ${x\in\Bbb R^n| \langle a, x \rangle \leq0, a\neq 0$}. Then $\langle a, z \rangle \leq0$. Let $\lambda \in [0,1]$. Then $\langle a, \lambda z\rangle = \lambda \langle a, z\rangle \leq \lambda * 0=0 $. So the set is a cone. – elsadd Oct 15 '18 at 14:46
  • Sorry to bother you all, I don't know why I just didn't follow the definitions, my bad. – elsadd Oct 15 '18 at 14:47
  • you should add your solution as an answer instead of a comment, and then accept the answer :) – LinAlg Oct 15 '18 at 19:53
  • The vector $a$ should be outside the set. There is ambiguity the way it is stated. Should it be true for all $a\neq 0$? For some $a\neq 0$ etc. – max_zorn Oct 16 '18 at 21:35

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