Convex cone question. Hoi, let $V$ be finite dimensional real vector space with inner product $\left\langle . \right\rangle$ and let $\Gamma \neq \{0\}$ be a closed convex cone.  Let $$\Gamma_0^{\perp}:=\{v\in V:\left\langle v,w\right\rangle > 0 \ \text{for all} \ w\in \Gamma\setminus \{0\}\}
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Take for granted:
(a) If $\Gamma$ is pointed and $\Gamma_0^{\perp} =\emptyset $ then there exists for all $v\in \Gamma\setminus \{0\}$ a $w \in \Gamma\setminus  \{0\}$ such that  $\left\langle v,w\right\rangle =0$. I want to show 
(b)$\Gamma$ is pointed $\Rightarrow \Gamma_0^{\perp}\neq \emptyset$. 
I got as hint: Use (a) repeatedly, reducing the dimensionality of a suitable cone in each step, until you reach a one-dimensional situation, leading to a contradiction. 
Ok, so we argue by contradiction. Since for $K\supset \Gamma$ we have $K_0^{\perp}\subset \Gamma_0^{\perp}$ it satisfies to show $K_0^{\perp}\neq \emptyset$  for a suitable cone $K$. So we can take a (pointed) polehydral cone, which is finitely generated by say $\{w_1,\cdots, w_n\}$. So we assume for a contradiction $K_0^{\perp} =\emptyset$.
How can we reduce the 'dimensionality'...i dont understand how fact (a) leads to a contradiction. ? Ofcourse on a one-dimensional cone $\Gamma$, which is a line, there exists no $v,w\in \Gamma\setminus \{0\}$ s.t. $(v,w)=0$. But how to make this reduction to a line? Thanks for any insights.  
Edit: Pointed means $\Gamma\cap -\Gamma = \{0\}$. 
 A: I have an idea of an inductive dimension reduction. If $\Gamma_0^\perp=\varnothing$, choose an arbitrary vector $v_0\in\Gamma_0\backslash\{0\}$ and put $\Gamma_1=\{w\in\Gamma_0:(v_0,w)=0\}$. Condition (a) implies that $\Gamma_1$ is nonempty. It should be easily to check that $\Gamma_1^\perp=\varnothing$. Choose an arbitrary vector $v_1\in\Gamma_1\backslash\{0\}\dots$.
A: Suppose to the contrary that $\Gamma_0^{\perp}=\varnothing$ i.e. $\left\langle v,w\right\rangle \le 0$ for every $v \in V$ and for all $w\in \Gamma\setminus \{0\}$. That means, if we fix a nonzero vector $s \in V$ then $\left\langle s,w\right\rangle \le 0$ and $\left\langle -s,w\right\rangle \le 0 $ for all $w\in \Gamma\setminus \{0\}$. Therefore, $\left\langle s,w\right\rangle \le 0 \le \left\langle s,w\right\rangle$ which implies $\left\langle s,w\right\rangle=0$. Hence, $\Gamma$ contains a hyperplane $\left\langle s,w\right\rangle= 0$ i.e. $\Gamma$ contains a straight line which is a not pointed cone. It contradicts to our assumption.
Is this solution fine?
