I have the following problem
Prove that if $X$ is a pointed cone then there exists a vector $c$ such that $c^Tx>0$ for all nonzero $x \in X$.
For this, I seem to have a simple solution as follows:
Suppose to the contrary that, $c^Tx \le 0$ for every $c \in \mathbb R^n$ and for all nonzero $x \in X$. That means, if we fix a nonzero vector $d \in \mathbb R^n$ then $d^Tx \le 0$ and $(-d)^Tx \le 0$ for all nonzero $x \in X$. Therefore, $d^Tx \le 0 \le d^Tx$ which implies $d^Tx=0$. Hence, $X$ contains a hyperplane $d^T x= 0$ i.e. $X$ contains a straight line which is a not pointed cone. It contradicts to our assumption.
Any comment is greatly appreciated.
Intuition, It suggests that the Cone "Opening" is less than 180 Degrees which means it is pointed.
Suppose to the contrary that $ \forall d \in \mathbb{R}^{n}, \; {d}^{T} x \le 0 $ for any $ x \in X $.
Specifically one could chose $ d \neq 0 $ which yields:
$$ {d}^{T} x \le 0, \; -{d}^{T} x \le 0 $$
Assuming the Cone $ X $ includes items besides the zero point then:
$$ {d}^{T} x \leq 0 \Rightarrow -{d}^{T} x \geq 0 $$
Yet by out assumption $ -{d}^{T} x \leq 0 $ hence $ {d}^{T} x = 0 $ which suggests the set $ \left\{ x \mid {d}^{T} x = 0 \right\} \subseteq X $, namely there is an Hyperplane inside $ X $ which suggests there is a line in $ X $ which is a contradiction to the assumption $ X $ is pointed Cone.