Let $a_1, \dots, a_p$ be $p$ vectors in $\mathbb{R}^n$ and $K$ the closed convex cone generated by $(a_1, \dots, a_p)$, i.e. $$K = \left\{ \sum_{j=1}^p \lambda_i a_i | (\lambda_i) \in \mathbb{R}^p_+ \right\}$$ Let's assume that $K\cap\mathbb{R}^n_+ = \emptyset$.
Show that (a) $K-\mathbb{R}^n_+$ is a closed convex cone (b) that does not contain $\mathbf{1} = (1, \dots, 1) \in \mathbb{R}^n$.
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For (a),
- $K-\mathbb{R}^n_+$ is convex ($\mathbb{R}^n_+$ is convex and the sum of two convex set is convex).
- $\mathbb{R}^n_+$ is closed and so is $K$ (as a cone generated by a finite number of elements of $\mathbb{R}^n_+$)
However the sum $K + (-\mathbb{R}^n_+)$ is closed only if either of the two sets is compact. Therefore I need to show that $K$ is compact, but is it the case?
For (b),
I am trying to prove by contradiction that if $\mathbf{1}$ can be written as $\mathbf{1} = k + x_+$
with $k \in K$ and $x_+ = -x_-, \: x_- \in -\mathbb{R}^n_+$
then $K\cap\mathbb{R}^n_+ \neq \emptyset$ (the contradiction).
But I cannot get there.
Any suggestion? Thank you.