generalized inequalities defined by proper cones

Question

The generalized inequality defined by a proper cone $K$ is that $x \ge_{K} y$ if $x-y \in K$ for $x,y \in K$. Does this means that for any $x \in K$, we have $x \ge_{K} 0$ since $x - 0 = x \in K$ ?

EDIT: My definitions of cones and proper cones follow the ones in "Convex Optimization" by S. Boyd, in which 0 is contained in cones. On page 53 of the book, there is a statement about generalized inequalities:

$x \le_{K} y$ if and only if $\lambda^T x \le \lambda^T y$ for all $\lambda \ge_{K^*} 0$

where $K^*$ is the dual cone of the proper cone $K$, so $K^*$ is also a proper cone.

The reason I ask this question is that if for any $\lambda \in K^*$, we have $\lambda \ge_{K^*} 0$, why not just using $\lambda \in K^*$ instead of $\lambda \ge_{K^*} 0$ in the above statement?

Answer 1

The answer to this question depends on how you define cones.

Wikipedia says that a cone is a subset of a vector space closed under scalar multiplication by positive constants. So a subset $K$ such that if $x \in K$ and $a > 0$ then $ax \in K$.

These cones need not contain $0$, in which case your reasoning does not hold.

However, Wikipedia also mentions:

A cone is said to be pointed if it includes the null vector (origin) $0$; otherwise it is said to be blunt. Some authors use "non-negative" instead of "positive" in this definition of "cone", which restricts the term to the pointed cones only. In other contexts, a cone is pointed if the only linear subspace contained in it is $\{0\}$.

So, according to some authors, the $>$ in the above definition should be $\ge$, from which it easily follows that $0 \in K$ and your reasoning is correct.