generalized inequalities defined by proper cones 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?
 A: From what I can tell, most of your question is adequately addressed in this other post, except perhaps for the last part:

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?

I can think of two good reasons to prefer the inequality notation here.
First, for aesthetic consistency. Which of these two stacks of constraints seems cleaner/easier to read (left or right)?
\begin{array}{ccc}
x \succeq_K y &\qquad& x\succeq_K y \\
y \succeq_K z && y\succeq_K z \\
z \succeq_K 0 && z\in K
\end{array}
I claim the left-hand side is cleaner. I see no reason to drop back to set notation $z\in K$ just in that last case just because the right-hand side happens to be zero.
Second, for conceptual clarity. The inequality notation $\lambda \succeq_{K^*} 0$ reminds the reader of the tie to traditional inequalities. Consider this linear program:
\begin{array}{ll}
\text{minimize} & c^T x \\
\text{subject to} & A x \leq b
\end{array}
The Lagrangian for this problem is
$$L(x,z) = c^T x - \langle z, b - Ax \rangle \qquad z\geq 0.$$
Now let's replace the inequality with a generalized inequality:
\begin{array}{ll}
\text{minimize} & c^T x \\
\text{subject to} & A x \preceq_K b
\end{array}
where $K$ is a proper cone. The new Lagrangian is
$$L(x,z) = c^T x - \langle z, b - Ax \rangle \qquad z \succeq_{K^*} 0.$$
where $K^*$ is the dual cone.
Yes, I could have written $b-Ax\in K$ and $z\in K^*$ instead; but doing so breaks the conceptual parallels to linear programming. Why do that?
A: 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.
