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$K$ in $\mathbb{R}^n$, and $K$ is a proper cone.

Partial Ordering of $K$ : $x \leq_K y$ iff $y-x\in K$ (S. Boyd p. 43)

My questions are:

  1. Does it require $x,y\in K$?

  2. If $x,y\in K$, it seems that $y-x$ and $x-y$ are in $K$ (just draw a cone and draw $x, y$ in the cone and draw $y-x$ or $x-y$); if it is correct, what is the relation between $x$ and $y$ ($x = y$)?

Thanks :)

DKal
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sleeve chen
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1 Answers1

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Take the usual order on $\mathbb{R}$. Then the relevant cone is $K=[0,\infty)$.

For 1. the answer is no. We have $-2 \le -1$ because $-1-(-2) = 1 \in K$, but neither $-1$ nor $-2$ lie in $K$.

For 2. take $x=0, y=1$. Then $x,y \in K$ and $y-x \in K$, but $x-y \notin K$.

copper.hat
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