The following is an excerpt from a Wikipedia article about ordered vector space:
Given an ordered vector space $(V,\geq)$, the subset $V^+$ of all elements $x$ in $V$ satisfying $x≥0$ is a convex cone, called the positive cone of $V$. Since the partial order $≥$ is antisymmetric, one can show, that $V^+∩(−V^+)=\{0\}$, hence $V^+$ is a proper cone. That it is convex can be seen by combining the above two axioms with the transitivity property of the (pre)order.
I don't see how one can get $V^+∩(−V^+)=\{0\}$. Following the definition, suppose $y\in V^+\cap (-V^+)$, then
- $y\geq 0$ since $y\in V^+$
- $y=-x$ for some $x\geq 0$ (i.e. $x\in V^+$ by definition) since $y\in -V^+$.
I can formally manipulate the inequalities to reach the conclusion but I cannot convince myself using just the definitions and axioms of the ordered vector space:
$y=-x$ and $x\geq 0$ imply that $-y\geq 0$ (one can deduce this by noting that $-y=(-1)y=(-1)(-x)=x$). Then $y\leq 0$ (formally). But we have $y\geq 0$ also. Therefore by antisymmetricity we have $y=0$.
Here is my question:
By what axiom can one deduce $y\leq 0$ from $-y\geq 0$?
It is because $\ge $ is compatible with addition: $-y\ge 0$ implies $y+(-y)\ge y+0$. As $y+(-y)=0$ be definition of additive inverse and $y+0=y$ by neutral element property, we obtain $0\ge y$.