When a vector space will be a complete lattice? Let $E$ a vector space, and let $P$ a strict cone in $E$ (i.e) $P\subset E$ verify: 
$$
\mathbb{R}^+ P\subset P \\
P+P\subset P\\
P\cap (-P)=\{0\}
$$
So we can easily construct a partial order on $E$ which is :
$$
x,y\in E \qquad x\leq_{P}y \iff y-x\in P
$$
My question is for which condition on $P$, $E$ will be a complete lattice ?

This happens if and only if $E=P=0$. If $x$ is the top element in $E$, then $x \geq 0$. This means that by definition $x = x-0 \in P$. Since $x$ is the top element $x \geq 2x$ meaning that $-x = x-2x \in P$ or equivalently $x \in -P$. It follows that $x=0$ since $P \cap -P = \{0\}$. For each $y$ in $E$ since $x$ is the top element $x\geq -y$ meaning that $y=x-(-y)$ is in $P$. It follows that $P=E$ and therefore that $E = E \cap -E = P \cap -P = \{0\}$. Trivially if $E=P=0$ then $E$ is a complete lattice.
