total order on finite dimensional vector space over $\mathbb{R}$ We know that, If we start with a basis of a finite dimensional vector space $V$ over $\mathbb{R}$ by using lexicographic ordering with respect to that basis, we have a total ordering $\lt$ on $V$ satisfying, 
i) $v \in V,   v_1 \lt v_2 \implies v_1 + v \lt v_2 + v$
ii) $ \alpha \in \mathbb{R}, v_1 \lt v_2 \implies $
a) $\alpha v_1 \lt \alpha v_2$ if $\alpha \gt 0$
b) $\alpha v_1 \gt \alpha v_2$ if $\alpha \lt 0 $
I just wondering, given a such a total ordering on $V$ , can we find a basis of $V$ with respect to which the lexicographic order will be the same as the one we start with?
is there any similar theory of infinite dimensional vector spaces? 
Thanks in Advance
 A: Let $<$ be such a total ordering on $V$ (which is, as in the question, finite dimensional). Then the set $C=\{v \in V: v\geq 0\}$ clearly is a convex cone. 
Since $C \cap (-C) =\{0\}$ and since $V=(-C) \cup C$ there is a $v_0 \in V$ which does not lie in the closure of $C$.
 Thus there is a nonzero linear form $l$ on $V$ such that $l \geq 0$ on $C$ (see below in "edit"). Now the zero set of $l$ is a hyperplane as Harald thought of:
Let $l(v)<0$ then $v \not\in C$ thus $v<0$.
Let $l(v)>0$ and assume $v \leq 0$ then $-v \in C$ which contradicts $l(-v)<0$.
edit: This follows from biduality of convex cones, see http://arxiv.org/abs/1006.4894 (section 2.1): Assume that $C^*=\{0\}$ (notation as in the paper), then we have $V=\{0\}^*=(C^*)^*=\textrm{cl}(C)$, but we have $V \neq \textrm{cl}(C)$ ($\textrm{cl}(C)$ is the closure of $C$).
I expect that you can find the proof of this biduality theorem (which uses the separation theorem that I mentioned in the first place) in a standard book about convexity, for example in Barvinok's "A First Course in Convexity".
A: In case of an infinite dimensional vector space there exists a basis that represents the order. 
Nevertheless the separating “Hyperplane” may be  the whole vector space itself. For example take $\mathbb{R}^{\mathbb N}$ with the natural orders on $\mathbb R$ and $\mathbb N$. In this vector space every vector is infinitesimal with respect to some other vector. If there were a hyperplane that separates the positive cone and the negative cone, then there would have been a vector that is not infinetesimal with respect to any other vectors.
This topic is discussed in this article.
