My lecture notes on real analysis mention that we cannot order the vector space $\mathbb{R}^n$ in a way that respects vector addition:
$\mathbb{R}^2$ has the structure of a vector space:
- Componentwise addition defines the additive Abelian group ($\mathbb{R}^n$; +), whose neutral element (0,...,0)^T we also denote by 0.
- Multiplication with a scalar $\lambda \in \mathbb{R}$ is also defined componentwise, $\lambda x := (\lambda x_1,...,\lambda x_n)^T$. For us a point $x \in \mathbb{R}^n$ is also a vector, and vice versa. For $n \geq 2$ there exists no ordering on $\mathbb{R}^n$ compatible with addition.
Naturally this is a proof by contradiction, but I just don't know where to start.
I have proven this for the complex field before, but I doubt that the same proof will work as it uses the fact that $i^2=-1$, which we don't have here. In addition, we don't even have multiplication of elements but only scalar multiplication.
Can anybody please point into the right direction?
Thanks very much!
I cannot think of any reasonable interpretation of the statement which would make this true. For instance, consider the following ordering on $\mathbb{R}^2$: we say $(a,b)\leq (c,d)$ iff either $a<c$ or $a=c$ and $b\leq d$. This is a total order, and it satisfies any reasonable compatibility with the vector space structure you could ask for. In particular, if $x\leq y$ then $x+z\leq y+z$ for any $z$ and $cx\leq cy$ for any scalar $c\geq 0$.
(This is known as the "lexicographic order". Intuitively, you can think of it as identifying $(a,b)$ with "$a+b\epsilon$" where $\epsilon$ is some infinitely small positive quantity.)