Complex Numbers form Preordered Vector Space

Theorem

Consider the complex numbers $\C$ as a vector space over itself.

Define the relation $\ge^\C$ by:

$z \ge^\C w$

if and only if:

$z - w \in \hointr 0 \infty$

for each $z, w \in \C$.

Then $\struct {\C, \ge^\C}$ is a preordered vector space.

Proof

From Characterization of Preordered Vector Spaces, it is enough to show that $\hointr 0 \infty$ is a convex cone.

Let $x \in \hointr 0 \infty$ and $\alpha \in \R_{\ge 0}$.

Then $\alpha x \ge 0$, so $\alpha x \in \hointr 0 \infty$.

So $\hointr 0 \infty$ is a cone.

Now let $x, y \in \hointr 0 \infty$.

Then we have $x + y \ge 0$, so $x + y \in \hointr 0 \infty$.

So $\hointr 0 \infty$ is a convex cone.

$\blacksquare$