"Ordering" of Complex Plane I have heard that the Complex Numbers do not form an ordered field, so you can't say that one number is larger or smaller than another. I've always been fine with this,  but I recently wondered what happens if one describes size as the distance along a space-filling curve?  If the real numbers can be ordered, and you can map a curve through all points in 2D space,  then what prevents you from ordering the points purely by how far along the curve the points are? Each point should be a unique distance,  and the distance continuously increases (as far as my understanding goes). Is there perhaps some difference between being "ordered" and being an "ordered field" that I am missing? I know the basics about why they differ,  but I lack formal math training above Algebra 2... Everything else (up to most of Real Analysis)  I have studied on my own, so my knowledge is spotty in areas like this. 
 A: You can definitely put an order $\prec$ on the complex numbers. In fact, the order can be a well-order, by the well ordering theorem. The question is whether such an order "is useful."
Here are some things we probably want in the order $\prec$:


*

*if $0 \prec \alpha, \beta$, then $0 \prec \alpha \beta$. 

*if $\alpha, \beta, z, w$ are complex with $\alpha \prec \beta$ and $z \prec w$, then $\alpha + x \prec \beta + y$. 


In fact, an order on a field (such as the complex numbers) that satisfies these properties gives us an ordered field. See Ross Millikan's comment for why you cannot put such an order on $\mathbb C$.
A: You can of course put a total order on the complex numbers (an example easier than your space-filling curve: simply take the lexicographic order on $\mathbb{C}\cong\mathbb{R}^2$). However, if you are studying the complex numbers as a field, then you'd like your order to be in some sense compatible with sum and multiplication. For example, you want


*

*if $a<b$, then $a+c<b+c$

*if $a<b$ and $c>0$, then $ac<bc$


and some others. However, this is impossible to obtain in $\mathbb{C}$. Sadly, I don't know the proof of this fact. I would be interested in seeing it though.
A: In an ordered field the order has to be compatible with the algebraic structure; in particular it should be translation invariant (if $a<b$ then $a+c<b+c$).
