Cardinality and order relations on $ \Bbb{C} $ and $ \Bbb{R} $. It has been demonstrated that complex numbers have cardinality $ \aleph_{1} $. However, it can also be shown that the complex numbers cannot be made an ordered field. How can these two facts coincide? If complex numbers exist in bijection with the real line, can we not define a bijection $ B $ that takes any given complex number to a real number, define an order relation such that if $ z $ and $ y $ are complex numbers, $ B(z) < B(y) \Rightarrow z < y $, and call it done?
 A: Not so, it has not been demonstrated that $|\Bbb C| = \aleph_1$. All we can say is $|\Bbb C| = 2^{\aleph_0}$. 
When it's said that $\Bbb C$ can't be made an ordered field, that means using the standard field operations on complex numbers. 
Of course you can transfer any structure from one set to another via a bijection, but the resulting operations of $+$ and $\times$ on $\Bbb C$ that arise from a bijection with $\Bbb R$ won't be the usual operations.
A: The problem is that we cannot find an ordering that is consistent with the natural algebra and topology on the complex numbers. In other words, no ordering that we might define is useful in most contexts. However, there are orders which are useful in some special circumstances. For example, we could order them by $x_1+y_1 i<x_2+y_2 i$ if $x_1<x_2$, or if $x_1=x_2$, then if $y_1<y_2$. This is called the dictionary order.It does not respect the algebra or topology of the complex numbers, but it can be used in constructing some fun counterexamples.
A: An ordered field is not just a set of elements of a field, together with an order relation: the relation must be compatible with addition and multiplication in the sense that
$$x\le y\quad\Rightarrow\quad x+z\le y+z$$
and
$$x\le y\,,\ 0\le z\quad\Rightarrow\quad xz\le yz\ .$$
In fact, it is impossible to define an order on $\Bbb C$ which satisfies these two properties and


*

*antisymmetry: if $x\le y$ and $y\le x$ then $x=y$;

*comparability: for all $x,y$, either $x\le y$ or $y\le x$.


Comment: there was a question virtually identical to this a couple of days ago but I can't find it.  If anyone can, please post.
