If any clarification is necessary, here is a little definition of what "dictionary" order encompasses. In Rudin's book it was stated in the following way.
Let $z=a+bi$ and $w=c+di$ where $z,w$ are arbitrary complex numbers and $a,b,c,d$ are reals.
$z<w$ if $a<c$ or if $a=c$ and $b<d$.
It is not hard to prove that this definition of ordering does indeed turn the complex field into an ordered set, but there is another question that is posed in the book and it is about the least upper bound property of this set. More specifically, I would like to know if the complex number set with order defined in this way does actually have the least upper bound property.
Intuitively speaking, I think that it should, because even the fact that the complex number set becomes an ordered set in this setting relies pretty much only on the fact that reals are an ordered set. I do not know, however, how to construct a fairly decent proof of this, or even how to approach this question a bit more formally.
Any help would be much appreciated.
Best Answer
The complex numbers $\mathbb C$ with dictionary order do not have the l.u.b. property.
Consider the following subset of $\mathbb C$: $\{a+bi|a\le0\}$.
In the dictionary order, any $x+yi$ with $x>0$ is an upper bound of this subset,
and any $x+yi$ with $x\le0$ is not an upper bound of this subset.
But you can't say there is any least upper bound $x+yi$ with $x>0$,
because $\dfrac x2+yi$ is a smaller upper bound.