Exercise 1.9 Suppose $z = a + bi, w = c + di$. Define $z<w$ if $a<c$, and also if $a = c$ but $b<d$. Prove that this turns the set of all complex numbers into an ordered set. (This type of order relationis called a dictionary order, or lexicographic order, for obvious reasons.) Does this ordered set have the least upper bound property?
The solution of this question is the procedure this set has an ordered property such as "$x<y$ or $x=y$ or $x>y$", "$x<y$ and $y<z$ then $x<z$".
Solution. We need to show that either $z<w$ or $z=w$, or $w<z .$ Now since the real numbers are ordered, we have $a<c$ or $a=c$, or $c<a .$ In the first case $z<w ;$ in the third case $w<z$. Now consider the second case. We must have $b<d$ or $b=d$ or $d<b .$ In the first of these cases $z<w$, in the third case $w<z$, and in the second case $z=w$.
We also need to show that if $z<w$ and $w<u$, then $z<u$. Let $u=e+f i$ Since $z<w$, we have either $a<c$ or $a=c$ and $b<d .$ Since $w<u$ we have either $c<f$ or $c=f$ and $d<g .$ Hence there are four possible cases:
Case 1: $a<c$ and $c<f$. Then $a<f$ and so $z<u$, as required.
Case 2: $a<c$ and $c=f$ and $d<g .$ Again $a<f$, and $z<u .$
Case 3: $a=c$ and $b<d$ and $c<f$. Once again $a<f$ and so $z<u$.
Case $4: a=c$ and $b<d$ and $c=f$, and $d<g .$ Then $a=f$ and $b<g$, and so $z<u$
But I think that this question already gives us that this set is ordered, so we don't need prove it. In addition, all of ordered sets have a least-upper-bound property, so we also need not prove they have a least-upper-bound property. I think this book asks us a wrong question. Is it right? ps. Why this solution doesn't prove there is a least upper bound?
Hint: consider the subset $\{(0,y)\,:\,y\in\Bbb R\}$.