Exercise 1.9

Question

Exercise 9: Suppose $z = a + bi$, $w = c + di$. Define $z < w$ if $a < c$ or $a = c$ and $b < d$. Prove that 
    this is an order. Does it have the least upper bound property?

ghostofgarborg · Accepted Answer

Let $z_i = a_i + b_i$.

Property 1.5(i): Assume $z_1 
eq z_2$. If $a_1 
eq a_2$, either $z_1 < z_2$ or $z_2 < z_1$. 
    Otherwise $b_1 
eq b_2$, in which case $z_1 < z_2$ or $z_2 < z_1$.

Property 1.5(ii): Assume $z_1 < z_2$ and $z_2 < z_3$. Then $ a_1 \leq a_2 \leq a_3 $. 
    If either of the inequalities is strict, $a_1 < a_3$, and $z_1 < z_3$. 
    Otherwise, $b_1 \leq b_2 \leq b_3$ and all inequalities are strict, so that 
    $z_1 < z_3$.

This proves that the order is well-defined.

The set does not have the least upper bound property. Consider the set $ \mathbb R i = \{ 0 + xi : i \in \mathbb R \}$. The set is bounded above by $1$. Assume for contradiction that $\alpha = a + bi$ is a least upper bound. It is clear that we must have $a \geq 0$. If $a > 0$, then $\frac 1 2 \alpha$ is another upper bound that is strictly smaller than $\alpha$, contradicting minimality. Therefore, we must have $a=0$. Then $\alpha + i \in \mathbb R i$ is greater than $\alpha$, contradicting $\alpha$ being an upper bound.

Amit Awasthi · Answer

\begin{proof}[Proof of ordering] First prove Definition 1.5(i) holds. Consider $z = a+bi$, $w = c+di$. Since $a,b,c,d \in \mathbb{R}$ which is an ordered set, there are 5 cases: \begin{enumerate} \item If $a < c$ then $z < w$. \item If $a > c$ then $z > w$. \item If $a = c$ and $b < d$ then $z < w$. \item If $a = c$ and $b > d$ then $z > w$. \item If $a=c$ and $b=d$ then $z = w$. \end{enumerate} So one and only one of the statements $z < w$, $z = w$, $z > w$ is true. \par Now prove Definition 1.5(ii) holds. Consider $z_1,z_2,z_3$ such that $z_1 < z_2$, $z_2 < z_3$. $z_1 < z_2$ implies either \begin{enumerate} \item $ ext{Re}(z_1) < ext{Re}(z_2)$, or \item $ ext{Re}(z_1) = ext{Re}(z_2)$ and $ ext{Im}(z_1) < ext{Im}(z_2)$. \end{enumerate} Similarly, $z_2 < z_3$ implies either \begin{enumerate} \item $ ext{Re}(z_2) < ext{Re}(z_3)$, or \item $ ext{Re}(z_2) = ext{Re}(z_3)$ and $ ext{Im}(z_2) < ext{Im}(z_3)$. \end{enumerate} Then, there are two cases for the relation between $z_1$ and $z_3$: \begin{enumerate} \item $ ext{Re}(z_1) < ext{Re}(z_3)$, so $z_1 < z_3$, or \item $ ext{Re}(z_1) = ext{Re}(z_2) = ext{Re}(z_3)$ and $ ext{Im}(z_1) < ext{Im}(z_3)$, so $z_1 < z_3$. \end{enumerate} So $z_1 < z_2$, $z_2 < z_3$ implies $z_1 < z_3$. \end{proof} \begin{claim} This ordered set of complex numbers does not have the least-upper-bound-property. \end{claim} \begin{proof} Let $A = \left\{ in : n \in \mathbb{N} ight\}$, which is bounded above by 1. Suppose $a + bi = \sup A$. Since if $a > 0$, we can find $a'$ such that $a > a' > 0$ by the archimedean property, $a = 0$, and $bi = \sup A$. However, we can always find $n > b$, which contradicts that $bi$ is an upper bound. Thus, $\mathbb{C}$ with order as defined above does not have the least-upper-bound property. \end{proof}

Exercise 1.9

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Exercise 1.9

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