There is no total order on $\mathbb{C}$ with the algebraic property $0\leq z_1\land0\leq z_2\Rightarrow0\leq z_1+z_2\land0\leq z_1z_2$

Question

This question is about the proof of Example 62 (d) on Page 27 of these notes. The statement is that there is no total order $\leq$ on $\mathbb{C}$ with the algebraic property $0\leq z_1\land0\leq z_2\Rightarrow0\leq z_1+z_2\land0\leq z_1z_2$.

The statement is wrong; Thomas Andrews's answer below shows that there is a total order on $\mathbb{C}$ such that $0\leq z_1\land0\leq > z_2\Rightarrow0\leq z_1+z_2\land0\leq z_1z_2$.

My attempt goes as follows:

1. Any total order $\leq$ has the following properties:
   1. $\forall _{z_1\in \mathbb{C}}(z_1\leq z_1)$ (reflexive)
   2. $\forall _{z_1,z_2\in \mathbb{C}}\left(z_1\leq z_2\land z_2\leq z_1\Rightarrow z_1=z_2\right)$ (anti-symmetric)
   3. $\forall_{z_1,z_2,z_3\in \mathbb{C}}\left(z_1\leq z_2\land z_2\leq z_3\Rightarrow z_1\leq z_3\right)$ (transitive)
   4. $\forall _{z_1,z_2\in \mathbb{C}}\left(z_1\leq \ z_2\lor z_2\leq z_1\right)$ (total order)
2. The question proposes a further algebraic property:
   5. $\forall _{z_1,z_2\in \mathbb{C}}\left(0\leq z_1\land \ 0\leq z_2\Rightarrow 0\leq z_1+z_2\land 0\leq z_1 z_2\right)$
3. We need to prove that there is no relation with all five properties.
4. The partial order $z_1\leq z_2\Leftrightarrow\left(z_1=z_2\lor\left|z_1\right|<\left| z_2\right|\right)$ has properties 1, 2, 3 and 5.
5. Hence, the most straightforward possible proofs are $4\Rightarrow\neg 5$ or $5\Rightarrow\neg4$.

An attempt at $4\Rightarrow\neg 5$:

1. Either $0\leq i$ or $i\leq 0$ but not both as $i\neq 0$ (due to anti-symmetry).

2. If $0\leq i$, then using $z_1=z_2=i$ in property 5 gives $0\leq 2i$ and $0\leq -1$.

   1. Using $z_1=2i$ and $z_2=-1$ in property 5 gives $0\leq -1+2i$ and $0\leq -2i$

   2. Alternatively, using $z_1=z_2=-1$ in property 5 gives $0\leq -2$ and $0\leq 1$

   3. In general, we generate $0\leq a+bi$ for various integer $a$ and $b$.

3. If $i\leq 0$, then property 5 gives no results.

Regardless of the initial $0\leq z$, property 5 only gives $0\leq P(z)$, where $P(z)$ are integer coefficient polynomials in $z$. How does this provide a contradiction? Furthermore, the $z\leq 0$ case seems entirely intractable.

An attempt at $5\Rightarrow\neg 4$:

1. Any violation of 4 requires $\neg 0\leq z\land \neg z\leq 0$
2. The contrapositive of 5 only generates $\neg 0\leq P(z)$ starting with $\neg 0\leq z$ and hence does not violate 4.
3. Starting with $\neg z\leq 0$ ends with nothing further and thus does not negate 4.

It seems that the two most straightforward approaches are not going to work. I have no idea how to proceed with the more complicated options.

Answer 1

I can give you a total order like your $4,$ but extended. Write all non-zero complex numbers as $z=re^{i\theta}$ where $r>0$ and $0\leq \theta <2\pi.$ Then $z_1\leq z_2$ is defined to be true if $r_1<r_2$ or if $r_1=r_2$ and $\theta_1\leq \theta_2.$ Add that $0\leq z$ for all $z\in \mathbb C.$ This is then a total order, and since $0$ is the minimal element of the order, (2) is satisfied by default.

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One condition you can add is that $z_1\leq z_2$ if and only if $0\leq z_2-z_1.$ This gives you translation invariance since it means $z_1\leq z_2$ if and only if $z_1+w\leq z_2+w.$