prove that any total order $R$ on $\mathbb{C}$ satisfies $\forall_{z_1,z_2,c\in\mathbb{C}}(z_1Rz_2\Rightarrow(z_1+c)R(z_2+c))$

Question

I am having some problems in the last stage of my proof:

1. Assume that $z_1Rz_2$. Then either $z_1=z_2$ or $z_1\neq z_2$.
2. If $z_1=z_2$, we have $z_1+c=z_2+c$ and hence $(z_1+c)R(z_2+c)$ because $R$ is reflexive over $\mathbb{C}$.
3. If $z_1\neq z_2$, we have $z_1+c\neq z_2+c$ for all $c\in\mathbb{C}$.
   • Since $R$ is anti-symmetric, we have $\neg(z_1+c)R(z_2+c)\lor\neg(z_2+c)R(z_1+c)$ for all $c\in\mathbb{C}$.
   • Since $R$ is also a total order, we also have $(z_1+c)R(z_2+c)\lor(z_2+c)R(z_1+c)$ if $c\in\mathbb{C}$.

Hence, one and only one of $(z_1+c)R(z_2+c)$ and $(z_2+c)R(z_1+c)$ is true for all $c\in\mathbb{C}$. We need to show that the former is true to complete the proof.

I verified the result for the Lexicographic total order $L$ where $z_1Lz_2\Leftrightarrow\Im\left(z_1\right)<\Im\left(z_2\right)\lor\left(\Im\left(z_1\right)=\Im\left(z_2\right)\land\Re\left(z_1\right)\leq\Re\left(z_2\right)\right)$ but that wasn't particularly instructive for the general case.

The $c=0$ case is straightforward, but I have no idea how to proceed with other values of $c$. It's also possible that the statement is false, but I cannot find counter-examples. I'd appreciate any proof or counter-example.

Answer 1

This is just false. Consider a translation-invariant total order $(\Bbb C,\le)$, two distinct elements whatsoever $a,b\in\Bbb C$ and the map $\Phi:\Bbb C\to \Bbb C$ $$\Phi(x)=\begin{cases}x&\text{if }x\ne a\land x\ne b\\ b&\text{if }x=a\\ a&\text{if }x=b\end{cases}.$$

Then the total order $(\Bbb C,\le_\Phi)$, $x\le_\Phi y\Leftrightarrow \Phi(x)\le\Phi(y)$ is not translation-invariant.