1. ## Linear order isomorphism

Hello,

I have a linear order relation defined on the product of set of complex numbers $\displaystyle \mathbb{C}. \times \mathbb{C}.$

$\displaystyle xRy \Leftrightarrow \Re x < \Re y \vee ( \Re x = \Re y \wedge \Im x \leq \Im y)$

I can prove that this relation is a linear order that's dense and without endpoints. But I have a problem with this question:

Is this order isomorphic with $\displaystyle <\mathbb{R},\leq>$? $\displaystyle <\mathbb{R},\leq>$ is a normal greater or equal relation on the set of real numbers.

For order to be isomorphic there has to exist a bijection between two sets that preserves the relation between the elements, meaning that if there's a bijection $\displaystyle f: \mathbb{C} \to \mathbb{R}$ then for some $\displaystyle x,y \in \mathbb{C}$ if $\displaystyle x R y \Rightarrow f(x) \leq f(y)$ . Do I have to create a bijection $\displaystyle f: \mathbb{C} \to \mathbb{R}$, because I have trouble coming up with one, or is there a simpler way to answer this question?

2. ## Re: Linear order isomorphism

Interesting. The theory of dense linear orders without endpoints is categorical only for ω and not for uncountable cardinals. So, 〈ℂ,R〉and 〈ℝ,≤〉can be non-isomorphic. In fact, they are. For a hint, see this Dr. Math question.