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Math Help - Linear order isomorphism

  1. #1
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    Linear order isomorphism

    Hello,

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

    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 <\mathbb{R},\leq>? <\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 f: \mathbb{C} \to \mathbb{R} then for some x,y \in \mathbb{C} if x R y \Rightarrow f(x) \leq f(y) . Do I have to create a bijection f: \mathbb{C} \to \mathbb{R}, because I have trouble coming up with one, or is there a simpler way to answer this question?
    Last edited by MachinePL1993; January 7th 2013 at 03:01 AM.
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  2. #2
    MHF Contributor
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    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.
    Thanks from MachinePL1993
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