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Let $(G,+,0,<)$ be an ordered divisible group of uncountable dimension. Consider the subset $G^{<0}$ of $G$.

Question: Are $G$ and $G^{<0}$ isomorphic as ordered sets? Does there exist an order-preserving isomorphism of $G^{<0}$ onto $G$? Intuitively I would say, this is true.

My ideas: 1. In the case where $(K,+, \cdot, 0,1,<)$ is an divisible ordered field we can give an explicit order-preserving isomorphism of $K$ onto $K^{<0}$, e.g. $f(x):= \begin{cases} x-1 \quad \text{if} \; x \leq0, \\ -\frac{1}{x+1} \quad \text{else}. \end{cases} $

  1. In the case where $G$ is in ordered set the statement is false.

How can we use the group structure and the divisibility of $G$ to construct an isomorphism or does anybody know a counterexample?

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Let $G=\mathbb{Q}^{\omega_1}$ with the lexicographic order where $\omega_1$ is the first uncountable ordinal. So elements of $G$ have the form $(x_\alpha)_{\alpha<\omega_1}$, and two elements are compared by looking at the first $\alpha$ on which the entries are not equal.

Then $G$ has a countable cofinal subset consisting of elements with $x_0=n$ and $x_{\alpha}=0$ for $\alpha\neq 0$ as $n$ ranges through $\mathbb{N}$, while $G^{<0}$ does not: if $A$ were a countable cofinal subset of $G^{<0}$ then extracting the first non-zero entry of each element of $A$ would give a countable cofinal subset of the ordinals less than $\omega_1$ which is impossible. Therefore $G^{<0}$ and $G$ are not isomorphic as ordered sets.

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  • $\begingroup$ I do not understand, why there is no countable cofinal subset in $G^{<0}$? $\endgroup$ – N.W. Nov 4 '16 at 21:31
  • $\begingroup$ It follows from the fact that $\omega_1$ has no countable cofinal subset (since it is the first uncountable ordinal). Now for a non-zero element a of A let $\alpha_a$ be its first non-zero entry. Since A is cofinal in $G^{<0}$, the set of such $\alpha_a$ is cofinal in $\omega_1$, a contradiction. $\endgroup$ – shane.orourke Nov 5 '16 at 9:58

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