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A group $H:=(\mathbb R,\boxplus)$, is given to be

  1. Ordered as per the canonical order of $\mathbb R$.
  2. Archimedean as per order in 1.
  3. Topological as per canonical topology of $\mathbb R$.

Can it be deduced that $H$ is group isomorphic to $(\mathbb R,+)$?

If not, what additional properties would be sufficient?

I have found that, by a theorem of Hölder, any Archimedean group is group isomorphic to a subgroup of $(\mathbb R, +)$. Hence the question above reduces to whether, as a result of the fact that the set itself is $\mathbb R$ and $H$ topological, the case of $H$ being isomorphic to a proper subgroup of $(\mathbb R,+)$ can be excluded.

Crispost
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    "isomorphic" in what sense ... group? ordered group? Topological group? ... – GEdgar Nov 20 '23 at 12:29
  • I thought that group isomorphism, in that particular case would automatically entail order and topological isomorphisms as well. Is that wrong?

    If it is wrong, then group isomorphism is good, all of them even better.

     – Crispost Nov 20 '23 at 12:52
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    The ordering on $H$ is complete so I think this answers your question strongly in the affirmative - there is an isomorphism of ordered groups between $H$ and $(\Bbb R, +, <)$. This isomorphism is also a homeomorphism automatically since the topology on both ends is the order topology. Indeed when asking questions about this some care is required in specifying what type of isomorphism you want. Just as a group, $(\Bbb R, +)$ is isomorphic to quite a lot of things - any continuum-sized torsion-free divisible abelian group in fact. – Izaak van Dongen Nov 21 '23 at 14:50
  • "(R,+) is isomorphic to quite a lot of things - any continuum-sized torsion-free divisible abelian group in fact."

    Could you please reference that somehow? Which theorem can I invoke?

     – Crispost Nov 21 '23 at 15:57
  • Responding to comments above, I have edited the post to clearly indicate that group isomorphism is the one of interest. – Crispost Nov 21 '23 at 16:06
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    In model theory this result is called "the uncountable categoricity of the theory of torsion-free divisible abelian groups". It's mentioned for example here and here. It's really a special case of the more general fact that "two vector spaces over $F$ of large enough cardinality must be isomorphic as vector spaces", which follows from taking bases and showing they have the same dimension. – Izaak van Dongen Nov 21 '23 at 16:49
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    So to show $H$ is isomorphic to $(\Bbb R, +)$ just as groups is a bit easier.. The assumptions are a bit "overkill" for this purpose. Since $H$ is an ordered group, it's torsion-free. Since it's Archimedean, it's abelian. For any $x \in H$ and $n \in \Bbb N$, there must be some $y \in H$ such that $ny = x$, because the map $H \to \Bbb R$ given by $y \mapsto ny$ is continuous and takes on arbitrarily small and large values, and $H$ is connected. ($ny$ always means "$n$ copies of $y$ added together in the group $H$"). – Izaak van Dongen Nov 21 '23 at 17:02

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