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Let's treat $\mathbf{No}$ as a group under addition, and forget its field structure for a little bit.

I will define a "maximally Archimedean subgroup" of $\mathbf{No}$ as a subgroup which is

  1. Archimedean, and
  2. not strictly contained by any larger Archimedean subgroup of $\mathbf{No}$.

Each of these groups is isomorphic to $\mathbb{R}$. Examples: $\mathbb{R}$, $\omega\mathbb{R}$, $\frac{1}{\omega}\mathbb{R}$, $\sqrt{\omega_1}\mathbb{R}$ etc.

Now consider the direct sum of all of these subgroups. Is this group isomorphic to $\mathbf{No}$?

Also, can anything useful be said about the direct product of the subgroup instead, other than that it contains things like the sum of all infinite cardinals, stuff like $\sum_{r\in\mathbb{R}} \omega^r$, $\sum_{r\in\mathbf{Ord}} \omega^r$, $\aleph_0+\aleph_1-\aleph_2+\aleph_3-...$, etc, and hence is just baffling all around?

You can formalize $\mathbf{No}$ however you like when dealing with the surreals to handle the foundational issues.

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The surreals are naturally a vector space over the real numbers, and so by a global axiom of choice (that would apply to class-sized structures), that vector space has a basis. This would give a decomposition of the underlying additive group of No as a direct sum of copies of $\mathbb{R}$. Any proper class indexed direct sum of copies of $\mathbb{R}$ would be isomorphic to No, I would think. –  Todd Trimble Aug 29 '13 at 11:07
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But to follow up on my previous comment: is it not true that maximally archimedean subgroups are the same thing as 1-dimensional $\mathbb{R}$-subspaces of No? For example, $(\omega + 1)\mathbb{R}$. So if you're asking whether the canonical map from the direct sum of such 1-dimensional subspaces to No (the one whose restriction to each summand is the subspace inclusion) is an isomorphism, then the answer is clearly "no". You'd want some sort of homogeneity requirement as well, to rule out examples like $(\omega + 1)\mathbb{R}$. –  Todd Trimble Aug 29 '13 at 13:04
    
Thanks Todd, that's a good counterexample to the question I asked. I'm not sure I understand the homogeneity requirement though - what do you mean by that? –  Mike Battaglia Sep 2 '13 at 0:29
    
Hm, I'm not sure I know what I mean by that. I can't locate relevant texts right now, but I seem to recall that surreals can be put into a Cantor normal form, and roughly speaking, homogeneous elements would be the individual summands of Cantor normal form. So: a real number times a power of $\omega$, if I remember correctly. But for now this is speculative. –  Todd Trimble Sep 2 '13 at 2:34
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