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Jun 17, 2016 at 3:38 comment added Twiffy @Nombre, thanks, that's very clear.
Jun 15, 2016 at 15:06 comment added Philip Ehrlich @Twiffy. Yes, it follows from the K-K-S result that there is no hyperreal ordered field $^*\mathbb{R}$ having an underlying ordered field isomorphic to a Hahn field. On the other hand, like the surreals, many hyperreal ordered fields have underlying ordered fields that are isomorphic to distinguished subfields of Hahn fields. This was pointed out in my: An alternative construction of Conway's ordered field No. Algebra Universalis 25 (1988), no. 1, 7–16.
Jun 15, 2016 at 9:01 comment added nombre @Twiffy: If $G$ is a set-sized ordered group, then $|\mathbb{R} \boxtimes G| \geq {|\mathbb{R}|}^{\kappa}$ for each cardinal $\kappa$ embedding in $(G,<)$. It is consistent (assuming the continuum hypothesis) that $|^*\mathbb{R}| = |\mathbb{R}|$ is such a cardinal for $G = ^*\mathbb{R}$, and thus that $|\mathbb{R} \boxtimes ^*\mathbb{R}| \geq 2^{|^*\mathbb{R}|} > |^*\mathbb{R}|$.
Jun 14, 2016 at 23:41 comment added Twiffy It seems that the hyperreals $^*\mathbb{R}$ admit an exponential map in this sense, where we take $\textrm{exp}(x) = e^x$ componentwise on $\mathbb{R}^\mathbb{N}$ and then mod out by an ultrafilter. So the above paper would seem to imply that $^*\mathbb{R}$ cannot be written as (in my notation) $\mathbb{R} \boxtimes G$ for any ordered abelian group $G$. But I would have guessed that the hyperreals (as with the surreals) are a fixed point of the $\mathbb{R} \boxtimes \mbox{--}$ functor. Is it otherwise known that this is not the case?
Jun 14, 2016 at 23:24 comment added Twiffy Thanks @PhilipEhrlich, that's perfect. I actually know Franz-Viktor so it's rather embarrassing that I hadn't found that paper myself.
Jun 14, 2016 at 23:14 vote accept Twiffy
Jun 14, 2016 at 14:26 history answered Philip Ehrlich CC BY-SA 3.0