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Nov 24, 2019 at 10:18 comment added Alec Rhea @GerryMyerson First order definitions, quantifying over individual members, allow these structures to be perceived and understood in a relatively simple manner. If you allow second or higher order quantification over the objects in play, I would argue that the surreals surmount any previously conceived totally ordered field in complexity.
Nov 23, 2019 at 2:55 comment added Gerry Myerson I don't know what any of that means. But I note that user Ultradark has just posted (a model of) ${\bf R}^3$ as an answer.
Aug 1, 2018 at 17:06 comment added Alec Rhea @GerryMyerson If we allow definitions of all order over $\mathbb{R}^3$ then I'd say yes; if we stick to first order definitions it may be simpler, since RCF is complete but PA isn't.
Aug 1, 2018 at 13:01 comment added Gerry Myerson ${\bf R}^3$ contains a copy of many objects already on the list – does that mean it surpasses them in complexity?
S Aug 1, 2018 at 2:23 history answered Alec Rhea CC BY-SA 4.0
S Aug 1, 2018 at 2:23 history made wiki Post Made Community Wiki by Alec Rhea