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Nov 27 at 5:06 answer added Elliot Glazer timeline score: 3
Jul 6, 2016 at 14:42 history edited Mikhail Katz CC BY-SA 3.0
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Apr 11, 2013 at 12:01 history edited Mikhail Katz
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Apr 11, 2013 at 11:03 comment added Emil Jeřábek @Joel: By your observation, Oz is a model of open induction, but that’s about it. Oz does not satisfy any of the usually considered stronger fragments of PA, as it is not even integrally closed. See mathoverflow.net/questions/72691 .
Apr 11, 2013 at 4:29 comment added Philip Ehrlich My pleasure, Vladimir. Glad to be of help.
Apr 11, 2013 at 4:01 comment added Vladimir Kanovei Philip: yes, and I am thankful for pinting on my historic misinterpretation.
Apr 11, 2013 at 3:28 comment added Vladimir Kanovei >all of nonstandard analysis can be done using the surreals or some particular initial subfield of the surreals as the underlying ordered field in a nonstandard model of analysis. To be exact, changing for whatever purpose the elements of R* with some surreals. The only foundational appeal of this is that the surreal fields are uniquely defined - but the field structure is too little to work with infinitesimals (further than "aga, 1/w is an infinitesimal"). The challenge has been stated: is No capable to run the Euler sine decomposition w/o borrowing *N to replace its own non-PA omni ones.
Apr 11, 2013 at 3:25 comment added Philip Ehrlich Vladimir: I'm delighted to see that you edited your remarks on Hausdorff, presumably in light of my answer. By the way, the intimate relation between Hausdorff's pantachie and the surreal numbers is given by Theorem 17 of my BSL paper named in the answer.
Apr 11, 2013 at 3:08 history edited Vladimir Kanovei CC BY-SA 3.0
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Apr 11, 2013 at 2:09 comment added Philip Ehrlich (Continued from above): By the way, most of the nonstandard analysts I have spoken with (including H.J. Keisler, David Ross and Mauro Di Nasso, ...) are great fans of the surreals and would welcome the discovery of useful connections.
Apr 11, 2013 at 1:40 comment added Philip Ehrlich David, don't despair, all of nonstandard analysis can be done using the surreals or some particular initial subfield of the surreals as the underlying ordered field in a nonstandard model of analysis. On the other hand, given the present state of knowledge, it would be inconvenient to do so. After all, there are highly visible nonstandard models of analysis that are easy to work with. Of course, none of them have the intuitive appeal of the surreals. Things may change with time, but the importance of the surreals has little to do with its relation to the great theory of nonstandard analysis.
Apr 11, 2013 at 0:47 comment added David Corwin Aww, and I was hoping that the NSA was using surreal numbers!
Apr 11, 2013 at 0:34 answer added Philip Ehrlich timeline score: 4
Apr 10, 2013 at 20:38 comment added Vladimir Kanovei But - well, my criterion is that a candidate for integers soundly defines all elementary real functions in one click - as N does in R and N* does in R*. A grade-2 task would be to support Euler sine factoring.
Apr 10, 2013 at 20:18 comment added Vladimir Kanovei You bet a discrete subset N of F cofinal in F and such that 0 belongs to N and for any $x \in N$ also $x+1\in N$ and $x-1\in N$, and (x,x+1) has no elements of N - can be defined for just any rcof F. Therefore the omni part of No - if characterized by these properties only - is just a banality.
Apr 10, 2013 at 13:49 comment added Joel David Hamkins I should also say $\gamma_\alpha\geq 0$, for membership in Oz.
Apr 10, 2013 at 13:47 comment added Joel David Hamkins The omnific integer part of No, denoted Oz, defined in theorem 16 of Ehrlich's paper ohio.edu/people/ehrlich/Unification.pdf seems to be ordinal-definable. These are just the surreal numbers whose coefficients $n_\alpha$ in their Conway names $\Sigma_{\alpha\lt\beta} \omega^{\gamma_\alpha}n_\alpha$ are integers. But I don't know if Oz satisfies PA or if you get the elementary property that you want. But they are an integer part of No, in the sense that every surreal number $x$ is between $a$ and $a+1$ for some $a\in\text{Oz}$. Could you elaborate on why this isn't what you want?
Apr 10, 2013 at 13:30 comment added Joel David Hamkins Great question!
Apr 10, 2013 at 8:02 history edited Vladimir Kanovei CC BY-SA 3.0
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Apr 10, 2013 at 7:14 comment added Mikhail Katz In the second line of your question you mention "superreals". Do you mean "surreals"?
Apr 10, 2013 at 5:16 history asked Vladimir Kanovei CC BY-SA 3.0