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The Wikipedia article on surreal numbers states that hyperreal numbers are a subfield of the surreals. If I understand correctly, both fields contain:

  • real numbers
  • a hierarchy of infinitesimal numbers like $\epsilon, \epsilon^2, \epsilon^3, \ldots$
  • a hierarchy of transfinite numbers like $\omega, \omega^2, \omega^3, \ldots$ where $\omega = 1/\epsilon$

and both allow the four standard arithmetic operations to be applied to any combination of real, infinitesimal, and transfinite numbers. So what is the difference, if any, between these number systems? If the Wikipedia statement is accurate, what numbers are surreal but not hyperreal?

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2 Answers 2

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There are many non-isomorphic non-standard models of reals; any of them can be called hyperreals, although one specific model (the ultrafilter construction on $\mathbb{R}^\mathbb{N}$) is often called "the" hyperreals.

Models are generally taken to be sets. The surreal numbers are a proper class: they are "too big" to be considered a non-standard model of the reals in this sense.

But to some extent, we don't really have to insist on models being sets: with suitable set-theoretic axioms, I believe the surreal numbers are also a non-standard model of the reals. In fact, they would be the largest model.

If we pick one particular (set-sized) non-standard model -- e.g. "the" hyperreals -- then we cannot compare its elements to surreal numbers directly. First, we'd have to choose a way to embed the hyperreals into the surreals. There isn't a unique way to do this. In fact, there is a vast number of ways -- an entire proper class of embeddings! (I believe) we can choose to make any particular surreal a hyperreal number by choosing an appropriate embedding.

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    $\begingroup$ Interesting. Could you elaborate on the embedding? I'm just trying to get some intuition for this, not formal details. Are there "typical" or "natural" embeddings, and given such an embedding, where would the extra surreals be? For instance, is it somewhat analogous to how the rationals are embedded in the reals, where the rationals are dense but there are extra real numbers "between" the rationals? $\endgroup$
    – Nathan Reed
    Oct 26, 2012 at 17:26
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    $\begingroup$ For any (set-sized) non-standard model of the reals, call it $F$, we can construct a new non-standard model of the reals extending $F$ that contains an $F$-infinitessimal $\epsilon$: that is, $\epsilon$ is a positive number that is smaller than every positive element of $F$. Therefore, $F$ can't be dense in this extension: e.g. in this new model, the interval $[3+\epsilon, 3+2\epsilon]$ doesn't contain any elements of $F$. $\endgroup$
    – user14972
    Oct 27, 2012 at 8:25
  • $\begingroup$ Things get a little hairy when dealing with class-sized hyperreal fields, but Keisler is quoted with a precise statement of "the surreals are the biggest version of the hyperreals" and proof sketch in Ehrlich's "The absolute arithmetic continuum and the unification of all numbers great and small" $\endgroup$
    – Mark S.
    Mar 29, 2015 at 14:59
  • $\begingroup$ If the surreal numbers cannot be considered a set, doesn't that by definition mean they can't form a field? $\endgroup$
    – BlueRaja - Danny Pflughoeft
    May 3, 2020 at 22:48
  • $\begingroup$ @Blue, no, if you look up the list of field axioms, you will notice that they make perfect sense without assuming the collection to be a set. $\endgroup$
    – Mikhail Katz
    2 days ago
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According to the recent work by Ehrlich, the surreals are also a subset of the hyperreals. More precisely, maximal class-size fields of the surreals and the hyperreals are isomorphic.

When one wishes to trim down one's model to set-size proportions, the advantage of the hyperreals over the surreals becomes obvious, because the hyperreals possess a transfer principle that makes them useful in analysis and other fields of mathematics.

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