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I have seen approaches at building hyperreal systems by using complicated notions like ultrafilters and the like.

Why not just postulate the existence of infinitesimal element $\varepsilon$ and infinite $\omega=1/\varepsilon$ like we do with complex numbers and build a field system around them?

Anixx
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  • Why is the downvote? – Anixx Mar 17 '15 at 04:34
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    Can you elaborate on exactly how you would "build a field system around them?" The devil is presumably in those details. – Nate Eldredge Mar 17 '15 at 05:19
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    Also, note that the hyperreals are an elementary extension of the reals: any first-order sentence that's true in the reals is also true in the hyperreals. I suspect that any simpler construction is not going to satisfy the full strength of that. – Nate Eldredge Mar 17 '15 at 05:24
  • @NateEldredge: I think the spirit of the question is the field $\Bbb R[x]$ with $x$ considered as an infinitesimal. This seems to be in the spirit of add an infinitesimal and close under the field operations. An appropriate answer would be either that you can't construct a field that way or here is a property that field doesn't have that we would like, not that I have such an answer. – Ross Millikan Mar 25 '15 at 01:25
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    @RossMillikan, that field is the field of Formal Laurent series. It doesn't have the "elementary extension" property that we would like. – Mark S. Mar 25 '15 at 11:17
  • If you find the construction of Hyperreal too hard try to read about Surreal, they are pretty the same proper class (someone calls them set but it's not correct) but they're building technique is easier to understand, it's a generalization of Dedekind's cut for building Reals. You can learn about on wikipedia page: http://en.wikipedia.org/wiki/Surreal_number#Relation_to_hyperreals – AlienRem Aug 28 '15 at 11:20

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There are ways to build a field by throwing in an infinitesimal to the reals and then whatever else you need. For example, the field of formal Laurent series (ordered so that monomials with positive coefficients are first ordered by the power of x, e.g. $0 <x <1<100<1/x$) is basically that. The Levi-Civita field is similar.

However, neither of those actually have the properties needed to answer questions about calculus. You need to throw a lot in in just the right way to get all simple statements to pass through, as Nate Eldredge alluded to.

As an aside, the basic idea behind of the construction isn't too complicated. I like Terry Tao's voting analogy. A hyperreal is a sequence of reals that vote each time you ask about a property (like "are you bigger than 5?"). How to determine which infinite collections of voters count as good majorities is handled by some technical stuff, but you don't have to worry about that to get the idea.

Mark S.
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    Thanks for pointing out the Levi-Civita field – Anixx Mar 25 '15 at 02:41
  • What questions about calculus remain unanswered? – Anixx Mar 25 '15 at 19:09
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    @Anixx, I think you misunderstood my meaning. The purpose of the hyperreals is to have a framework for derivatives and integrals that let's you answer all the same questions as regular calculus, but with infinitesimal numbers, rather than epsilon delta definitions. – Mark S. Mar 26 '15 at 19:50
  • Well what questions still remain unanswered with the definition from the question? – Anixx Mar 26 '15 at 20:04
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    @Anixx For one thing, you can't define differentiability properly for all real functions without the transfer principle of the hyperreals. I suppose "is this function differentiable at x=17?" Would be an unanswered question. – Mark S. Mar 26 '15 at 20:18
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    Differentiability: $$(f(x+\varepsilon)-f(x-\varepsilon))/(2\varepsilon)$$ What needs more definition with it? – Anixx Mar 26 '15 at 20:21
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    @Anixx, Those expressions needn't even make sense for arbitrary real functions f. I can't explain this issue in a comment, but maybe I or someone else will write a new long answer to this question that clears this stuff up. – Mark S. Mar 26 '15 at 21:26
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One good reason to use the hyperreals $\mathbb{R}^*$ as constructed using a nonprincipal ultrafilter is that they satisfy a transfer principle (as do other versions of NSA, for example IST). This means in particular that every function $f$ or relation $R$ on the reals can be "enlarged" to a function $f^*$ or relation $R^*$ on the hyperreals in such a way that a first order statement about $\mathbb{R}$ is true if and only if the corresponding enlarged statement is true of $\mathbb{R}^*$.

Existence of enlargements is useful when you have some standard function which you want to extend to a nonstandard setting. If you want to work in $\mathbb{R}((\epsilon))$ for example, the "right" definition of $\cos(\epsilon)$ is clearly $\sum_{n=0}^\infty (-1)^n \epsilon ^{2n} / (2n)!$ (though what $e^{-1/\epsilon}$ should be isn't so clear). But if I have some real function that can't be expressed by a power series, how should it be defined on $\epsilon$? The transfer principle allows you to dodge this problem.

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Details matters!

As an example, if you just postulate the existence of infinitely big naturals you will have troubles because

$\{\omega \in \Bbb{N} \ : \ \omega \ \mbox{is infinitely big}\}$

is a non-empty set without minimum.

Those complicated constructions arised to avoid this and a lot of other gaps that could appear if we are not careful enough.

  • I think what you intended is a true statement about the hyperreals, so that's not really an avoided problem. – Mark S. Mar 24 '15 at 20:48
  • The transfer principle states that any first order sentence that works for the reals works for the hyperreals. The statement $\forall A\subseteq\Bbb N(A\text{ has a minimum})$ isn't first order, I believe. ($x\in\Bbb N$ is first order, I believe, because it's equivalent to $x=\lfloor x\rfloor$. The hyperreal translation is $x\in\Bbb N^*$, or $x=\lfloor x\rfloor$ where $\lfloor\cdot\rfloor$ is the natural extension of the floor function.) – Akiva Weinberger May 15 '15 at 20:37
  • "First order" meaning it doesn't talk about sets. – Akiva Weinberger May 15 '15 at 20:45