5
$\begingroup$

Several times I have seen questions/answers here about using the correct definition of derivatives. There are also questions about whether or not $1/0$ is defined. Sometimes there is a discussion about the concept of infinitesimal and infinite numbers which I guess are related to hyperreal numbers ${}^*\mathbb{R}$ (correct me if I am wrong).

My question is

How is is the set of hyperreal numbers constructed/defined?

I am specifically interested in understanding what the hyperreals are as a set and how the hyperreals become an ordered field (This would necessarily include specifying what the inverse of the "extra" elements in ${}^*\mathbb{R}$ are). How, for example, does $\mathbb{R}$ sit inside ${}^*\mathbb{R}$?

Edit: I see that one question already asks about this, but I am, wondering if the construction of the set might be explained clearly.

$\endgroup$
  • $\begingroup$ Several constructions are given here: en.wikipedia.org/wiki/Hyperreal_number $\endgroup$ – vadim123 Mar 16 '14 at 20:15
  • $\begingroup$ Perhaps math.stackexchange.com/questions/315768/… might help? $\endgroup$ – user122283 Mar 16 '14 at 20:16
  • $\begingroup$ @SanathDevalapurkar: Would you say that my question is a bit more general since I am not asking about a specific method? $\endgroup$ – Thomas Mar 16 '14 at 20:18
  • $\begingroup$ @Thomas Yes, I would agree. However, have you tried looking at the links in the answers there? $\endgroup$ – user122283 Mar 16 '14 at 20:20
  • $\begingroup$ @SanathDevalapurkar: No, I haven't. I will. $\endgroup$ – Thomas Mar 16 '14 at 20:20
7
$\begingroup$

There are a couple of Monthly articles that give reasonably accessible introductions. See

Wm. Hatcher. Calculus is Algebra, AMM 1982.

D.H. Van Osdol. Truth with Respect to an Ultrafilter or How to make Intuition Rigorous. AMM, 1972..

For a much more comprehensive introduction to ultraproducts see

Paul Eklof. Ultraproducts for Algebraists, 1977.

$\endgroup$
1
$\begingroup$

To motivate the construction of the hyperreals, one could compare it with the construction of the real numbers as equivalence classes of Cauchy sequences of rational numbers. Namely, hyperreals can be similarly constructed as equivalence classes of Cauchy sequences of real numbers. To motivate the construction, note that there is a lot of "collapsing" going on when one passes from a Cauchy sequence to its equivalence class. Namely, one loses all information about anything related to the rate of convergence of the sequence.

In fact, the construction of the hyperreals can be achieved by refining the Cauchy sequence construction. The refined equivalence relation will declare two sequences $(u_n)$ and $(v_n)$ to be equivalent if they agree on a "dominant" set of indices; i.e. the subset of $\mathbb{N}$ given by $\{n\in\mathbb{N} : u_n=v_n\}$ is "dominant". I will comment on the nature of "dominant" sets of indices in a moment. Now taking the set of equivalence classes will give you only finite (more precisely, limited) hyperreals. This includes infinitesimals, i.e. hyperreals represented by null sequences $(u_n)$ (i.e., sequences that tend to zero).

To make this into a field, one also needs the elements represented by sequences $(\frac{1}{u_n})$ where $(u_n)$ represents an infinitesimal. Note that such sequences $(\frac{1}{u_n})$ are no longer Cauchy. The punchline is that if one assumes CH then the result is the full hyperreal field (given an appropriate choice of ultrafilter, namely a P-point filter). Moreover a hyperreal field constructed via the ultrapower is unique up to isomorphism assuming CH (in particular, independent of the ultrafilter used).

To comment briefly on the notion of a "dominant" set of indices: here a finite set is never "dominant", and a cofinite set is always dominant. To get the job done one needs to choose, out of every pair of complementary infinite subsets of $\mathbb{N}$, precisely one that will be declared "dominant" in a coherent fashion. For the details you would need to look up the notion of an ultrafilter.

Edit 1. To respond to your question concerning the way $\mathbb{R}$ sits inside ${}^{\ast}\mathbb{R}$, note that the constant sequences $(u_n)$ where $u_n=r$ give an imbedding, namely a real number $r$ goes to the constant sequence $(u_n)$ as above.

$\endgroup$
  • $\begingroup$ if one assumes CH then the result is the full hyperreal field (given an appropriate choice of ultrafilter), which is moreover unique up to isomorphism. I don't think this is right. CH is irrelevant, and uniqueness up to the choice of ultrafilter is no uniqueness at all. See mathoverflow.net/questions/88292/… $\endgroup$ – Ben Crowell May 19 '14 at 22:30
  • 1
    $\begingroup$ @BenCrowell, that discussion did not relate to the role of CH. The ultrapower $R^N/U$ is known to be unique up to isomorphism under the hypothesis of CH, and not unique otherwise (there are some SE posts on this as well). Here it is the field that's unique up to isomorphism, but the isomorphism is not unique as Kanovei points out there $\endgroup$ – Mikhail Katz Jul 28 '14 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.