I know that it is possible to construct the hyperreal number system in ZFC by using the axiom of choice to obtain a non-principal ultrafilter. Would the non-existence of a set of hyperreals be consistent with just ZF, without choice? Let me be conservative, and say that by a "set of hyperreals," I just mean a set together with some relations and functions such that the transfer principle holds, and there exists $\epsilon > 0$ smaller than any real positive real number.

  • 5
    $\begingroup$ Abraham Robinson suggested that ZF and ZFC were in some sense constructed exactly so as to allow us to do analysis on the reals. From that point of view, it's not surprising that in ZF(C) the reals exist and are unique, whereas ZF doesn't make the hyperreals exist, and ZFC doesn't make the hyperreals unique. $\endgroup$ – Ben Crowell Dec 4 '14 at 2:45
  • $\begingroup$ @BenCrowell, I saw a remark of this sort in a paper by Keisler but I don't recall seeing it in Robinson. Do you have a source for this? $\endgroup$ – Mikhail Katz Dec 5 '14 at 9:33
  • 2
    $\begingroup$ @katz: Just because you added a comment about the philosophical implications of not assuming the axiom of choice, and just because you added that as an answer instead of a comment, doesn't mean that the question itself is about mathematical philosophy. This is not a question for "What are the philosophical arguments in favor of assuming a hyperreal field exists" or something similar. This is just a simple question, can we prove in $\sf ZF$ that a hyperreal field exists with the transfer principle? The answer to which has nothing to do with philosophy, or mathematical philosophy in particular. $\endgroup$ – Asaf Karagila Dec 5 '14 at 9:42
  • $\begingroup$ What are you trying to ask? On possibile definition of a hyperreal number system is a system satisfying all the requirements for a complete ordered field but the requirement of completeness and replaces that requirement with the two requirements that for any Dedekind cut where you can take a number in each part to be arbitrarily close, the cut has a boundary number position, and there exists a number that exceeds all natural numbers. Another possible definition is the same definition with the additional requirement that it includes the real numbers and its operations on the real numbers are $\endgroup$ – Timothy Jan 15 '19 at 7:24
  • $\begingroup$ the same as the ones defined for the real numbers. Are you asking whether it's consistent with ZF that it's the case that no system satisfies the additional requirement because for every system that satisfies the other requirements for a hyperreal number system, there does not exist a way to pick one member from each equivalence class of infinitesimally close hyperreal numbers such that it's closed under addition and subtraction because the axiom of choice is false? $\endgroup$ – Timothy Jan 15 '19 at 7:28

The answer is yes, provided ZF itself is consistent. The reason is that the existence of the hyperreals, in a context with the transfer principle, implies that there is a nonprincipal ultrafilter on $\mathbb{N}$.

Specifically, if $N$ is any nonstandard (infinite) natural number, then let $U$ be the set of all $X\subset\mathbb{N}$ with $N\in X^*$. This is a nonprincipal ultrafilter on $\mathbb{N}$, since:

  • If $X\in U$ and $X\subset Y$, then $N\in X^*\subset Y^*$, and so $Y\in U$.
  • If $X,Y\in U$, then $N\in X^*\cap Y^*=(X\cap Y)^*$ and so $X\cap Y\in U$.
  • If $X\subset\mathbb{N}$, then every number is in $X$ or in $\mathbb{N}-X$, and so either $N\in X^*$ or $N\in(\mathbb{N}-X)^*$ and thus $X\in U$ or $\mathbb{N}-X\in U$.
  • For any particular standard natural number $n$, the set $X=\{m\in \mathbb{N}\mid n\leq m\}$ is in $U$, because $n^*\leq N$.
  • The empty set $\emptyset$ is not in $U$, since $N\notin\emptyset=\emptyset^*$.

So $U$ is a nonprincipal ultrafilter on $\mathbb{N}$. The way that I think about $U$ is that it concentrates on sets that express all and only the properties held by the nonstandard number $N$. (See also my answer to A remark of Connes, where I make a similar point, and explain that, therefore, nonstandard analysis with the transfer property implies that there must be a non-measurable set of reals.)

Thus, in a model of ZF with no nonprincipal ultrafilter on $\mathbb{N}$ (and as Asaf mentions in the comments, there are indeed such models if there are any models of ZF at all), there is no structure of the hyperreals satisfying the transfer principle.

  • 2
    $\begingroup$ It might be worth pointing that it is consistent with $\sf ZF$ that there are no free ultrafilters on $\Bbb N$. And for that matter, on any set. $\endgroup$ – Asaf Karagila Dec 4 '14 at 10:42
  • 3
    $\begingroup$ A small remark: in ZF the existence of a proper elementary extension of the structure $(\Bbb{N}, X)_{X\in\cal{P}(\omega)}$ is equivalent to the existence of a nonprincipal ultrafilter on $\Bbb{N}$. Joel has explained one direction, the other direction uses a standard ultrapower argument (the Łoś-theorem goes through in the absence of choice in this case since $\Bbb{N}$ is well-orderable). $\endgroup$ – Ali Enayat Dec 8 '14 at 23:46

In response specifically to the title of the question: "Is non-existence of the hyperreals consistent with ZF?", technically speaking the answer is NO. Kanovei and Shelah constructed a definable model of the hyperreals in ZF; see http://arxiv.org/pdf/math/0311165.pdf

Therefore ZF is not "consistent with the nonexistence of the hyperreals". Of course, to prove any of their properties (such that that they are actually a proper extension, satisfy transfer, etc) one needs AC, but the same goes for many other crucial mathematical results (see below).

Goldblatt in his book "Lectures on the hyperreals" takes countable additivity to be part of the definition of measure (see M1 on page 206). Then he uses hyperfinite partitions, outer measures, and transfer to show that express the Lebesgue measure in terms of the Loeb measure (page 217). In particular countable additivity of Lebesgue measure follows.

Note that the failure of countable additivity of the Lebesgue measure (such countable additivity is taken for granted in analysis) is also consistent with ZF; see Is sigma-additivity of Lebesgue measure deducible from ZF?

Similarly, it is consistent with ZF that the Hanh-Banach theorem (arguably foundation of functional analysis) fails.

  • 4
    $\begingroup$ Seems like a reasonable comment. Not quite an answer to the question at hand. $\endgroup$ – Asaf Karagila Dec 4 '14 at 18:17
  • 4
    $\begingroup$ Yes, and this is a very good comment. Answers should answer the question, not make remarks on its background. $\endgroup$ – Asaf Karagila Dec 5 '14 at 9:39
  • 6
    $\begingroup$ @AsafKaragila I find your criteria to be unnecessarily restrictive. We should welcome any post that makes, as here, a substantive and interesting mathematical contribution bearing on a question. I believe that expert and insightful remarks on the background of a question, in general, make fine answers. $\endgroup$ – Joel David Hamkins Dec 5 '14 at 12:14
  • 2
    $\begingroup$ @Joel: This is a discussion for the meta site, rather than here. But I think that answers should be answers. Of course, a question asking for something, and you want to post an answer showing that certain assumptions are really needed, is some sort of answer. But if you are giving it as an answer, you should at least give some substantial explanation. Here, on the other hand, there is just a remark equivalent to "Hey, by the way, the axiom of choice is needed elsewhere in analysis", and not much longer either. I agree this makes an excellent comment, it makes a bad answer, though. $\endgroup$ – Asaf Karagila Dec 5 '14 at 12:41
  • 2
    $\begingroup$ Asaf, my personal view is that we should welcome any mathematically substantive, interesting post that sheds light on the issues of a question. So I don't really agree to have the kind of formal rules that you are advocating. (And my comments here object to your criterion only in so far as you have put it forth as a general rule, apart from the merits of this particular case.) $\endgroup$ – Joel David Hamkins Dec 6 '14 at 22:34

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.