It is a well-known fact that the real numbers are the only complete totally ordered field. So, if we perform the Dedekind cut construction on the hyperreals, then the result must either be:

  • The real numbers, or
  • Something that isn't a complete totally ordered field.

I tried doing this construction to see what goes wrong, and I couldn't seem to prove that additive inverses exist (without using the Archimedean principle, which of course doesn't hold for the hyperreals). But perhaps I'm just not clever enough. So, my question is: is this the axiom that fails, or is it another one, or do we get a complete ordered field, ie. the real numbers?


If you use only internal Dedekind cuts, then you get a field -- in fact, you just get the hyperreals themselves, as the hyperreals are already internally Dedekind complete.

But if you allow external Dedekind cuts, then you do get problems as you observed. One cut that, IMO, makes things easy to see is the cut:

  • L = all negative numbers and finite numbers
  • R = all positive transfinite numbers

The (external) Dedekind completion of the hyperreals has just the one element filling this gap between the positive finite and transfinite numbers. However, to be a field, if there are any numbers in this gap, we clearly need many numbers -- e.g. adding one to any number in this gap should give yet another number in this gap.

This is similar to your observation that additive inverses don't seem to exist; if they did, then we could use the above observation to show, e.g., that 1=0 fairly easily.

  • $\begingroup$ Thank you - this example does clarify things a lot for me. Could you explain what you mean by internal Dedekind cuts and internally Dedekind complete? $\endgroup$ – Aaron Tikuisis Aug 17 '13 at 15:04
  • $\begingroup$ @Aaron: The intuitive explanation is that anything that anything invoking standard analysis is external. The external Archimedean principle compares numbers with the standard integers -- there are hyperreal numbers larger than every standard integer, and thus the hyperreals are externally non-Archimedean. The internal Archimedean principle instead compares with the nonstandard integers -- for every hyperreal $x$, there is a hyperinteger larger than $x$, and so the hyperreals satisfy the internal Archimedean principle. $\endgroup$ – user14972 Aug 17 '13 at 15:25
  • $\begingroup$ The simplest technical explanation, IMO, is simply to take "internal" as a primitive notion and understand it via the transfer principle. There is a standard power set operation $\mathcal{P}$, and $\mathcal{P}(\mathbb{R})$ is the set of all subsets of the real numbers. By transfer, there is a non-standard power set operation ${}^\star \mathcal{P}$, and the internal subsets of ${}^\star \mathbb{R}$ are precisely the elements of ${}^\star \mathcal{P}({}^\star \mathbb{R)}$. There are other explanations, but this (and others like it) are the only ones that are really meaningful to me. $\endgroup$ – user14972 Aug 17 '13 at 15:31
  • $\begingroup$ Thank you, I think I understand. What you mean is that, while the reals satisfy "every bounded set has a least upper bound", the hyperreals also satisfy this, except with "internal set" in place of "set"? $\endgroup$ – Aaron Tikuisis Aug 17 '13 at 16:04
  • $\begingroup$ @Aaron: That's right. $\endgroup$ – user14972 Aug 17 '13 at 16:17

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.