There is an aspect of Non-Standard Analysis that for quite some time I cannot get my head around. I have done some research on this platform to avoid an exact double answer, or one that could be derived, but I still find my question unanswered. I'm happy to receive links to the threads I missed.

So, one can construct Non-Standard reals, say via an ultrafilter. Then, infinitesimals in the nonstandard realm are the equivalence classes of null sequences. Infinite nonstandard reals are equivalence classes of real sequences that are unbounded (positive and negative sign) For alternating sequences like (-1, +1, -1, +1,...), there is an infinite set in the ultrafilter that does determine which of the +1, -1 it is. So all hyperreals are (representatives of equivalence classes of) sequences.

Now, as so many other questions ask, what happens with $\mathbb{C}$? It is clear that the hyperreals are elementary equivalent to the reals, and categoricity for the complex numbers make them isomorphic to any completion of the hyperreals in that same sense. So what happens to all the infinitesimals? Do the hyperreal non-zero infinitesimals suddenly become hyperreal complex null-sequences? While when going the 'standard way' they should be zero all along? In particular, what is the function $f$ in the following diagram:

$\require{AMScd}$ \begin{CD} \mathbb{R} @>N>> \mathbb{R}^*\\ @Vc V V @VV CV\\ \mathbb{C} @>>f> \mathbb{C}\equiv\mathbb{C}^{*} \end{CD}

Here, $N$ is the usual ultrapower embedding, $c$ and $C$ are the respective algebraic completions.

The particular question is, what happens to $(1/n)_{n\in\mathbb{N}}$? That is a non-zero infinitesimal, equivalent to zero taking standard parts, so it should be a zero non-standard complex number, but cannot be going the non-standard complex way. What is wrong with this? Or rather, what implicit projection are we missing? And for the set theorists, what kind of cardinal do we need to exist for this to work?

  • $\begingroup$ Your question is unclear. What do you mean by "the categoricity for the complex numbers"? The first-order theory of $\Bbb{C}$ is not categorical. $\endgroup$
    – Rob Arthan
    Aug 30 '18 at 22:23
  • $\begingroup$ A complex number can be defined as $a + ib$ where $a$ and $b$ are real numbers. A nonstandard complex number will then be $a+ib$ where $a$ and $b$ are nonstandard real numbers. For this to be infinitesimal, $a$ and $b$ should both be infinitesimal. For it to be infinite, at least one of $a$ and $b$ should be infinite. $\endgroup$ Aug 30 '18 at 22:25
  • $\begingroup$ Your characterization of infinitesimal and infinite elements in the ultraproduct model is wrong. For example, suppose the even natural numbers form a small set. Then the sequence $(1,0,2,0,3,0,\ldots)$ is unbounded, but it represents zero, and $(1,1/2,1,1/3,1,1/4,1,\ldots)$ doesn't converge to zero, but it represents an infinitesimal. Similar things can be said if the odd natural numbers form a small set. (one of "even" and "odd" must be a small set) $\endgroup$
    – user14972
    Aug 30 '18 at 22:46
  • $\begingroup$ @Hurkyl But those sequences are equivalent to the the identically 0 sequence and to a null sequence, just as claimed in the post. The description given is correct. $\endgroup$ Aug 31 '18 at 1:02

The story for $\mathbb{C}$ is exactly like the story for $\mathbb{R}$, and nothing weird happens like you are asking about. We have elements of $\mathbb{C}^*$ which are "infinite" or "infinitesimal". A nonzero infinitesimal hyperreal is still a nonzero infinitesimal when considered as an element of $\mathbb{C}^*$.

Your point of confusion seems to stem from the fact that $\mathbb{C}$ and $\mathbb{C}^*$ are isomorphic as fields. This is true, but is pretty much irrelevant. Yes, there exists an isomorphism between them. So what? There is no canonical choice of such an isomorphism, and such an isomorphism is just an isomorphism of fields and does not respect any other structure about $\mathbb{C}$ which we might care about (such as the subfields $\mathbb{R}$ and $\mathbb{R}^*$).

It is true that you can choose some field isomorphism $g:\mathbb{C}^*\to\mathbb{C}$, and then ask what happens to $\mathbb{R}^*$ (or to the standard copy of $\mathbb{C}$ inside $\mathbb{C}^*$) under this isomorphism. The answer is that you can't really say much, since this $g$ is highly non-unique. Indeed, it is possible to choose $g$ to map any transcendental element of $\mathbb{C}^*$ to any transcendental element of $\mathbb{C}$ (where "transcendental" means "not algebraic over $\mathbb{Q}$"). So, for instance, the element of $\mathbb{C}^*$ represented by the sequence $(1/n)$ could map to any transcendental number in $\mathbb{C}$ at all, depending on the choice of isomorphism $g$.

  • $\begingroup$ Question about notation: what is $\Bbb{C}^*$? $\endgroup$
    – Rob Arthan
    Aug 30 '18 at 22:29
  • 2
    $\begingroup$ Following OP's notation, $\mathbb{C}^*$ refers to an ultrapower of $\mathbb{C}$ by some nonprincipal ultrafilter on $\mathbb{N}$. $\endgroup$ Aug 30 '18 at 22:29
  • $\begingroup$ I see: it's in the OP's commutative diagram, where the isomorphism $\Bbb{C} \equiv \Bbb{C}^*$ presumably follows by considering the cardinality of transcendence bases of the two extensions of the field of algebraic numbers. And this isomorphism, as you say, will play havoc with the embedding of $\Bbb{R}$. $\endgroup$
    – Rob Arthan
    Aug 30 '18 at 22:37
  • $\begingroup$ Are $\mathbb R$ and $\mathbb R^*$ isomorphic as fields (not ordered)? $\endgroup$
    – mr_e_man
    Jun 24 '20 at 1:50
  • 1
    $\begingroup$ @mr_e_man: No: the order is uniquely determined by the field structure, since the nonnegative elements are just the squares. $\endgroup$ Jun 24 '20 at 2:19

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.