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In the context of Robinson's framework, or more precisely its reformulation by Ed Nelson, one of the practitioners in the field expressed the sentiment something like "the naive counting numbers don't exhaust $\mathbb{N}$." Who was this mathematician and what was the precise wording?

For a follow-up question see What's Reeb's take on naive integers?

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Mikhail Katz
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    That's a great quote! – goblin GONE Feb 14 '16 at 09:35
  • Could someone explain what this means? – thedude Feb 14 '16 at 13:08
  • @thedude, a good starting point is Joel David Hamkins' summary in his answer to this question. I would be happy to explain further if you are interested. – Mikhail Katz Feb 14 '16 at 13:38
  • The answer you mentioned is way over my head. Ultrapowers? Show me an element of $\mathbb{N}$ which is not a naive integer – thedude Feb 14 '16 at 18:45
  • @thedude, very good question. I would like to answer it with a question: show me an element of $\mathbb{N}$ that cannot be expressed even by a computer the size of the universe, in the entire time alloted to our civilization. I gave a lower-level introduction to the variety of approaches to Robinson's framework here. You could also consult the wiki page for Internal Set Theory. Let me know what you find. – Mikhail Katz Feb 15 '16 at 07:47
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    You can exhibit a non-naive integer like this: take some formulation of Peano arithmetic that has a variable $x$. Now add infinitely many axioms axioms $x\ne0$, $x\ne1$, $x\ne2$, and so on. If these introduce a contradiction, then some finite subset of them must introduce a contradiction, and that would imply PA is inconsistent. $x$ is the element of $\mathbb{N}$ that you seek. – Dan Piponi Feb 21 '16 at 15:20
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    @thedude: Consider Peano arithmetic, on top of which you add a constant $x\in\mathbb N$, and the sequence of axioms $x>1$, $x>2$, $x>3$, ... Clearly the system of axioms cannot lead to any contradiction (a contradiction is a proof of $0=1$, and proofs are finite, so it would have to do so using only finitely many of those extra axioms.) Now call $\mathbb N$ any model of that theory. Then $x$ is an element of $\mathbb N$ which is not a naive integer. – André Henriques Feb 21 '16 at 15:20
  • @AndréHenriques Synchronicity? – Dan Piponi Feb 21 '16 at 15:21
  • I think that we were less than 7 seconds off from each other. – André Henriques Feb 21 '16 at 15:22
  • @DanPiponi, I have the impression user:thedude may have been looking for a more intuitive argument in favor of what seems like a puzzling statement against the background where the intended interpretation, naive integers, and $\mathbb{N}$ are all one and the same thing. Reeb argued that there is a difference, as I tried to analyze at http://mathoverflow.net/questions/231599 – Mikhail Katz Feb 21 '16 at 15:35
  • @AndréHenriques, see comment above. – Mikhail Katz Feb 21 '16 at 15:35
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    Thanks every one for all your effort in trying to illuminate me. Sadly, its not working for several (related) reasons. One, I don't have the background (I'm a physicist). Two, I can't really process what you are saying. The comment of @DanPiponi, for example, reads to me like this "Suppose $x$ is in $\mathbb{N}$, but is different from all integers, then..." which seems to assume what it wants to prove. Three, I lack motivation to invest much effort into understanding this, since I can't see the point in these weird new numbers (my fault, no criticism). Anyway, thanks again! – thedude Feb 21 '16 at 16:06
  • @thedude, you should have mentioned right away that you are a physicist. If so you should be comfortable with infinite numbers that come up all the time in physics (as well as infinitesimals). If so, adding one to one to get 2 and then 3 etc. will obviously never reach that infinite number, which can be taken to be the non-naive one. – Mikhail Katz Feb 21 '16 at 16:08
  • @goblin, can I ask you to take a look at the closure here: http://math.stackexchange.com/questions/1648838/is-it-to-the-students-advantage-to-learn-the-language-of-infinitesimals if possible? – Mikhail Katz Feb 21 '16 at 16:14
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    @thedude. What is here called $\mathbb N$ isn't the integers. It's something that contains the integers (condescendingly called the "naive integers"), and also contains an extra thing called "$x$", along with everything that this new $x$ generates. Now, the catch is that there isn't a single way to let this new $x$ generate stuff. There are infinitely many way of doing so. Even worse, there is no way of singling out one among those infinitely many ways. So this thing which is here called $\mathbb N$ (and for which the common notation is $\mathbb N^*$) is highly non-unique. – André Henriques Feb 21 '16 at 16:55
  • @André, if you denote this object $\mathbb{N}$ then you are apparently adopting Nelson's viewpoint as popularized in Alain Robert's book for example. But then $\mathbb{N}$ is just the ordinary system of natural numbers that has all the uniqueness properities one expects it to have in ZFC. On top of that Nelson points out that there is a richer syntax that we haven't used before; but your nonuniqueness claim seems to confuse Nelson's and Robinson's framework. – Mikhail Katz Feb 21 '16 at 17:14
  • @thedude Try assuming a number that's bigger than 3 and less than 2. You'll quickly hit a contradiction. Nonetheless you can assume the existence of an integer not equal to any of the integers you can name and you won't get a contradiction. The axioms of PA don't exclude this possibility. – Dan Piponi Feb 21 '16 at 20:22
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    @katz, it seems like a reasonable question to close, for the simple reason that it would be very hard to give that question an "objective" answer. Surely there's mathematics forums around where these kinds of issues can be discussed (as opposed to "answered"). – goblin GONE Feb 22 '16 at 01:54

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Georges Reeb, I believe. At least that was his slogan, and C. Lobry made it the title of an uproarious book: Et pourtant... ils ne remplissent pas N! Reeb would go around conferences and confront random attendees with the quote, insisting that even Bourbaki said that.

Francois Ziegler
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  • Thanks, I thought it was perhaps one of the Dieners, good to know. – Mikhail Katz Feb 14 '16 at 09:46
  • I think your added comment is unnecessarily sarcastic. – Mikhail Katz Feb 14 '16 at 13:35
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    @katz Why? I agree with Reeb! He would look at you and say: "One is a naive integer; two is a naive integer; three is a naive integer; and so on. Now, is every member of N a naive integer." And if you said so, "Really? What makes you so sure?" (The point being, "naive integers" need not make a set.) – Francois Ziegler Feb 14 '16 at 13:43
  • I was just pointing out that the way his comments at conferences are described is not particularly respectful. Reeb is certainly making an interesting point (if I didn't think so I wouldn't be interested in it enough to post the question in the first place). – Mikhail Katz Feb 14 '16 at 13:44