9
$\begingroup$

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?

Improve this question
$\endgroup$
18
  • 4
    $\begingroup$ That's a great quote! $\endgroup$
    – goblin GONE
    Feb 14, 2016 at 9:35
  • 2
    $\begingroup$ 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. $\endgroup$
    – Dan Piponi
    Feb 21, 2016 at 15:20
  • 3
    $\begingroup$ @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. $\endgroup$
    – André Henriques
    Feb 21, 2016 at 15:20
  • 2
    $\begingroup$ 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! $\endgroup$
    – thedude
    Feb 21, 2016 at 16:06
  • 1
    $\begingroup$ @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. $\endgroup$
    – André Henriques
    Feb 21, 2016 at 16:55

1 Answer 1

Reset to default
4
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks, I thought it was perhaps one of the Dieners, good to know. $\endgroup$
    – Mikhail Katz
    Feb 14, 2016 at 9:46
  • $\begingroup$ I think your added comment is unnecessarily sarcastic. $\endgroup$
    – Mikhail Katz
    Feb 14, 2016 at 13:35
  • 3
    $\begingroup$ @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.) $\endgroup$
    – Francois Ziegler
    Feb 14, 2016 at 13:43
  • $\begingroup$ 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). $\endgroup$
    – Mikhail Katz
    Feb 14, 2016 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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