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Edit : I think that this article is the answer i was waiting for, i stubble upon it by chance while investigating automatic proof verification and lambda calculus : http://plato.stanford.edu/entries/mathematics-constructive/

Original post :

Due to negative feed back, as of 15/04/2016, i delete the original post, only leaving collected links and article. Thank you for reading, i guess mathematic community is a closed one, not very open to self learners.

edit : I found this related question Proving that the set of natural numbers is well-ordered

edit : another related question : Proving the so-called "Well Ordering Principle"

edit : here is another one, that point towoard Z being suffiscient Prove the principle of mathematical induction in $\sf ZFC $

edit : here is a discussion about constructives approaches https://mathoverflow.net/questions/30643/are-real-numbers-countable-in-constructive-mathematics

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user3286435
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  • What theory adds an axiom declaring that any infinite set of integers is enumerable? No theory does that, none has to: it's a theorem of ZFC. Of course the enumeration won't in general be computable. Of course it's false that "any subset of the integers is recursively enumerable". The article doesn't state that at all, not even "somehow". – BrianO Apr 12 '16 at 06:14
  • Your algorithm relies on the existence of the order restricted to the set you want to enumerate. I doesn't help that you know the order on all the naturals. At any step you don't want to list the least element not listed thus far, you want to list the least element in your subset that hasn't been listed thus far. – Stefan Mesken Apr 12 '16 at 08:02
  • In Peano arithmetic (PA), the induction axiom is equivalent to the principle you cite, but first-order PA doesn't (can't) talk about all subsets of $\Bbb N$, only about each subset definable in PA (by a formula). In ZFC, however, the full statement is simply a theorem: the integers are well-ordered. In fact I don't think choice is needed; ZF should prove this, thanks to Regularity. Any subset of a well-ordered set is itself well-ordered (by the same/superset's ordering, restricted to the subset). – BrianO Apr 12 '16 at 11:50
  • http://plato.stanford.edu/entries/mathematics-constructive/ I think my only problem is that ZFC is overwhelming in all google search. There exists alternatives, but it seems that for some reason very few people have the ability to hear about them. – user3286435 Apr 14 '16 at 18:07
  • Also i found an interesting discussion here : http://mathoverflow.net/questions/30643/are-real-numbers-countable-in-constructive-mathematics – user3286435 Apr 14 '16 at 18:15

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You are conflating recursively enumerable (r.e.) and enumerable (countable). There are lots of sets which are countable but not r.e. - for example (as you mention) the set of indices of Turing machines which do not halt on input 0 (say). This set is countable, even though it is not r.e. You write e.g.

There is no bijection from the set of all N, and . . . any non recursively enumerable set

and cite the diagonal argument as proof - however, all a diagonal construction shows is that there is no recursive bijection between $\mathbb{N}$ and the set of all programs that do not terminate. There are lots of such bijections, they just aren't recursive.

I suspect that you are implicitly assuming some principle like "every thing which exists is recursive." While philosophically interesting, this is deeply at odds with classical mathematics, which indeed proves that there is a rich universe of non-recursive entities.

Noah Schweber
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