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In this wikipedia article, natural numbers are extracted from the infinite set which includes all the natural numbers as $\forall n(n\in N \Leftrightarrow ([n=\emptyset \lor \exists k(n=k\cup \{k\})]\land \forall m\in n[m=\emptyset \lor \exists k\in n (m=k\cup \{k\})]))$.

Let the extracted set be $N'$ and $n_1\in N'$.

  1. If $n_1\neq \emptyset$, $\exists n_2(n_1=n_2\cup \{n_2\}).$
  2. If $n_2\neq \emptyset, \exists n_3\in n_1(n_2 = n_3\cup \{n_3\}).$
  3. If $n_3\neq \emptyset, \exists n_4\in n_1(n_3= n_4\cup \{n_4\}).$
  4. ...

How to prove that this can't continue infinitely?

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closed as unclear what you're asking by Andrés E. Caicedo, José Carlos Santos, Namaste, The Phenotype, GNUSupporter 8964民主女神 地下教會 Jan 25 '18 at 0:36

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  • $\begingroup$ Not clear... What are you trying ? The starting point is the def of Inductive set, i.e. a set that contains $0$ and, for very element $a$, contains also $a \cup \{ a \}$. The issue is that e.g. the set $I'$ that "starts" with $0$ and $\phi$ and is closed with respect to the operation $x \cup \{ x \}$ also satisfy the def. $\endgroup$ – Mauro ALLEGRANZA May 8 '17 at 7:58
  • $\begingroup$ The proposal is to carve out exactly the set $\mathbb N$; how to do this ? With a new def that says: $\mathbb N$ is the inductive set (and the only one, by extensionality) that contains only $0$ and it is closed with respect to the successor operation. In this way, unwanted elements (as $\phi$ above) cannot "crop in". $\endgroup$ – Mauro ALLEGRANZA May 8 '17 at 8:00
  • $\begingroup$ You cannot prove it, unless ZF is inconsistent, because otherwise there are models with "infinite" natural numbers, ones where the given process would keep on going indefinitely (you can prove it using the compactness theorem if you know ir). But you would still be able to do induction and so on. $\endgroup$ – Max May 8 '17 at 8:26
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The def'n in the article, and the "more formal" def'n immediately following it, are flawed. Without the axiom of Foundation (a.k.a Regularity) it is consistent that there exists $x$ with $x=\{x\}.$ It is even consistent that there is an uncountable non-empty set $S$ such that $\forall x\in S\;(x=\{x\}).$ But according to the def'ns in the article, any such $x$ belongs to $N.$

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    $\begingroup$ I have removed the flawed definition from the Wikipedia article. The version Daniel was commenting on here is en.wikipedia.org/w/… $\endgroup$ – Henning Makholm Jan 24 '18 at 21:32
  • $\begingroup$ ... but the removal was reverted. Sigh. $\endgroup$ – Henning Makholm Jan 25 '18 at 16:05
  • $\begingroup$ @HenningMakholm. Unfortunate. So my A still stands but I would rather see the removal removed. $\endgroup$ – DanielWainfleet Jan 27 '18 at 2:09