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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?

Andrés E. Caicedo
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  • 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. – Mauro ALLEGRANZA May 08 '17 at 07:58
  • 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". – Mauro ALLEGRANZA May 08 '17 at 08:00
  • 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. – Maxime Ramzi May 08 '17 at 08:26

1 Answers1

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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 $N.$

DanielWainfleet
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