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Recall that if x is a set, S(x) is defined to be the set $x\cup\{x\}$. A set x is called inductive if $0\in x$ and for all $y\in x$, $S(y)\in x$.

Lemma. If x is a set of inductive sets then $\cap_{y\in xy}$ is an inductive set.

Axiom. Inductive Sets. There is an inductive set.

Theorem. There is a smallest inductive set. In other words, there is an inductive set which is a subset of all inductive sets. This smallest inductive set is (Why?) the intersection of all inductive sets. -

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  • $\begingroup$ Assume not. What does happen? $\endgroup$ – N.Raki Nov 24 '15 at 12:17
  • $\begingroup$ I would rather call this a part of the definition of $\mathbb N$ (containing $0$). Let $a$ be some inductive set. Then the set of natural numbers is defined as: $\{x\in a: x\text{ is element of every inductive set}\}$. This set is inductive and is a subset of each inductive set. So it deserves the label "smallest inductive set". Here a set $a$ is "inductive" if $\varnothing\in a$ and $x\in a\implies x\cup\{x\}\in a$. $\endgroup$ – drhab Nov 24 '15 at 12:20
  • $\begingroup$ Recall that if x is a set, S(x) is defined to be the set x ∪ {x}. A set x is called inductive if 0 ∈ x and for all y ∈ x, S(y) ∈ x. Lemma. If x is a set of inductive sets then ∩y∈xy is an inductive set. Axiom. Inductive Sets. There is an inductive set. Theorem. There is a smallest inductive set. In other words, there is an inductive set which is a subset of all inductive sets. This smallest inductive set is (Why?) the intersection of all inductive sets. $\endgroup$ – N.Raki Nov 24 '15 at 12:29
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An inductive set must have the smallest element ($0$ or $1$, depending whether $\mathbb N$ starts with $0$ or $1$) and with any element it must contain its successor.

Hence, the set must contain $\mathbb N$.

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  • $\begingroup$ Hah! Yes, thanks. $\endgroup$ – N.Raki Nov 24 '15 at 12:31
  • $\begingroup$ "Hence, the set must contain $\mathbb N$". This conclusion can only be drawn on base of some definition of $\mathbb N$. I have the feeling that we are going in circles here. $\endgroup$ – drhab Nov 24 '15 at 12:48

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