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I identify the natural number $0$ with the empty set $\emptyset$, $1$ with $S(0)$, $2$ with $S(1)$, etc, etc.

The axiom of infinity says $\exists x (\emptyset\in x\wedge \forall z\in x\space z\cup\{z\}\in x)$ and the Axiom schema of specification says $\forall y_0,...,y_n\exists x\forall z (z\in x\leftrightarrow (z\in y_0\wedge \phi(z,y_1,...,y_n)))$.

My question now: Why is there now a smallest element $x$ which can be identified with the natural numbers?

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  • $\begingroup$ @Alex: I agree that it doesn't merit both set theory tags; but at least one of them should be kept. This question is definitely set theoretic. It's also unclear why the elementary one is better than the non-elementary one; but I flipped an imaginary coin and it came out "elementary". $\endgroup$ – Asaf Karagila Apr 17 '14 at 1:33
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Let $y$ be an inductive set whose existence follows from the axiom of infinity, then consider $\{x\subseteq y\mid x\text{ is inductive}\}$. This is a definable collection of members of the power set of $y$, so it is a set, and $y$ is there so it's not an empty set.

Now take the intersection of all those sets. This is an inductive set as well (you have to prove this, of course). Call this inductive set $N$. Now prove that if $x$ is any inductive set then $N\subseteq x$, by considering $M=N\cap x$. Show that $M$ is inductive as well, and $M\subseteq y$, now use the property which defined $N$ to conclude $N=M$ and therefore $N\subseteq x$.

And now we're done.

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  • $\begingroup$ Thanks, its clear for me that the inetrsection of all sets is again an inductive set, but could you explain me the last part a little bit more in detail, that $N\subseteq x$? $\endgroup$ – Babla May 1 '13 at 15:58
  • $\begingroup$ If you already know that, then it's slightly easier, and I have updated my answer. $\endgroup$ – Asaf Karagila May 1 '13 at 16:03
  • $\begingroup$ There is one more question I have (it came to mind 1 minute ago): It is obvious that the set contains all natural numbers, but is there a simple argument why the set cannot contain more elements or is this a non-trivial problem? $\endgroup$ – Babla May 15 '13 at 20:15
  • $\begingroup$ @Babla: It's not a very difficult point. You just have to note that the set of natural numbers is itself inductive. Therefore this intersection is a subset of the natural numbers, therefore it must be equal to the natural numbers. $\endgroup$ – Asaf Karagila May 15 '13 at 21:34

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