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I'm trying to find $\mathbb N$ in ZFC.

I understand that the axiom of infinity guarantees the existence of an inductive set which has no inductive proper subsets (a set $A$ is inductive if $\emptyset \in A$ and $n\in A \implies n \cup \{n\}\in A$). I think this could be defined to be $\mathbb N$. But now I would like to prove Peano axioms. Namely that there exists $0\in \mathbb N$ and $\sigma\colon \mathbb N\to \mathbb N$ such that:

  1. $\sigma(x)=\sigma(y)\implies x=y$;
  2. $\sigma(x) \neq 0$;
  3. $(A \subset \mathbb N, 0\in A, n \in A \implies \sigma(n)\in A) \implies A = \mathbb N$.

I define $0=\emptyset$. I can then define $\sigma=\{(n, m) \in \mathbb N \times \mathbb N\colon m= n \cup \{n\} \}$. I would like to prove that $\sigma$ is an injective function. I failed at this. I think I first need to prove that given $n,m\in \mathbb N$ then $n\subset m \lor m \subset n$. Is this the correct path? How do I prove the seemingly simple fact that $\subset$ is a total order on $\mathbb N$?

Emanuele Paolini
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2 Answers2

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There's a more direct route to showing that $\sigma$ is injective - not just on $\mathbb{N}$, but on all sets.

Suppose $n\cup\{n\}=m\cup\{m\}$. Then:

  • $n\in m\cup\{m\}$, so either $n=m$ or $n\in m$.

  • Similarly, either $m=n$ or $m\in n$.

Assuming $m\not=n$, this means that $m\in n$ and $n\in m$; this contradicts Foundation.

Noah Schweber
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  • Actually foundation is not needed. – drhab Sep 23 '17 at 14:37
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    @drhab To argue that $\sigma$ is injective on all sets, it is: without foundation we can have distinct sets $a, b$ with $a={b}$ and $b={a}$ - then $\sigma(a)=\sigma(b)={a, b}$. To show that $\sigma$ is injective on $\mathbb{N}$ specifically, we don't need foundation, but why not prove the stronger result? – Noah Schweber Sep 23 '17 at 14:39
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It is direct that:$$\sigma(n)=\sigma(m)\wedge n\neq m\implies n\in m\wedge m\in n\tag1$$

The correct path now is to prove that every $n\in\mathbb N$ is transitive.

For that define: $$T:=\{n\in\mathbb N\mid n \text{ is transitive}\}$$ and prove that it is inductive, so that $T=\mathbb N$.

Then $n\in m\wedge m\in n$ implies that $n\subseteq m\wedge m\subseteq n$ hence $n=m$.

drhab
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  • You don't even need to show that every element of $\mathbb{N}$ is transitive for this problem, you can just argue directly through Foundation. – Noah Schweber Sep 23 '17 at 14:37
  • @NoahSchweber Transitivity of naturals will come along once anyway, so why not doing it before injectivity of $\sigma$. Also it can happen that foundation comes in later (that was my route in sets). Finally: it it is always good to know what axioms can or cannot be missed. – drhab Sep 23 '17 at 14:47