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For a system to be Peano's you requite 3 things: 1) First element isn't a succesor of any other element 2) "Successor" function is injective. 3) If $A \subseteq P$, First element is in $A$, and $S(A)\subseteq A \Rightarrow A=P$

But i've got a sucesor function which makes impossible to find this $A$ set. The fucntion is $S:P\rightarrow P$.

$S(a) = -a$ with $a>0$

$S(a)=-(a-1)$ with $a<0$

$S(a)=1$ for a=0

P is the set of integers, also the first element of my system is $0\in P$ (not $1$)

José Osorio
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  • I don't understand - what do you mean by "impossible to find this set $A$"? The point is that if $A$ is such a set, then $A$ is in fact all of $P$ (and in fact this is true of your example!). – Noah Schweber Jun 06 '16 at 23:12
  • I understand now! thank you. In my first try i didn't have this doubt, so I tried to use an $A$ set, but I coulnd't prove $P\subseteq A$, can you give me a hint? – José Osorio Jun 06 '16 at 23:19
  • Suppose $A\subseteq P, 0\in A$, and $S(A)\subseteq A$. Do you see why $3\in A$? How about why $-4\in A$? Once you see the picture, it's an induction argument (or you can build an explicit isomorphism with $\mathbb{N}$). – Noah Schweber Jun 06 '16 at 23:20
  • woah i see it now, so for any $x\in P$, there is a line of succesors which lead to $0$, which we know it's an element of $A$. Since the third requirement (called induction principle in my book) is the only thing you have for now: Can I actually use induction to prove this statement? – José Osorio Jun 06 '16 at 23:31
  • You can't use induction for this $P$, since you're trying to prove that; but presumably you already know induction for $\mathbb{N}$. That's what you can use. – Noah Schweber Jun 06 '16 at 23:33
  • Introducing a notation: Let $S^3(0)=S(S(S(0)))$ and so on... so if $x=S(0)$, then $x\in A$. (for the third condition) supose it holds for $k, x=S^k(0)$, it holds for k+1 since $S^k(0)=x\in A$.

    Hence it works for any n, now we can supose $x=S^h(0)$, for an h in $\mathbb{N} $.

     – José Osorio Jun 06 '16 at 23:44
  • BUt i think the real proof there is to prove that S(a) is an integer! – José Osorio Jun 07 '16 at 00:03

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