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I am having a discussion about Peano Arithmetic and I was told that there is actually nothing in Peano Arithmetic to prevent one from defining an element a with S^k(a)=a.
I am only able to derive that a is not S^j(0) for any j, but I don't see how that is a problem for introducing a in the first place.
If introducing those elements is actually possible, there are some neat properties with those.

Matthew Conroy
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Thomas Bartscher
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    Nothing except the induction axiom scheme. – André Nicolas Dec 15 '13 at 22:42
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    As to non-reachability, which is very different from invariant under some iterated successor, certainly the axioms of first-order PA have models that, beside the ordinary natural numbers, have "non-standard" elements. – André Nicolas Dec 15 '13 at 22:56
  • Ah, good, those "non-standard" elements I am interested in. Are there any sources on that? I was told about one such element a for which S(a)=a that can be used as one would expect infinity to work. – Thomas Bartscher Dec 15 '13 at 23:02
  • That is second-order PA. Second-order PA is categorical, it has up to isomorphism only one model, so the comment above does not apply to it. – André Nicolas Dec 15 '13 at 23:03
  • We cannot get $Sa=a$ holding in any model of PA (either version). For "non-standard models of arithmetic" Google that and relatives, you will get lots of hits. – André Nicolas Dec 15 '13 at 23:06
  • Why? I don't see why S(a)=a should be forbidden. I mean, which axioms does it contradict and how? Is there a proof for ￢(S(a)=a) in first-order PA? – Thomas Bartscher Dec 15 '13 at 23:15
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    Yes, there is a proof. The proof given in the answer by universal set can be adapted to show that for any fixed positive integer $k$, $S^k(a)=a$ has no solution. That's beasically because for any fixed $k$, the assertion of non-solvability can be stated as a sentence of first-order PA, and first-order induction gives a proof. – André Nicolas Dec 15 '13 at 23:23

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To prove $\lnot \exists a S(a)=a$ you appeal to the induction hypothesis. We have $\lnot S(0)=0$ Then assuming $\lnot S(n)=n$ we have $\lnot S(S(n))=S(n)$ From that we conclude $\forall n \lnot S(n)=n$ This is typical of the non-standard elements. Any property shared by all the standard elements (or even all sufficiently large ones) is shared by all the non-standards. This style of proof shows that.

Ross Millikan
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It's probably worth expanding on André Nicolas's comment above. For any (standard) natural number $k>0$, one can show in Peano Arithmetic that there is no element $a$ such that $S^k(a) = a$. Why? Certainly this is true for $a=0$, since $0$ is not a successor, and since successor is injective, if $S^k(b) \neq b$, then $S^k(S(b)) = S(S^k(b)) \neq S(b)$, so we have a proof that if $b$ satisfies $\phi(x) = (x \neq S^k(x))$ then $S(b)$ does as well. This, of course, is exactly what is needed to apply the induction axiom for $\phi$ and hence there is no such $a$ with $S^k(a) = a$.

There are, however, plenty of non-standard models of Peano Arithmetic; in such models there are "unreachable" elements, that is, elements $a$ such that $a$ is not $S^n(0)$ for any (standard) natural number $n$.

universalset
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  • Well, it is exactly these non-standard models I am interested in. I wrote: "I am only able to derive that a is not S^j(0) for any j, but I don't see how that is a problem for introducing a in the first place". – Thomas Bartscher Dec 15 '13 at 23:21
  • The proof above (in the first paragraph) is a proof in Peano Arithmetic and does not rely on being a standard model; it applies just as well to non-standard models of PA. – universalset Dec 15 '13 at 23:23