Are elements not "reachable" from 0 prevented in Peano Arithmetic? 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.
 A: 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.
A: 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 $n$.  
