In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 26 Mrz., 22:07, Virgil <vir...@ligriv.com> wrote: > > > > > One acceptable form of induction is: > > > > > > There exists a set of objects, N, and a zero object, 0, such that > > > > 1. 0 is a member of N. > > > > 2. Every member of N has a successor object in N. > > > > 3. 0 is not the successor object of any object in N. > > > > 4. If the successors of two objects in N are the same, > > > > then the two original objects are the same. > > > > 5. If a set, S, contains 0 and the successor object of every > > > > object in S, then S contains N as a subset. > > > > > That is a definition of a sequence, not a proof by induction. It is > > > not even a definition of the natural numbers, because even the ordered > > > set > > > N = (0, pi, pi^2, pi^3, ...) > > > obeys your five points. > > > > Mathematical Induction does not require use of natural numbers, but only > > of a set which is as inductive as (is order-isomorphic to) the set of > > natural numbers, and my form satisfies that requirement. > > Your mess is not induction and is obviously not a proof. A proof by > induction runs as I said:
That WM said something is very nearly proof that it is false.
And induction requires only an inductive set, as I described above, which need not be the set of positive naturals. --