In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 11 Apr., 02:02, Virgil <vir...@ligriv.com> wrote: > > > Technically, WH is not wrong, but quite correct, since induction can > > only prove that something is true for EACH MEMBER of the relevant > > inductive set, but does not prove it true true for that set itself. > > The proof holds for all members. Therefore for infinitely many > members. The set itself may do what matheologians, who think that it > differs from the collection of all its members, like.
Consider the infinite set of all FISONs, Finite Initial Sets Of Naturals. One can easily prove by induction that every member of that set is finite, so that by WM's argument the set of all FISONs is also finite, even though it can be proved not to be finite. > > > Given a set of natuals, E, such that 2 is a member of E and for each > > member m , m + 2 is also a member, and that E is a subset of any other > > set with that property, then one can correctly say that every member of > > that set is even, but one cannot properly claim that the set is even. > > Who did so?
WM tries to do so repeatedly. But when he tries to make it hold outside of Wolkenmuekenheim, he always fails. --