In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 9 Dez., 04:24, William Hughes <wpihug...@gmail.com> wrote: > > On Dec 8, 5:31 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 8 Dez., 13:39, William Hughes <wpihug...@gmail.com> wrote: > > > > > > On Dec 8, 4:33 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > <snip> > > > > > > > That requires, in the present case, to find two FISONs A, B > > > > > that are necessary: > > > > > > Piffle. You need an infinite *number* of FISONs. > > > > > 1) Every FISON is finite. > > > 2) There are never two FISONs that contain more than one of them. > > > > True but irrelevant. > > > > Let T be a non-empty sequence of FISONs with no last FISON. > > a) Then every element of T is a proper subset of some other element of > > T. > > But T has to contain at least one FISON because it is non-empty. > > > > T contains N. > > Every sensibly defined set of natural numbers or FISONs has a least > element. (The set |N is really well-ordered.)
Except the empty set!
> The set required to contain aleph_0 naturals has no least element.
If some fison has a least element that element is also the least element of the set of all naturals, at least outside of Wolkenmuekenheim.
Proof: all fisons have the same least element and every fison is a proper subset of the union of the set of all fisons,
> So > thge definition is meaningless. What can be meaningless in this > definition?
It has plenty of meaning outside of Wolkenmuekenheim, but unforunately, WM goes dim whenever he tries to step out of Wolkenmuekenheim. > > > > So despite the fact that T does not contain two "necessary" > > FISONs, T still contains FISONs. > > No. It is a meaningless definition
It means plenty outside of Wolkenmuekenheim.
> and an approach doomed to fail to > obtain aleph_0 naturals by unioning FISONs.
When one already has the set of all naturals naturally without ever resorting to WM's fisonsery, one can ignore fisonery and still have all the naturals in one set. --