|
Re: Matheology 203
Posted:
Feb 7, 2013 3:54 AM
|
|
On 7 Feb., 09:27, Virgil <vir...@ligriv.com> wrote: > In article > <30e6f8dd-f487-4335-ba77-35f182b79...@e10g2000vbv.googlegroups.com>, > > > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > On 7 Feb., 08:53, Virgil <vir...@ligriv.com> wrote: > > > In article > > > <457d0429-33c1-46ad-9151-4f5d9dc96...@fv9g2000vbb.googlegroups.com>, > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > > > On 7 Feb., 05:21, fom <fomJ...@nyms.net> wrote: > > > > > > Or as Virgil would write > > > > > in a very lucid moment: > > > > > > What Cantor proved was that no list of accessible real numbers > > > > > (accessible because listable) can include all accessible numbers, > > > > > because any such list itself proves the existence of numbers not listed. > > > > > That is, Cantor proved the countable set of accessible numbers being > > > > uncoutable. > > > > That may well be WM's misunderstanding but it is not an understanding. > > > > A number being accessible does means that it can appear in some list, > > > but does not at all mean that all accessible numbers can appear together > > > in a single list. > > > All elements of countable sets can be counted by definition, i.e., > > they can appear in a list. > > But some subsets of a set may be countable even though the set itself is > not.
Of course, the algebraic real numbers for instance, or the definable real numbers.
> Certainly SOME sets of reals can be counted but not every set of > reals can be counted.
My question has been this: Why are there countable sets that are uncountable?
Regards, WM
|
|