On 8 Feb., 23:05, Virgil <vir...@ligriv.com> wrote: > In article > <43d2d64e-7641-4f96-bbe6-59fe20991...@e11g2000vbv.googlegroups.com>, > > > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > On 8 Feb., 12:13, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote: > > > WM <mueck...@rz.fh-augsburg.de> writes: > > > > On 7 Feb., 20:17, William Hughes <wpihug...@gmail.com> wrote: > > > ... > > > >> In classical set theory the accessible numbers are listable > > > > >> Note from the Wikipedia quote > > > > >> > Constructively it is consistent to assert the > > > >> > subcountability of some uncountable collections > > > > > Of course, the intuitionists accepted this nonsense, perhaps forced by > > > > the matheologians. > > > > What a joker! > > > > You tell us that you do not know Brouwer's opinion on this question, > > > but here you are telling us what intuitionists accept. > > > I know Brouwer's opinion very well But I do not discuss with you about > > that opinionb because you turn every word in my mouth. > > That words seem to turn in WM's mouth does not man that anyone other > than WM himself is responsible for such turnings. > > > > WM is inconsistent. > > > > As for intuitionists being "forced" into taking up a > > > position inconsistent with classical mathematics by classical > > > mathematicians ... > > > a classic absurdity. > > > No. Hilbert fired Brouwer from his most prestigious position with the > > Annalen. > > How would that force Brouwer into taking up a position INCONSISTENT with > classical mathematics?
Vice versa. Brouwer had a position inconsistent with Hilbert's mathematics. Therefore he was fired.
> > Could an intelligent man or woman who observes that all levels of the > > Binary Tree are crossed by a finite number of distinct paths really > > believe that there are uncountably many, where uncountable means much > > more than infinitely many? > > While each "level" individually may be only finitely crossed, it is > wrong, or at least deliberately misleading, to say that all of > infinitely many of them are collectively only "finitely crossed".
It is simple to prove it in mathematics. Every term of the sequence 2^n is finite. And there is no infinite level. > > And uncountable does not mean more that infinitely many, but only more > that countably many. Infinite includes both countable and uncountable.
That is purest nonsense (what you talk as well as what you talk about).