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Topic: Matheology § 210
Replies: 24   Last Post: Feb 12, 2013 1:12 PM

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mueckenh@rz.fh-augsburg.de

Posts: 16,037
Registered: 1/29/05
Re: Matheology § 210
Posted: Feb 8, 2013 5:28 PM
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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).

Regards, WQM



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