Virgil
Posts:
4,482
Registered:
1/6/11
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Re: Matheology � 210
Posted:
Feb 7, 2013 7:33 PM
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In article
<99dea8f7-e1b2-49e0-9b09-88683dba6730@z9g2000vbx.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 7 Feb., 19:14, William Hughes <wpihug...@gmail.com> wrote:
> > On Feb 7, 5:57 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 7 Feb., 15:56, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Feb 7, 3:25 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > <snip>
> >
> > > > >... a subset S of the countable set F of finite words bijects with
> > > > > the set D of definable numbers
> >
> > > by definition.
> >
> > > > Nope. Every D corresponds to some finite word.
> >
> > > No, D is a set or at least a collection. A definable number is an
> > > element of D.
> >
> > > > However, S,
> > > > the collection of all the correspondences, may not be a subset
> > > > of F (subsets must be computable).
> >
> > > Need not be a subset. It is sufficient to know that there are not more
> > > than countably many correspondences,
> >
> > There is no set of correspondences thus there is no number
> > of correspondences. You cannot know anything about
> > the number of correspondences.-
>
> You are in error again. There is the axiom of power set. For any F,
> there is P such that D e P if and only if D c F. According to it every
> subset of the countable set F exists. Will you dispute that the finite
> definitions of numbers are a subset of F? Not even the axiom of choice
> is required to prove that.
A mere set of words, without even a specified order or grammar is not a
definition of anything. Thus your claim fails of its own idiocy.
>
> Regards, WM
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