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Topic: Matheology § 233
Replies: 20   Last Post: Apr 4, 2013 10:16 AM

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Virgil

Posts: 9,012
Registered: 1/6/11
Re: Matheology � 233
Posted: Apr 2, 2013 5:46 PM
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In article
<7e6e8c91-ba4b-47fe-adcb-08fca8118d3f@y9g2000vbb.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 2 Apr., 02:06, Virgil <vir...@ligriv.com> wrote:
>

> >
> > > Nevertheless Cantor has given a finite formula to construct the list
> > > of all rationals between 0 and 1. From that formula we can find every
> > > entry and the anti-diagonal up to every digit d_n. From that we can
> > > easily prove that for every FIS d_1, d_2, ..., d_n of d there are
> > > infinitely many rationals with the same FISs. For every finite number
> > > n - and there are no other lines than such enumerated with a finite
> > > number!

> >
> > But as there is no end to the set/list of natural/finite numbers, there
> > is also no end to the set/list of such lines.

>
> What shall that "argument" be good for?


WM is continually claiming a "last" natural must always exist, even
though it is always also indeterminate, but now any of his arguments
that require such a last natural automatically fail.




> Why should there be an end?

That's what we keep asking WM.

> If I say every natural number is divisible by 1 without remainder.
> Would you doubt that on the grounds that there is no end to the
> sequence of natural numbers? Such that possibly a natural number would
> follow that is not divisible by 1?


WM has long been claiming that every and any set of naturals must have
an inaccessible last member, i.e., that every set of naturals is
necessarily a subset of a FISON (finite initial set of naturals),
just as he always claims that in a Complete Infinite Binary Tree every
path is a FISON (finite initial sequence of nodes), but he is wrong on
both counts..
--





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