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Re: Matheology § 162
Posted:
Nov 28, 2012 1:43 AM
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On 28 Nov., 00:03, William Hughes <wpihug...@gmail.com> wrote:
> On Nov 27, 5:17 pm, WM <mueck...@rz.fh-augsburg.de> wrote:> On 27 Nov., 21:48, William Hughes <wpihug...@gmail.com> wrote:
>
> <snip>
>
> > > > but that does not mean that all
> > > > positions are unoccupied:
>
> > > But this is not what I am saying.
>
> This is an important point. Clearly "each n in |N has
> property P" does not mean "all n in |N have property P"
> if we assume that "all n in |N" might not exist.
> However from "each n in |N has property P" we can
> conclude that no n in |N has property (not P).
But even this seemingly innocent conclusion can lead to
contradictions. Compare the sequence:
> > 01.
> > 0.1
> > 010.1
> > 01.01
> > 0101.01
> > 010.101
> > 01010.101
> > 0101.0101
> > ...
No term has more digits left than right. This includes, according to
analysis (potential infinity = always finite although not limited),
the limit.
>
> > > Compare
>
> > > Each natural number >1 is either prime or composite.
> > > The set of natural numbers >1 that are neither prime
> > > nor composite is empty.
>
> > Yes, but here we have a property that is a property of the set like:
> > Every natural number is a natural number.
>
> Induction works for any property P. E.g Each natural number
> is a natural number can be proved by induction.
>
> 1 is a natural number.
> If n is a natural number n+1 is a natural number
> Each n in |N is a natural number.
Induction proves also: Every set of natural numbers is finite.
Why do you overlook this simple proof?
>
> One gets the same "contradictions" if one uses potential
> infinity rather than actual infinity.
No. Only by actual infinity one is forced to accept that the limit of
my sequence has no digits (neither 1 nor 0) left of the point and thus
is less than 1.
Regards, WM
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