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Topic: Matheology � 300
Replies: 6   Last Post: Jul 14, 2013 6:01 PM

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Posts: 8,833
Registered: 1/6/11
Re: Matheology � 300
Posted: Jul 12, 2013 5:02 PM
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In article <>, wrote:

> On Friday, 12 July 2013 17:47:28 UTC+2, Zeit Geist wrote:
> > On Friday, July 12, 2013 6:46:42 AM UTC-7, wrote:
> > > On Friday, 12 July 2013 01:12:23 UTC+2, Zeit Geist wrote: > > >> > > > >
> > > >> Try to find some individuals that are not in one and the same line.
> > Fail. Recognize - or, most probably, not. > > > > > > > > Ok, you win. > >
> > > > There exist a line in the list, k, such that all n e |N, n e k. > > > >
> > No!!! It is simply absurd and stupid to talk about all n in |N.

> > You keeps saying there is line that they are all on.
> I say that all that can be used as individuals, must be in one and the same
> line.

But having any line as an individual line implies the existence of a an
individual natural NOT in that line.

So which line can be used as an individual line without thereby defining
(as an individual natural) the successor of its largest member?

> All that can be there, must be in one and the same line.

As soon as it is "there", so is the successor of its largest member and
the successor of the line's index number.

> But there are
> never more than finitely many, although their number is not limited.

Unless there are all natural numbers, induction cannot be said to prove
statements of the form: For all n in |N, f(n)
> > " Try to find some individuals that are not in one and the same line.
> > Fail."

> Obviously, you will fail.

The index of a line and the successor of that index are never both in
that indexed line.
> > If that does fail, then they must be all on one line.

WM's quantifier dyslexia strikes again.
For all P there is an L with P in L A(P) E
There is an L such that for all p p in L

TRUE:For every pair of naturals there is a line containing both.
FALSE:There is a line containing every pair of naturals.

> Therefore the set of aleph_0 natural numbers does not exist.

If it does not then inductive arguments are invalid.
"P(1) and for all n in |N, P(n) implies P(n+1)"
is no longer possible unless one can truthfully say
"for all n in |N"

Which WM now claims is not possible.

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