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Topic: Matheology § 203
Replies: 202   Last Post: Feb 10, 2013 3:34 AM

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 Virgil Posts: 8,833 Registered: 1/6/11
Re: Matheology � 203
Posted: Jan 29, 2013 7:10 PM

In article
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 29 Jan., 09:54, William Hughes <wpihug...@gmail.com> wrote:
> > On Jan 29, 9:33 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >

> > > "All" and "every" in impredicative statements about infinite sets.
> >
> > > Consider the following statements:
> >
> > > A) For every natural number n, P(n) is true.
> > > B) There does not exist a natural number n such that P(n) is false.
> > > C) For all natural numbers P is true.

> >
> > > A implies B but A does not imply C.
> >
> > Which is the point.  Even though A
> > does not imply C we still have
> > A implies B.
> >
> > Let  L be a list
> >      d the antidiagonal of L
> >      P(n),  d does not equal the nth line of L
> >
> > We have (A)
> >
> >    For every natural number n, P(n) is true.
> >
> > This implies (B)
> >
> >   There does not exist a natural number n
> >   such that P(n) is false.
> >
> > In other words, there is no line of L that
> > is equal to d.

>
> And how can C be correct nevertheless? Because "For all" is

It is no contradictory if properly stated, analogously to A an B, as
C) For all natural numbers n, P(n) is true.

WM's misrepresentation of how statements C should be expressed to be
comparable to A and B shows how dishonest WM really is, in trying to
sell us his gold bricks and Brooklyn Bridges.
>
> There is no natural number that finishes the set N.

That |N is is well ordered but without a last element in that ordering
is what distingishes it from its FISONs.

> There is no finished set N.

There is everywhere outside of Wolkenmuekenheim.
>
> There is, in the list of all reminating decimals, no anti-diagonal,
> that differs from all terminatig decimals at digits belonging to at
> least one of these terminating decimals. Reason: The list is complete.

There is no terminating decimal that differs from a complet list f al
all terminating decimals but for every list of not necessarily
terminating decimals the is a not-necessarily terminating decimal not in
that list.

> Again, the only solution is, there is no complete set Q.

Perhaps not in Wolkenmuekenheim, but there is in standard analysis,
along with a complete set of reals, and loads of other complete sets
that Wolkenmuekenheim has exiled.
>
> There is, in the construction of the complete Binary Tree, no node
> that adds more than one path to the tree. Nevertheless the completely
> constructed tree contains uncountably many paths. No reason to be
> taken aback, at least a little bit?

There is, in the construction of the complete set or decimals for reals
between 0 and 1, no last nonzero digit that adds more than one number to
the set. Nevertheless the completely constructed set of posbily infintie
decimals contains uncountably many reals.
At lest outside of Wolkenmuekenheim.
>
> Nevertheless, the steps of construction can be enumerated and
> therefore can be considered as a list. In no line you find any
> infinite path. But the complete list contains uncountably many
> infinite paths

Yup! Infinitely many finite steps can produce more than finitely many
finite steps can produce.
--

Date Subject Author
1/29/13 mueckenh@rz.fh-augsburg.de
1/29/13 William Hughes
1/29/13 mueckenh@rz.fh-augsburg.de
1/29/13 William Hughes
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1/29/13 mueckenh@rz.fh-augsburg.de
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