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Topic: Uncountable Diagonal Problem
Replies: 52   Last Post: Jan 6, 2013 2:43 PM

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 ross.finlayson@gmail.com Posts: 2,720 Registered: 2/15/09
Re: Uncountable Diagonal Problem
Posted: Dec 30, 2012 9:50 PM
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On Dec 30, 6:01 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> <b933563c-4654-4759-b964-c3cd27e0a...@lb9g2000pbb.googlegroups.com>,
>  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
>
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>
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>

> > On Dec 30, 3:27 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <4036660e-9527-479d-9c47-a1adf9d34...@px4g2000pbc.googlegroups.com>,
> > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:

>
> > > > On Dec 30, 1:33 pm, Virgil <vir...@ligriv.com> wrote:
> > > > > In article
> > > > > <2fc759b9-3c22-4f0b-83e0-bf9814a3f...@y5g2000pbi.googlegroups.com>,
> > > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:

>
> > > > > > Formulate Cantor's nested intervals with "mega-sequences" (or
> > > > > > transfinite sequence or ordinal-indexed sequence) instead of sequences
> > > > > > of endpoints. Well-order the reals and apply, that the sequences
> > > > > > converge yet have not emptiness between them else there would be two
> > > > > > contiguous points, in the linear continuum.

>
> > > > > Not possible with the standard reals without violating such properties
> > > > > of the reals as the LUB and GLB properties:
> > > > > Every non-empty set of reals bounded above has a real number LUB.
> > > > > Every non-empty set of reals bounded below has a real number GLB.
> > > > > --

>
> > > > Those are definitions, not derived. Maybe they're "wrong", of the
> > > > true nature of the continuum.

>
> > > if false for your "continuum" then that continuum is not the standard
> > > real number field.

>
> > > > A well ordering of the reals doesn't have uncountably many points in
> > > > their natural order.

>
> > > But, if one could find an explicit well-ordering of the reals, it would
> > > have to contain all those uncountably many reals in SOME order.
> > > --

>
> > So, the mega-sequences of the nested interval endpoints would end with
> > side-by-side endpoints?  Or, does any ordinally-indexed sequence of
> > all of a segment of reals necessarily contain duplicates?

>
> I see no reason why either need occur even in an explicit well-ordering
> of the reals. Why do you?
> --

This is from that, the constructed sequences (here ordinally-valued)
of nesting endpoints, of an interval, either meet or don't. If they
meet, they're either contiguous on the line, or identical. If they
don't, then the result is that there is an unmapped element, beyond
the sequence, or: Cantor's first, extended to the transfinite.

That's why - is that clear enough?

One of:
a) nesting leaves an empty interval (set between two points or two
copies of a point), and the mapping is onto, or
b) it doesn't, and there's an unmapped element to the sequence

ZFC has that the reals are well-orderable.

Simply in the course of passage of transfinite induction, the next
lower endpoint is the next in the well-order that is between the last
endpoints, as is the next upper. If a countable ordinal's worth can
have that intersection be empty, can it? Are there not points between
any two points in the Archimedean complete ordered field?

There are quite long discussions on this some years ago, in a variety
of details on structural consequences of well-ordering the reals vis-a-
vis Cantor's first.

http://groups.google.com/groups/search?hl=en&q=well-order+%22nested+intervals%22+group:sci.*+author:Finlayson

So, well-order them. Are there not enough ordinals to map to them in
order? Because, each subset of the reals is well-orderable, too.

Regards,

Ross Finlayson

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