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

 Messages: [ Previous | Next ]
 Virgil Posts: 8,833 Registered: 1/6/11
Re: Uncountable Diagonal Problem
Posted: Jan 1, 2013 6:56 PM

In article
Graham Cooper <grahamcooper7@gmail.com> wrote:

> On Dec 31 2012, 9:27 am, Virgil <vir...@ligriv.com> wrote:
> > In article
> >  "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.
> >

>
>
> LETS TRY!
>
> LIST
> R1 0.11111111...
> R2 0.22222222...
> R3 0.01010101...
> R4 0.99999999...
> ...
>
>
> DIAGONAL = 0.1209....
>
> WHAT ARE ALL THE MISSING REALS VIRGIL?
>
>
> HINT: you should be able to calculate 9*9*9*9 of them?

Way more than that!

As long as the digit replacement rule does not replace any digit with
either a 0 or a 9, one can have as many as 8*7 = 56 different rules for
any digit position, giving 4*56^8 nonmembers of your list.

And if your 4 listed elements are as periodic as they appear to be,
there are uncountably many non-periodic others, though one cannot, of
course, list them all..
>
> JUST FROM THAT LIST!
>
> WOW! THERE REALLY ARE A LOT OF UNCOUNTABLE REALS!!
>
> Herc

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