Date: Jan 1, 2013 6:56 PM Author: Virgil Subject: Re: Uncountable Diagonal Problem In article

<8733dfe2-163a-4e34-b402-2f018fcac874@i2g2000pbi.googlegroups.com>,

Graham Cooper <grahamcooper7@gmail.com> wrote:

> On Dec 31 2012, 9:27 am, 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.

> >

>

>

> 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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