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Re: Uncountable Diagonal Problem
Posted:
Jan 1, 2013 6:20 PM
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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?
JUST FROM THAT LIST!
WOW! THERE REALLY ARE A LOT OF UNCOUNTABLE REALS!!
Herc
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