
Re: Uncountable Diagonal Problem
Posted:
Jan 3, 2013 12:07 PM


On Jan 2, 12:48 am, Virgil <vir...@ligriv.com> wrote: > In article > <de9ee3af08234a9982167b6033235...@po6g2000pbb.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > > > > > On Jan 1, 11:22 pm, Virgil <vir...@ligriv.com> wrote: > > > In article > > > <ef09c567163746b8932abcb856e41...@r10g2000pbd.googlegroups.com>, > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > On Jan 1, 8:59 pm, Virgil <vir...@ligriv.com> wrote: > > > > > In article > > > > > <5e016173aa1b48349d700c6b08f19...@jl13g2000pbb.googlegroups. > > > > > com>, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > On Jan 1, 7:29 pm, Virgil <vir...@ligriv.com> wrote: > > > > > > > In article But in that proof Cantor does not require a well > > > > > > > ordering of the reals, only an arbitrary sequence of reals > > > > > > > which he shown cannot to be all of them, thus no such > > > > > > > "counting" or sequence of some reals can be a count or > > > > > > > sequnce of all of them.  > > > > > > > Basically > > > > > > Nonsense deleted!  > > > > > Nonsense deleted, yours? > > > > Nope!  > > > Great: from demurral to denial. >
Seems clear enough: in ZFC, there are uncountably many irrationals, each of which is an endpoint of a closed interval with zero. And, they nest. Yet, there aren't uncountably many nested intervals, as each would contain a rational. To whit: in ZFC there are and there aren't uncountably many intervals. Then, with regards to Cantor's first for the wellordering of the reals instead of mapping to a countable ordinal, there are only countably many nestings in as to where then, the gap is plugged (or there'd be uncountably many nestings). Then, due properties of a well ordering and of sets defined by their elements and not at all by their order in ZFC, the plug can be thrown to the end of the ordering, the resulting ordering is a wellordering. Ah, then the nesting would still only be countable, until the plug was eventually reached, but, then that gets into why the plug couldn't be arrived at at a countable ordinal. Where it could be, then the countable intersection would be empty, but, that doesn't uphold Cantor's first proper, only as to the finite, not the countable. So, the plug is always at an uncountable ordinal, in a wellordering of the reals. (Because otherwise it would plug the gap in the countable and Cantor's first wouldn't hold.)
Then, that's to strike this: "So, there couldn't be uncountably many nestings of the interval, it must be countable as there would be rationals between each of those. Yet, then the gap is plugged in the countable: for any possible value that it could be. This is where, there aren't uncountably many limits that could be reached, that each could be tossed to the end of the wellordering that the nestings would be uncountable. Then there are only countably many limit points as converging nested intervals, but, that doesn't correspond that there would be uncountably many limit points in the reals. " Basically that the the gap _isn't_ plugged in the countable.
Then, there are uncountably many nested intervals bounded by irrationals, and there aren't.
Regards,
Ross Finlayson

