Virgil
Posts:
8,833
Registered:
1/6/11


Re: Uncountable Diagonal Problem: Ross is WRONG, again!
Posted:
Jan 3, 2013 10:02 PM


In article <b7e06477b83641b1be03c4d0fe3c29c9@q16g2000pbt.googlegroups.com>, "Ross A. Finlayson" <ross.finlayson@gmail.com> wrote:
> On Jan 3, 9:07 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> > wrote: > > 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.
Yes there are, as I pointed out in a posting that Ross has carefully snipped entirely.
The set of intervals { [x,x] : x is a positive irrational} is one such set of uncountably many nested intervals bounded by irrationals.
A simple, and obvious, example of what Ross claims does not exist. > > > > Point being there are uncountably many disjoint intervals defined by > the irrationals of [0,1]: each nonempty disjoint interval contains a > distinct rational. Thus, a function injects the irrationals into a > subset of the rationals.
This too is false. { [x,1x] : x is an irrational between 0 and 1/2} being an explicit counterexample. And as there are way more such intervals than rationals in their union, no such injection from intervals as Ross claims to rationals can exist.
And Ross is totally wrong again!!!
And Ross will, no doubt, snip all of this proof of his errors too, just as he did the last one, if he repies at all. 

