
Re: Uncountable Diagonal Problem
Posted:
Jan 1, 2013 8:08 PM


On Jan 1, 4:05 pm, Virgil <vir...@ligriv.com> wrote: > In article > <d0cf5fff92d84229aec3499754ae6...@r10g2000pbd.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > > > > > On Jan 1, 2:57 pm, Virgil <vir...@ligriv.com> wrote: > > > In article > > > <659c05ff5b344ebe96174d54292a9...@pp8g2000pbb.googlegroups.com>, > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > The transfinite courseofpassage in wellordering the reals sees > > > > a diminishing interval. Do the endpoints of the interval meet, > > > > in the wellordering? A critical point of Cantor's first is that > > > > the intersection is nonempty. > > > > It is a well known property of the real number line, at least among > > > Mathematicians, that a nested sequence of closed intervals has > > > nonempty Intersection. > > > > Does Ross claim otherwise?  > > >http://www.tikilounge.com/~raf/finlayson_injectrationals.pdf > > That paper, even if it were valid, would not invalidate that a nested > sequence of closed intervals in R necessarily has nonempty intersection. > 
So, is the intersection in the proof for the first statement non empty? That is the emphasized note starting "It is so...".
Here, the consideration is of the nested intervals, formed a la Cantor's first, from a wellordering of the reals. From limit ordinal to limit ordinal, the endpoints may converge. The intervals as you claim would be nonempty. Then there are two cases with no duplicates in the range, a) the interval is degenerate and a later value plugs the gap, b) the interval is standardly realvalued in measure and only some later value, beyond the _next_ limit ordinal, fills all the gap. Yet, it is never complete until the gap is plugged.
So, how can the gap ever be plugged? There is the dichotomy that the convergent sequences from above and below have values closer to the limit than each epsilon, but that there is an epsilon for the limit bounding an empty neighborhod for each value in the sequence.
The sequences can converge in the countable, and be plugged at a consequent countable limit ordinal. The only reason they don't is as per "uncountability".
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.
So, you agree the intersection of nondisjoint closed intervals is: nonempty?
Regards,
Ross Finlayson

