
Re: Uncountable Diagonal Problem
Posted:
Jan 1, 2013 11:19 PM


On Jan 1, 7:29 pm, Virgil <vir...@ligriv.com> wrote: > In article > <7b68ec0a22f747238f6845bdcc5ff...@gg5g2000pbc.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > Here, the consideration is of the nested intervals, formed a la > > Cantor's first, from a wellordering of the reals. > > 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 within some countably many points, where the wellorder's mapping is not monotone in the normal ordering of the reals, the nested intervals have countably many nesting, that converge.
This is where, appending or concatenating wellorderings of disjoint sets, is a wellordering of their union. The real line in ZFC can't be broken into uncountably many disjoint segments, each would contain a rational. Yet, in ZFC it has uncountably many points. The real line has only countably many disjoint segments.
That where transfinite recursion would not allow the axiom of choice to apply to each subset of irrationals between zero and one: because then for each irrational, there are uncountably many irrationals less than it in the normal ordering, as there is for each of those courtesy density of irrationals, there can only be countably many nested intervals, but there are uncountably many points to be endpoints of those intervals.
Closed intervals are defined by their endpoints, sets are defined by their elements. Then, there couldn't be uncountably many different closed intervals, because any subset of them if disjoint would be segments and if not disjoint would be nested or have a nonempty intersection, where only countably many could be the other. In ZFC, there are uncountably many irrational points, each possibly an endpoint of a closed interval.
Then, the above note describes, and proves, due the density of the rationals, there is at least one for each of their disjoint in the reals, then here, the consideration is that due the uncountability of the irrationals, there would be uncountably many segments or nestings, yet each of those, due the density of the rationals, has a unique rational.
Regards,
Ross Finlayson

