```Date: Jan 1, 2013 11:19 PM
Author: ross.finlayson@gmail.com
Subject: Re: Uncountable Diagonal Problem

On Jan 1, 7:29 pm, Virgil <vir...@ligriv.com> wrote:> In article> <7b68ec0a-22f7-4723-8f68-45bdcc5ff...@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 well-ordering 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 well-order'smapping is not monotone in the normal ordering of the reals, thenested intervals have countably many nesting, that converge.This is where, appending or concatenating well-orderings of disjointsets, is a well-ordering of their union.  The real line in ZFC can'tbe broken into uncountably many disjoint segments, each would containa rational.  Yet, in ZFC it has uncountably many points.  The realline has only countably many disjoint segments.That where transfinite recursion would not allow the axiom of choiceto apply to each subset of irrationals between zero and one:  becausethen for each irrational, there are uncountably many irrationals lessthan it in the normal ordering, as there is for each of those courtesydensity of irrationals, there can only be countably many nestedintervals, but there are uncountably many points to be endpoints ofthose intervals.Closed intervals are defined by their endpoints, sets are defined bytheir elements.  Then, there couldn't be uncountably many differentclosed intervals, because any subset of them if disjoint would besegments and if not disjoint would be nested or have a non-emptyintersection, where only countably many could be the other.  In ZFC,there are uncountably many irrational points, each possibly anendpoint of a closed interval.Then, the above note describes, and proves, due the density of therationals, there is at least one for each of their disjoint in thereals, then here, the consideration is that due the uncountability ofthe irrationals, there would be uncountably many segments or nestings,yet each of those, due the density of the rationals, has a uniquerational.Regards,Ross Finlayson
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