Date: Jan 2, 2013 3:37 AM
Subject: Re: Uncountable Diagonal Problem
On Jan 1, 11:22 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
> > On Jan 1, 8:59 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <5e016173-aa1b-4834-9d70-0c6b08f19...@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?
Great: from demurral to denial.
There are uncountably many intervals in the reals, that intersect or
don't, each contains rationals.
In ZFC, how many intervals in the reals are there? Uncountably many.
How many disjoint intervals? Only countably many.
How many nested? Only countably many.
Intervals are disjoint, nested, or intersecting.
Then there are uncountably many intersecting intervals, but their
intersections nest in them and their disjoints are disjoint. For each
intersection there's a nest, and a disjoint.
So, in ZFC, how many intervals in the reals are there? Any two points
define an interval, and in ZFC, there are uncountably many of those.
Disjoint or nested Intervals in the reals inject into the rationals.