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Topic: Uncountable Diagonal Problem
Replies: 52   Last Post: Jan 6, 2013 2:43 PM

 Messages: [ Previous | Next ]
 ross.finlayson@gmail.com Posts: 2,720 Registered: 2/15/09
Re: Uncountable Diagonal Problem
Posted: Jan 1, 2013 8:08 PM

On Jan 1, 4:05 pm, Virgil <vir...@ligriv.com> wrote:
> In article
>  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
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> > On Jan 1, 2:57 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > >   "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:

>
> > > > The transfinite course-of-passage in well-ordering the reals sees
> > > > a diminishing interval.  Do the endpoints of the interval meet,
> > > > in the well-ordering?  A critical point of Cantor's first is that
> > > > the intersection is non-empty.

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> > > It is a well known property of the real number line, at least among
> > > Mathematicians, that a nested sequence of closed intervals has
> > > non-empty Intersection.

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> > > Does Ross claim otherwise? --
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> >http://www.tiki-lounge.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 non-empty 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 well-ordering of the reals. From limit ordinal
to limit ordinal, the endpoints may converge. The intervals as you
claim would be non-empty. 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 real-valued 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
well-ordering 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 non-disjoint closed intervals is:
non-empty?

Regards,

Ross Finlayson