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

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 ross.finlayson@gmail.com Posts: 2,630 Registered: 2/15/09
Re: Uncountable Diagonal Problem
Posted: Jan 3, 2013 9:32 PM

On Jan 3, 9:07 am, "Ross A. Finlayson" <ross.finlay...@gmail.com>
wrote:
> On Jan 2, 12:48 am, Virgil <vir...@ligriv.com> wrote:
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> > In article
> >  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:

>
> > > 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:

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> > > > > > > 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. --

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> > > > > > > Basically
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> > > > > > Nonsense deleted! --
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> > > > > Nonsense deleted, yours?
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> > > > Nope! --
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> > > Great:  from demurral to denial.
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> Seems clear enough:  in ZFC, there are uncountably many irrationals,
> each of which is an endpoint of a closed interval with zero.  And,
> they nest.  Yet, there aren't uncountably many nested intervals, as
> each would contain a rational.
> To whit:  in ZFC there are and there aren't uncountably many
> intervals.
> Then, with regards to Cantor's first for the well-ordering of the
> reals instead of mapping to a countable ordinal, there are only
> countably many nestings in as to where then, the gap is plugged (or
> there'd be uncountably many nestings).  Then, due properties of a well-
> ordering and of sets defined by their elements and not at all by their
> order in ZFC, the plug can be thrown to the end of the ordering, the
> resulting ordering is a well-ordering.  Ah, then the nesting would
> still only be countable, until the plug was eventually reached, but,
> then that gets into why the plug couldn't be arrived at at a countable
> ordinal.  Where it could be, then the countable intersection would be
> empty, but, that doesn't uphold Cantor's first proper, only as to the
> finite, not the countable.  So, the plug is always at an uncountable
> ordinal, in a well-ordering of the reals.  (Because otherwise it would
> plug the gap in the countable and Cantor's first wouldn't hold.)
>
> Then, that's to strike this:
> "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. "
> Basically that the the gap _isn't_ plugged in the countable.
>
> Then, there are uncountably many nested intervals bounded by
> irrationals, and there aren't.
>

Point being there are uncountably many disjoint intervals defined by
the irrationals of [0,1]: each non-empty disjoint interval contains a
distinct rational. Thus, a function injects the irrationals into a
subset of the rationals.

Regards,

Ross Finlayson