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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: 2,690 Registered: 2/15/09
Re: Uncountable Diagonal Problem
Posted: Jan 3, 2013 12:07 PM

On Jan 2, 12:48 am, Virgil <vir...@ligriv.com> wrote:
> In article
>  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
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> > On Jan 1, 11:22 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <ef09c567-1637-46b8-932a-bcb856e41...@r10g2000pbd.googlegroups.com>,
> > >   "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.
>

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.

Regards,

Ross Finlayson