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Re: Uncountable Diagonal Problem
Posted:
Jan 1, 2013 11:19 PM
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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's
mapping is not monotone in the normal ordering of the reals, the
nested intervals have countably many nesting, that converge.
This is where, appending or concatenating well-orderings of disjoint
sets, is a well-ordering of their union. The real line in ZFC can't
be broken into uncountably many disjoint segments, each would contain
a rational. Yet, in ZFC it has uncountably many points. The real
line has only countably many disjoint segments.
That where transfinite recursion would not allow the axiom of choice
to apply to each subset of irrationals between zero and one: because
then for each irrational, there are uncountably many irrationals less
than it in the normal ordering, as there is for each of those courtesy
density of irrationals, there can only be countably many nested
intervals, but there are uncountably many points to be endpoints of
those intervals.
Closed intervals are defined by their endpoints, sets are defined by
their elements. Then, there couldn't be uncountably many different
closed intervals, because any subset of them if disjoint would be
segments and if not disjoint would be nested or have a non-empty
intersection, where only countably many could be the other. In ZFC,
there are uncountably many irrational points, each possibly an
endpoint of a closed interval.
Then, the above note describes, and proves, due the density of the
rationals, there is at least one for each of their disjoint in the
reals, then here, the consideration is that due the uncountability of
the irrationals, there would be uncountably many segments or nestings,
yet each of those, due the density of the rationals, has a unique
rational.
Regards,
Ross Finlayson
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