Re: There are uncountably many irrationals was Re: abundance of irrationals

From: Virgil (ITSnetNOTcom#virgil_at_COMCAST.com)
Date: 01/11/05 

Date: Mon, 10 Jan 2005 17:16:01 -0700

In article <1105367570.099684.249250@c13g2000cwb.googlegroups.com>,
 mueckenh@rz.fh-augsburg.de wrote:

> Jesse F. Hughes wrote:
> > mueckenh@rz.fh-augsburg.de writes:
> >
> > > Robin Chapman wrote:
> > >> mueckenh@rz.fh-augsburg.de wrote:
> > >
> > >> As I explained before, the most that your method will produce
> > >> is a sequence (q_n)_{n in N} of rationals, not necessarily
> > >> containing all rationals in (0,1) and
> > >
> > > that is correct!
> > >
> > >> a sequence (x_n)_{n in N} of irrationals, certainly not
> > >> containing all irrationals in (0,1).
> > >
> > > Then simply continue! Or is there a step "in the infinite", a
> > > threshold, one cannot pass? What about the well-ordering theorem.
> What
> > > is it good for?
> >
> > Er, for well-ordering uncountable sets?
>
> For countable sets no well-ordering theorem is required!
> >
> > But what good does it do here? Pass through all the countable
> > ordinals and you're sure to exhaust the rationals and you are also
> > sure you have not exhausted the irrationals.
>
> Continue! There is always a rational between two reals.

There are always countably many rationals and uncountably many
irrationals between any two distinct reals.
> >
> > Disagree with the second claim? Then show which step in any (well,
> > properly, in each) of the proofs |N| < |R| is invalid.
>
> Cantors first proof (1873) uses the fact that limits do not belong to
> sequences.
 The limit of f(n) = 1 belongs to the equence {1,1,1,...}

> Nevertheless his proof fails, if applied to the set of
> irrational numbers alone.

Not if one reads the proof carefully enough. That first proof applies
only to sets which have aproperty that the set of irrationals lacks, so
the theorem says nothing about the countability of the set of
irrationals. It does not fail because it does not even apply.

> The reason is, that "any infinite sequence"
> need not converge to an irrational limit. Already the absence of a
> single number, zero for instance, cannot be tolerated, because it is
> the limit of several sequences.
>
> This situation, however, is the same if only the set of all rational
> numbers is considered. Therefore both sets have the same status with
> respect to this uncountability proof. And we are not able, based on
> this very proof, to distinguish between them.

The first Cantor proof demostrated that any open interval of reals is
uncountable, which, together with the countability of the rationals,
demostrates the uncountability of the irrationals.
>
> On the other hand, if applied to the alternating hamonic sequence
> alone, it would show that the rational numbers are uncountable, because
> 0 is never included.

It is clear that you do not understand Cantor's first proof if your
misunderstanding leads you to this conclusion.
>
> Cantor's diagonal method is wrong,

Even if it were, his first proof is not, so his conclusion that the set
of reals is uncountable is not.

> because it is not completed as long
> as only diagonal digits in lines enumerated by natural numbers are
> exchanged. In a line enumerated by omega, on the other hand, the
> position of the diagonal cannot be determined.
>
> The third proof (reported or conceived by Hessenberg) uses the set of
> nongenerators in IN -> P(IN). This set simply does not exists. (Like
> the set of all sets doesn't exist.)

This neither exhausts the set of proofs of the uncountability of the
reals nor disproves the result.

Relevant Pages

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