Re: RAF: Rational numbers, irrational numbers: each dense in real numbers

On Oct 22, 10:06 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:
Then, it is shown that given a well-ordering
of the reals, a well-ordering of a subset of the reals exists, such
that it is strictly descending/decreasing in the normal ordering
(alpha < beta => r_alpha > r_beta > 0), and not countable, else via
contradiction: the original object wasn't a well-ordering of the
reals.

OK, I've been trying to follow RAF and Mr. Hughes's
discussion over the last couple of days.

In the part that I quoted, it appears that what
RAF is trying to prove is,

For every wellordering r on the set of reals R,
there exists an uncountable subset C of R such
that the restriction of r to C is identical to
the restriction of > to C.

Obviously, the proof of this is not as trivial
as RAF was hoping. I tried to come up with a
proof myself, but failed.

Is RAF's claim true? I don't know. It may be
provable in ZFC or its negation may be provable
in ZFC. Or it may even be _independent_ of ZFC!

Indeed, RAF's claim reminds me a little of
Freiling's Axiom of Symmetry:

http://en.wikipedia.org/wiki/Freiling's_axiom_of_symmetry

Not only is this axiom, called AX, independent
of ZFC, but it is equivalent to ~CH. I'm
starting to wonder whether RAF's claim is also
related to CH.

Indeed, I actually made an attempt to use AX
to prove RAF's claim. Suppose RAF's claim
does not hold. So its negation holds, viz.:

There exists a wellordering r on the set of reals
R such that for every uncountable subset C of R,
the restriction of r to C is not identical to
the restriction of > to C.

In other words, for this particular wellordering
r, every set on which the restriction of r _does_
agree with > is countable. So perhaps we can come
up with a function f that assigns these countable
sets to the real numbers.

.

Relevant Pages

  • Re: Review of Mueckenheims book.
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