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

On Oct 24, 7:57 pm, lwal...@xxxxxxxxx wrote:
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.

No surprise. C would be an uncountable
strictly decreasing sequence. It is easy to
show that no such sequence can exist.
(The proof is the standard analysis proof that
a sum over an uncountable set can only be finite,
if all but a countable number of elements are 0).

- William Hughes



.

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