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

On Oct 22, 9:52 am, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On Oct 22, 12:25 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:


....

That's like saying: don't just prove there exists a well-ordering of
the reals, give an explicit well-ordering of the reals.

No. There is no requirement that the algorithm be contructible,
and you can certainly assume a well ordering of the reals.
The important part is that your method has to set the value
of r_x for "every x", not just some values of x. The way
to insure this is to start with an arbitrary value of x and
show how a value is assigned to r_x. Note that if x is a limit
ordinal, nothing using the concept of "next" will work.

- William Hughes

Hi,

It seems you accept that a well-ordering of the reals, if not
"constructible", exists. 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.

In the first example, given a well ordering <R, <{ >, above called
X(r), another well-ordering, say Y(r), is constructed as follows: y_0
is the x_alpha for the least ordinal alpha such that x_alpha is
greater than zero. Then, there exist as many ordinals beta less than
the (initial ordinal of) the cardinality of the continuum and greater
than alpha such that 0 < x_beta < x_alpha. Then y_1 is x_beta. So,
then alpha, as a free variable, is redefined as beta and the beta as
the next ordinal such that 0 < x_beta < x_alpha. That proceeds, then
there is consideration of limit ordinals. If there are only finitely
many decreasing members within the course-of-values of ordinals to the
next limit ordinal, then they're appended to the current (countable)
segment between limit ordinals, else the well-ordering's next limit
ordinal's value is defined.

In the second example, a well-ordering is defined in terms of
generated well-orderings, and, if its domain were countable, in making
a well-ordering of the unit interval of reals, then the reals would be
countable, otherwise it's not, and, it's strictly decreasing.

In the third, if there are uncountably many strictly decreasing
sequences, then, elements from basically each of them are selected,
then there's an uncountable strictly decreasing sequence.

So, there is described three different ways to illustrate the
existence of an uncountable decreasing transfinite "sequence" (well-
ordered by the reverse of the natural ordering, decreasing) given a
well-ordering of the reals.

Ross

--
Finlayson Consulting

.

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