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

On Oct 21, 6:54 pm, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On Oct 21, 9:05 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:

With regards to the proof (largely in sketch, but assembled piece by
piece) of the existence via choice of a transfinite "sequence" of
descending irrationals, I would like to hear more arguments as to why
that is not possible, so they can be addressed.

Let X be an ordinal, and for every x in X, let r_x
be a real number. 0 < r_x < 1

Let C={r_x| x in X}

Note that if x is in X, then x has
a successor, call it x+1.

Assume that 0<r_(x+1)<r_x.

C is a strictly descending "sequence".

Let N be the set of natural numbers.
Let n be an element of N.

Let U_n = {r_x| x in X, 1/n >= (r_x-r_(x+1)) > 1/(n+1)}

U_n has fewer than n+1, elements, so U_n is finite.

C = union n in N U_n

C is the union of a countable number of finite sets, so
C is countable.

Thus any strictly descending "sequence" is countable.

Note that all we need for this proof is that every element
of the "sequence" is strictly greater than its successor.
Since every element has a successor, we do not have
to distinguish betweeen limit and non-limit ordinals.

- William Hughes

OK, well-order the reals. Given the well-ordering X(r), construct
another X_-(r) in this manner: Let the first element of the well-
ordering X_-(r_0) be the first element greater than zero. Then, let
the next element of the well-ordering X_-(r_alpha+) be the next
element X(r_beta) such that 0 < r_beta < r_alpha. There are
uncountably many r_beta's between 0 and r_alpha, there are as many as
are in the initial ordinal of c, because there are that many real
numbers and they are well-ordered, and the r_alpha's are well-
ordered. X_-(r) is strictly descending/decreasing. (Worse, it's
natural.)

How about this, construct the well-ordering in this manner, via
construction of many intermediate well-orderings of subsets (of which
many exist). Let the first element be r_0 > 0. Then, let the next
element between r_0 and zero be the second element. Then, for the
next element r_beta from the well-ordering of R, if it's between
r_alpha and 0, then, it becomes r_alpha+. Otherwise, put it in order,
and, shift all the then following elements by one, ie make r_alpha =
r_alpha+1. That doesn't work, because it would put all the elements
of the reals in a countable sequence, but only in the same order as
the equivalency function (reversed).

Well-order the reals between zero and one. Then, there are
uncountably many descending sequences, one for each limit ordinal < c,
pairwise disjoint, with the elements between zero and one. Start with
any value, one of the sequences contains an element between it and
zero, another between that and zero.

Basically, in ZFC with the axiom of choice, for any subset of the
irrationals comprising an open (and thus non-degenerate) interval, a
particular element can be selected between the bounds of the interval,
partitioning the interval into two non-empty and equivalent
subintervals.

That's a nice, strong argument. I don't have an immediate direct
counterexample, instead a counterargument via construction. That's a
nice argument, I'll try and develop further arguments that a
reasonable person skilled in consideration of the objects of
discussion could use to mathematically rationalize (prove) the
existence of a counterexample. (Then there would exist infinitesimals
in the reals.)

Now I would like to hear more about how ~Con(ZFC) => ~Con(ZF).

Thanks,

Ross

--
Finlayson Consulting

.

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