Re: On Well-Ordering(s) and Sets Dense in the Reals, Infinity

From: Ross A. Finlayson (raf_at_tiki-lounge.com)
Date: 01/21/05 

Date: 20 Jan 2005 17:30:24 -0800

In ZFC, there is at least one well-ordering of the reals.

A well-ordering is necessarily a total ordering so the well-ordering(s)
are a subset of the total ordering(s).

I'd like to know more about the well-orderings of the reals because for
any well-ordering of the reals, it is a well-ordering of each subset of
the reals, and thus, for example, a well-ordering of the rationals.

One of the points I want to make clear to you is that it's possible to
consider that the only total ordering of the reals that is a
well-ordering is a monotonic function from the naturals to each element
of an interval of the reals, the normal or "usual" ordering. That's
because if you use infinite ordinals to apply the total ordering, then
the finite ordinals of those as a subset of the indicies of the total
ordering would have to be well-ordered for the complete total ordering
to be a well-ordering, and the only way known to me for the naturals to
biject to the reals on the unit interval is a monotonic mapping or
undecomposable composition of monotonic mappings.

That is to say, because by the definition of well ordering each subset
so ordered is itself well-ordered by that well-ordering, in application
to infinite ordinals the subset restricted to indices of finite
ordinals would as well have to be a well-ordering, and the only way to
biject the naturals and reals necessarily must avoid the consequence of
the Cantor/Megill style theorems. Here, Cantor means about his first
proof with the nested intervals and infinite convergent sequences, and
Megill means Norm Megill and his computer-verified theorem that the
reals are uncountable unless considered by their normal ordering that
he showed to us here.

A theorem of ZFC set theory is that there is a well-ordering of any
set.

A well-ordering of the reals is in your estimation R x O for some set
of ordinals O including an "uncountable" ordinal and all lesser
ordinals. Some subset of O is N, and thus for some subset of R R_n the
well-ordering R_n x N is a subset of R x O and R_n x N is a
well-ordering of R_n and each subset of R_n. Here, the Cartesian
product X x Y is indicating a function from Y to X.

As the reals are restricted to the unit interval, then R_n is some
infinite subset of the unit interval that is well-orderable because of
a bijection between it and N, and because R_n is a subset of R and N is
a subset of O.

So, when there is a well-ordering of the unit interval then some subset
of the unit interval is well-ordered by a function between that set and
the natural integers. That could be f(n)=1/n, and for n in N there is
always a greatest element in the normal ordering, it is well-ordered as
f', the inverse. What I'm trying to figure out is why it couldn't be
that or what other considerations there would be, yet thus far in this
progression R_n could be a random scattering of points on the unit
interval. Indeed, from what I understand of the Pi and Sigma Goedelian
complexity definitions a well-ordering of the reals would be some
random scattering of points, in terms of the normal ordering.

While that's so, with using the normal ordering it's possible that R_n
= R, where infinite sets are equivalent. Infinite sets are equivalent.

There are many discussions and questions about why the normal ordering
of the reals, a total ordering, is _not_ a well-ordering. I wondered
it myself some years ago, here's a link to another consideration, with
the result among the participants that the normal ordering of the reals
is _not_ a well-ordering.

http://mathforum.org/discuss/sci.math/m/587599/588095
http://groups-beta.google.com/groups?q=%22well-ordering%22+reals

Here's a consideration, the normal ordering is the well ordering for
the set of all positive integers x of x/n for finite integer n, eg
integral multiples of a half, a third, etcetera. That is where the
domain is x={1, 2, ..., n}, for some finite integer n, the normal
ordering is a well-ordering for f(x)=x/n, and it is so inductively for
each n+1 in N. That is not the same as that it is true for the
complete set of natural integers, it is thus in that way a
non-inductive result.

I guess I'm back to looking for a way that the set of non-negative
reals is as well a contiguous sequence of points from zero, indefinite
except for zero, and showing that each non-empty interval on the reals
contains at least one real number.

Yeah, that's what I need: showing that each non-empty interval on the
reals contains at least one real number. I think that's a pun, but it
could be one of those "plain language multiply-implicative" statements.
It does because it's non-empty.

So, basically I'm taking a theory that defines the numerical system up
to the reals, and then adding that the normal ordering is a
well-ordering, then considerations what further definitions on the
reals make that true without invalidating other consequences of the
regular or standard definition of the reals that are known to hold true
for reasons of ubiquity and real-world applicability.

To that end, I wonder what I should avoid. What happens if,
non-inductively, the normal ordering is used as a well-ordering? Here,
non-inductively means, similarly to the sense as it is used above,
that it is still true that our example 1/N has a lesser element than
1/n for 1/(n+1), but that that doesn't show that the sequence of the
normal ordering doesn't contain only elements of the range of that
function.

What are the consequences of well-ordering?

Regards,

Ross Finlayson

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