Date: Jan 17, 2005 1:50 AM
Author: Ross Finlayson
Subject: Re: On Well-Ordering(s) and Sets Dense in the Reals, Infinity

Timothy Little wrote:
> Ross A. Finlayson wrote:
> >  Any well-ordering is a total ordering.  If you can well-order the
> > reals is not that well-ordering a total-ordering?
>
> That is correct.  Every well-order is a total order, by definition.
> However, not every total order on an infinite set is a well-order.
>
>
> > As there are only two main total orderings on the reals "<" and ">"
> > is not thus one or both of them a well-ordering?
>
> No.  There are infinitely many total orderings.  Under the Axiom of
> Choice, at least one of those is a well-ordering.
>
>
> > do you have a point?
>
> Hopefully, showing you how to fill the hole in your logic.  It is up
> to you whether that purpose is pointless or not.
>
>
> > Also, the above is not nonsense,
>
> Many of the phrases and propositions individually made sense.  Some
of
> them were true.  However, they were logically disconnected in a
manner
> that made nonsense of the post taken as a whole.
>
>
> > For the reals on the unit interval, a total ordering is a
> > well-ordering.
>
> At least one total ordering is, of the infinitely many total
> orderings, but not all of the total orderings are well-orderings.
>
>
> >  For any semi-infinite interval, there is a total ordering that is
a
> > well-ordering, one of "<" and ">".
>
> Why one of those two?  Why not one of the infinitely many other total
> orderings?
>
>
> > It's simple to say that there's a total ordering that is a
> > well-ordering for any set because all well-ordering are total
> > orderings and all sets are well-orderable.
>
> It's simple to say, and in ZFC it's true.  It doesn't imply that all
> total orderings are well orderings.
>
>
> - Tim


The concept I want to promote is that the total ordering "less than" or
"<" for any finite open, closed, or half-open interval is a
well-ordering of the reals.

That is counter to the general argument that for a given real there is
no "previous" nor "next", instead, there is a previous, and next, and
then the idea is to assign names to the elements of the sequence that
are derived from the well-ordering of [0,1] as follows:  0, iota,
2iota, ....

When there is a bijection f between the naturals and set X then it is
simple to determine a total ordering between N and X, if the inverse of
f is f' then for x and y in X then if f'(x)<f'(y) then x "<" y in that
total ordering where N has a natural total ordering with ordinals.
When people consider a total ordering of the rationals they generally
consider that if x<y then x "<" y.  Here I could use better usage of
the conventional term R to denote a relation.  Anyways the perhaps most
obvious total ordering of the rationals is based upon the numeric
magnitude, or scalar value, due to trichotomy of the reals.  Another is
defined by any bijection from the naturals to the rationals.  Now for
some when they get a finite, open interval of the rationals, they might
have stated that the linear ordering was not a well-ordering, where
here it is seen that it is.

When there is a notion to well-order the reals, then a given total
ordering of the open unit interval of the reals that has a least
element or "avoids infinitely descending chains" of some form causes a
dilemna for some:  if you combine the claim that there is no "least"
real on the open unit interval with that there is no bijection between
the naturals and reals on the open unit interval, then there is not a
total ordering that is a well-ordering, and thus there could be no
well-ordering, because a well-ordering is a total ordering.  Yet, a
contradiction ensues from that there is a well-ordering for any set.

If instead the linear or trichotomous total ordering is deemed a
well-ordering, via the notion that the least element of the unit
interval is iota because the open unit interval is a subset of the
closed unit interval, where the total ordering is a well-ordering
simply with zero being the last element and proceeding elements named
integral multiples of iota, then it is seen that the linear total
ordering of the closed unit interval is a well-ordering for it and any
of its subsets.  The open unit interval is a subset of the closed unit
interval.

In that sense it is a provisioned example of a well-ordering of any
finite interval of the reals:  the linear or trichotomous total
ordering.  Here trichotomous could mean "greater than" or "less than."

With a semi-infinite interval, that is infinite to the left or right on
the number line but not both, then the trichotomous total ordering with
one or the other of ">" or "<" is a well-ordering.

With the infinite interval or set of all reals, consider the two
semi-infinite intervals of [0,oo) and (-oo, 0].  Each is well-orderable
by a "main" or trichotomous total ordering, eg (0, iota, 2iota, ...)
and (0, -iota, -2iota, ...).  Then, what seems to be the obvious
well-ordering of (-oo, oo) is (0, iota, -iota, 2iota, -2iota, ...).

That arose in earlier discussion as a function from the naturals to the
reals that would not engender contradiction in terms of "Cantor's first
proof" or "Cantor/Megill style", nested intervals, in the construction
of a bijection between the naturals and reals.  That is to say, if f(0,
1, 2, ...)= (0, iota, 2iota, ...) and that is a bijection from the
naturals to reals, then Cantor's first proof does not say otherwise,
because that function would not fit the assumptions of the proof, thus
that it does not apply in its otherwise contradiction.

Consider Megill's computer "verified" proof about the rationals, with
the focal "for any q>0 in Q (the rationals), there exists x such that
q>x>0."   Also consider that that focal statement is falsifiable.

It would seem then that from the closed unit interval's trichotomous
total ordering being a well-ordering, that it is also a well-ordering
of the open unit interval.

Do you know any total orderings of the reals besides those using
trichotomy?  If there are none, is not at least one of those a
well-ordering?

Regards, 

Ross Finlayson