Re: Well Ordering the Reals

David R Tribble wrote:
> Ross A. Finlayson wrote:
> >> The reals are a set, or, they aren't objects of a set theory, so, if
> >> they're a set and sets are well-orderable then they're well-orderable.
> >> [snip]
> >
>
> David R Tribble wrote:
> >> Your prose is hard to follow, but I think you are assuming that some
> >> well-ordering of the reals must imply a well-ordering under their
> >> normal order.
> >>
> >> It is my understanding that Cantor' first proof of the uncountability
> >> of the reals starts with the assumption that the reals are countable
> >> and uses countably infinite nested intervals to prove that those
> >> intervals do not include all reals, thus proving by contradiction that
> >> the reals are uncountable.
> >>
> >> But Cantor's proof appears to operate only within the normal ordering
> >> of the reals. So I fail to see how a different ordering (one that is
> >> well-ordered) is subject to the same contraints (and thus the same
> >> conclusion) as the normal ordering.
> >
>
> Virgil wrote:
> > Two nits to pick:
> > (1) The first proof is a *direct* proof that any mapping from the
> > naturals to the reals omits some reals, just as is the second(diagonal)
> > proof, and therefore proving there are no surjections from the naturals
> > to the reals.
> >
> > (2) A different ordering of the reals does not destroy the properties of
> > the standard ordering, so does not destroy the validity of a proof based
> > on the standard ordering. And such a proof for the standard reals with
> > the standard ordering will also be valid for any set that the reals can
> > be injected into, regardless of how it is ordered.
>
> Corrections noted. I was confused.
>
> If Cantor's first proof shows that the reals are uncountable under
> their normal ordering, then it can also show that they are uncountable
> under any other ordering. This is _obviously_ true.
>
> So even if the reals can be well-ordered, they are still uncountable.
>
> Or as, Daryl says:
> > There is no connection between Cantor's first and well-ordering
> > of the reals. Cantor's proofs show that the reals cannot be
> > enumerated using the natural numbers, but that doesn't mean
> > that they can't be well-ordered.
>
> But Ross appears to be arguing for the opposite, that well-ordering
> the reals would render them countable (although it's not always clear
> to me what he is arguing).

Hi,

No, determining that the normal ordering is the well-ordering is the
result, not the assumption.

A well-ordering of the reals will have a least element, like an
enumeration has a least element. Then, the disjoint of that element
from that set, the remaining piece of the set, will have a least
element. Those will form an interval that basically represents a pair
(a_0, b_0), the initial elements of the sequences a and b in Cantor's
first. Then, in the progression of the well-ordering, elements that
fall within that interval will form another, (a_1, b_1). Then, the set
of those endpoint pairs is well-ordered. Then, for whatever cardinal,
where each cardinal is well-ordered, it being an ordinal, for whatever
cardinal the reals may have, then unless there is (a_n, b_n) with those
being adjacent points in the reals, there is no well-ordering of the
reals.

The argument, Cantor's first, is based upon the completeness of the
reals, the complete ordered field, the gaplessness property of the
reals. That does not change regardless of what enumeration, or
well-ordering, or perhaps well-ordering's initial segment, is used,
unless there are adjacent points in the reals.

For example, if you can conceive of the reals as integral
iota-multiples, basically half-point or one-sided point widths, where
the (non-negative) reals could be addressed as points in the normal
ordering, ie zero, 1 iota, 2 iota, 3 iota, ..., then Cantor's first
does not preclude that from being a well-ordering or even an
enumeration.

There is obviously the problem with that that the reals are the
complete ordered field, so you want to divide 1 iota by two or average
1 and 2 iota, the result is undefined. That's a problem, I agree.
While that is so, the alternative leads to a variety of conclusions
that would be contradictory.

These are basically notions of a result of infinite divisibility that
is not zero, Zeno in reverse.

As has been noted, the notion of these reals as points on a line is not
new. Spinoza for exampled compared the natural integers to the
continuum. The natural integers are an interesting set, not {0, 1, 2}
except in extension but having the properties of compactness and so on
that their compactification is trivial, implicit, and a variety of
extensions of decidability related results may lead to the notion that
N E N, in discussion of a variety of number-theoretic and
metatheoretical related issues. Anyways, Leibniz for example has that
they're infinitesimals, and that lead to centuries of consternation and
the constraints of limit, which are useful but not necessarily
unrestricted enough in comprehension. Then, many modern analysts have
heard of varieties of nonstandard analysis of the reals, most
prominently Robinso(h)n with his hyperreals, Nelson's IST, and less
mainstream Schmieden and Laugwitz and other various constructions,
including deductive constructions, of the real numbers.

Well-order the reals, unless there is a least real greater than zero,
that is not a well-ordering of the reals.

V = L.

Ross

.

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