Re: Well Ordering the Reals


Tony Orlow wrote:
> imaginatorium@xxxxxxxxxxxxx said:
> > Tony Orlow wrote:
> > > David R Tribble said:
> > > > Tony Orlow wrote:
> >
> > <snip a bit: you seem to be in a more conciliatory mood today, so...>
> >
> > > > This same mapping also forms the basis for Cantor's diagonal proof,
> > > > which proves that the members of N cannot map to all the members
> > > > of P(N). Part of the reason for this is that real fractions are
> > > > infinitely long bitstrings but naturals are not, which means that there
> > > > are more real fractions than naturals. Thus the naturals are countable
> > > > but the real fractions are not.
> >
> > > Well, you know that I advocate the use of infinitely long bit strings on both
> > > sides of the digital point, contrary to most mathematicians.
> >
> > This is the problem. Mathematics is not about "advocating the use of
> > this or that". It's about proving things from axioms. If you want to
> > investigate the structure of objects with infinite strings of digits to
> > left and right, go ahead - no-one will object, and there may be
> > interesting things to discover. But when a bunch (a rather large
> > bunch!) of people have created and worked on a different structure (the
> > normal real numbers), then trying to "advocate" that they are not doing
> > it quite right is just silly.
> I am advocating the consideration, rather than flat-out rejection, of this
> approach. The standard attitude is that non-standard reals and naturals simply
> don't exist and it's foolish to even consider them, but that's foolish in the
> minds of others.

You've already been told this countless times, but your claim is simply
wrong. No competent mathematicion says (for example) that Conway's
surreals "don't exist". The two major differences between what Conway
did and what you are trying to do are:

(1) Conway does not think "Cantor's proof" (etc) is wrong
(2) Conway knows how to do mathematics.

> > > Personally, I
> > > don't see anything especially wrong with Ross ER function of inverting the bits
> > > and bijecting the naturals with the reals in [0,1].
> >
> > It's a standard crank fallacy. It doesn't work. Otherwise it's fine.
> It works if you allow the infinite naturals that are required for the set to be
> actually infinite. But then, your naturals become "uncountable", which seems
> rather odd for counting numbers. "Uncountable" is a bad term in my opinion.

The problem is that your so-called "infinite naturals" do not behave as
you think. Once again, people have studied the maths of infinite
bitstrings (the n-adics), which is interesting, but nothing like the
sort of "extending through the twilight zone" nonsense that you keep
talking about.

Another major problem is that you seem to hope to rewrite the whole of
mathematics in a few weekends, so you keep jumping from one thing to
another, and it's hard to keep up. So I apologise for getting your
current set of infinite strings the wrong way round, but hey I've got
other things to do, too.

The basic point about the infinite-leftward bitstrings (ILBs), which do
represent n-adics and which you are trying to use to represent the
index for an enumeration, is that they *do not have* a lexicographic
ordering.

(OK, by "infinite-leftward bitstrings" I mean bitstrings which extend
unendingly leftward, and do not have a left end. Can you understand
that?)

Anyway, consider the two repeating decimal ILBs:

.......642642642642
.......330033003300

Which is before the other in your imagined "lexicographic ordering"?
(Lexicographical ordering has to start at the most significant end
(MSE), so it works well when there are two ends, MSE and LSE, works ok
when there is one MSE, but does not work when there is an LSE but no
MSE.)

> > > It's not a bad idea to consider N naturals and N reals in [0,1].
> >
> > It might not be a "Bad Idea", but what does it mean, to "consider"? If
> > you have axiomatically defined something called the naturals, and
> > something called the reals, you then have to decide whether something
> > called "N" can reasonably be associated with them. (It's certainly
> > possible: consider the system of set magnitude classification using the
> > following labels - "empty", "count n elements" (for any pofnat n), and
> > "unlimited". Then both the naturals and the reals in [0,1] get the
> > "unlimited" label.)
>
> To consider, from the latin, cum sideris, or something like that, meaning "with
> the stars". To ponder, imagine, think about. Now, we can certainly consider
> that dividing a finite segment like [0,1] into any infinite number of equal
> pieces will yield infinitesimal pieces with zero measure. The points on the
> segment corresponding to the real numbers will be indistinguishable from the
> segments which result from this subdivision. In a simple way, this seems to
> work very well as a set of the reals. I do think, however, that it fails in
> certain areas of the integration of these ideas. It can be problematic to think
> of things that way, though it can solve some problems too.

Be interesting to have a list of the problems it has solved. But more
importantly, any such scheme must obliterate the most basic expected
properties: of integers that each one has a next one, and of reals that
any two have one between them, so none has a "next one".

> > > > So while any given real has a representation consisting of a countably
> > > > infinite number of bits, there are an uncountably infinite number of
> > > > such bitstrings.
> > > Right, because 2^n marks the boundary, somehow, between countable and
> > > uncountable, by standard theory. That makes no sense to me, but I am trying to
> > > understand the precepts of your system.
> >
> > What do you mean *precisely* by "2^n"? What is 'n'? I don't think
> > talking of a "boundary" between countable and uncountable is a helpful
> > idea, because it suggests something like
> >
> > ccc.. cc...c..cc.cccc|uuuu...u..u.uuu.uuu
> >
> > where | is the boundary. But this picture does not correspond to
> > anything meaningful, I think.
> I am simply commenting on the significance given to the power set relation, as
> if 2^n is somehow super-infinite when n is simply infinite, or where the power
> set is assumed to have infinite indexes whereas the set has only finite
> indexes. I am sorry, but if n is finite, then so is 2^n.

There is no need to be sorry, no-one disputes that. Unending is
unending, ending is ending. But it turns out that the equivalence
relation provided by the existence or not of a bijection shows that a
set and its power set are not in the same class. There is a very simple
proof that no set, under the normal and fairly intuitive ideas of set
theory, can be bijected to its power set. You can't seem to grasp the
proof; remember that people like Conway or Robinson have *no difficulty
whatsoever* in understanding it. (Remember they laughed at Bozo the
Clown, but they probably never even laughed at Conway, though I
remember him as the sort of person very likely to give a (friendly)
laugh.)


> > > > Any well-ordering of the reals has to provide ordering and successor
> > > > operations for an uncountably infinite amount of numbers (or subsets).
> > > Does a well ordering need to have a successor to every element? What element is
> > > omega successor to in the well ordering of the ordinals?
> >
> > Here is a classic demonstration of how you have not learned to be
> > careful in a mathematical way. The first sentence asks (rhetorically)
> > if every element must have a successor. Yes, it must. What does that
> > mean *precisely*? It means that for any one element I give you, you can
> > produce its successor. Does it say everything must be the successor to
> > something? No it doesn't. Therefore you cannot use your intuitive
> > experience with finite sets to assume that it does.
>
> I understood that when I wrote it. It wasn't a mistake. It just seems strange
> that this ordered set has an element which isn't a root element, and yet, isn't
> a successor. Doesn't that seem a little disjoint to you?

No it doesn't. If you read the first chapter of ONAG you would see
exactly the same thing happening in the context of non-standard number
systems. Conway's ordinals go:

{0|} = 1
{0,1|} = 2
{0,1,2|} = 3
{0,1,2,3|} = 4
....
{0,1,2,3,...|} = omega

The L set can include all pofnats up to n, some pofnat n, in which case
{L|} represents the ordinal n+1 (i.e. the set can END), *or* the L set
can include all (that's ALL) pofnats. Every set of of pofnats up to n
has a successor, which is another set of of pofnats up to n+1, but none
of them can have the whole set of pofnats as a successor. So the set of
all pofnats has no predecessor.

But we've said that before.

Brian Chandler
http://imaginatorium.org

.

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