Re: Well Ordering the Reals

Tony Orlow wrote:
> imaginatorium@xxxxxxxxxxxxx said:
> >
> > 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.
> Okay.
> >
> > > > > 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.
>
> Well, the numbers I was using are not n-adics precisely. The n-adics have no
> left end, do they? I don't think they consider precise infinite digit
> positions, do they? Who was it, Brian, that challenged me on why my numbers
> were all divisible by 7, which turned out to be a very clever but unrelated
> trick about 7 in decimal? That was with n-adics, but didn't work in my system.
> In my system, one could not tell which of 6 values N mod 7 was, just that it
> wasn't 0. That result made perfect sense, so maybe you are still thinking about
> adic numbers. I used to before I knew what they were called. I don't much any
> more.

No, I understand that your "system" is not exactly the n-adics.
Actually your "system" isn't sufficiently clearly defined to be
anything. You have never (and cannot, of course) defined what you mean
by "precise infinite digit" positions. Your definition of "Tinfinite"
is totally vacuous, since it depends on defining "finite" as "You know,
well, like 6, or 23, or 100^163", then defining "infinite" as "larger
than that". Since perfectly finite numbers can be arbitrarily large,
it's a mystery what "more than arbitrarily large" means. Anyway, people
keep demonstrating inconsistencies: like the captain of the Titanic,
you maintain full speed ahead, because you believe that any minute now
the icebergs will be behind you.

> > 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.
> Understandable. I have jumped around quite a bit. There's so much to talk
> about. This isn't the result of a few weekends. It's kind of like I have a lot
> of mathematical baggage to unlodad. Ha ha. Sorry about that. Still, I don't
> think your objection holds, since there is always a rightmost bit, and since
> there need be only countably many bits to begin with, so if the bits are
> numbered as the naturals, you cannot have bits in infinite positions. You can
> only a[s]cend infinitely.

Here's a typical piece of jumble. ("only ascend infinitely"??!) What do
_you_ mean be "countably"? I don't think you can possibly mean anything
like what mathematicians mean, since you can't do basic set theory. You
can't grasp the concept of an infinite set - any minute now you'll be
talking about how some set is "unending" and therefore "can't be
completed": well, this is *exactly* what the axiom of infinity says in
set theory. We consider the total ("completed") set of pofnats or
anything else that is unending, but we do not assume that this unending
set has an end. There is no "last member". You simply cannot understand
that, right?


> > 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.
> You must be refrerring to my use of ".....11111" to represent the "last element
> of the set", or "the element which maps to the set of numbers which are not in
> the set to which they map."

No. There is no element that maps to the set of numbers which are not
in
the set to which they map. There can be no such number, because it
would imply a contradiction. Therefore we conclude there is, cannot be,
such a number. You are frankly just babbling when you insist on saying
"X does not exist, I know, but I will suppose that it does". It's just
hopeless, and will never be part of anything that anyone else agrees is
mathematics.

> Notice that this number is not well-formed in my
> system of infinity-base digital numbers, since no infinite point is specified.
> It's not a well formed infinite number because it represents a poorly formed
> concept of a number. In this well ordering, we need only consider the infinite
> set of finitely long bitstrings. This is the standard approach, is it not?

Of course not. Standard mathematics is about sets, about numbers, about
well-defined things, not rambling sequences of digits-n-dots. And is
this an appropriate point to remind you (500 posts into the thread)
that your initial question was wrong. The axiom of choice implies that
a well-ordering exists. A very simple proof shows that, however, no
enumeration exists. Of course, if there were an enumeration, this would
show well-ordering, but an enumeration is not required to show
well-ordering, which is just as well, since no enumeration exists.


> >
> > (OK, by "infinite-leftward bitstrings" I mean bitstrings which extend
> > unendingly leftward, and do not have a left end. Can you understand
> > that?)
> Oh yes that was obvious.
> >
> > 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.)
> One really cannot say for sure. I suppose, if not for the fact that the 3 in
> the second is greater than the 2 in the first, one could argue that every digit
> in the first is at least equal to every digit in the second, so the first must
> be larger, but that's not the case. Indeed, without some definition as to
> whether we have a multiple of 3 digits or 4, or both, or the value modulo 12,
> we can't really compare the two. his is why such infinite digital numbers
> require the specification of common digits at the infinite level. So, you're
> right.

I think I'm right, but I'm not encouraged by your saying so. If
something goes on without ending, it means there is no end. If there is
no end, it is nonsense to ask whether the end is a multiple of 3 digits
away. Etc.

> > > > > It's not a bad idea to consider N naturals and N reals in [0,1].
<snip>
> > 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".
>
> I think Ross is right in saying that looking at things this way contributed to
> the development of calculus, and was somewhat common in variations before it
> was agreed on that math follow a more rigorous approach toward infinity, using
> limits and finite values. I don't see that this affects integers at all. It
> discretizes the real continuum, which is a fine thing for some purposes, but
> doesn't exactly seem to convey the continuity in the line. Of course, perhaps
> the universe ISN'T continuous.

Mathematics is not about the universe. Can't you understand that? (No
wonder cranks cross-post to sci.math and sci.physics - they can never
seem to understand the difference.)

> Perhaps the continuum cannot truly exist, in
> which case the infinitesimal view is worth entertaining.

Nonsense. If you mean "mathematically exist", then of course it does,
because it's defined consistently. If you mean "physically exist", then
of course it doesn't.

> I believe I explained how the proof regarding the bijection with the powerset
> specifically cites the "last element" as the one missing from the bijection.

Please quote the proof as written by someone who understands it, and
show us the expression "last element" (or equivalent). Of course it
doesn't.

(Well, hmm, philosophy puzzle: the proof shows that a putative element
mapped to the subset containing all those elements mapped to a subset
not including themselves [a statement of the proof in one line again,
btw; no mention of "last" anything] does not exist. Of course we also
know that the "last element" does not exist. So if P does not exist and
Q does not exist, does that mean that P and Q are the same? Who
knows... )

> the bijection I offered, no element was a member of the set to which they
> mapped, so the set of all elements not in the set to which they mapped was the
> entire set, and the element in question was thus that element that maps to the
> unending set of elements, and was thus an unending string of 1's. That was one
> of the times I referred to "...1111". It's an ill-formed number referring to
> the entire set which never ends. It's a non-answer to a trick question.

Nonsense. Notice how one minute you talk of the entire set, then
suddenly start mumbling about "unending something or other". Well, it's
the cookery trick: ah, if these elements aren't enough, I've got some
more in my desk drawer...

> You cannot name the point at which my bijection fails.

No. I don't need to, and do not intend to waste much time looking at
it. Learn some elementary maths then try again.

> > > > > > 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.

> Doesn't that sound amazingly like what I said concerning the power set, above?
> Uou say " none of them can have the whole set of pofnats as a successor". I say
> "none of them can map to the whole set of pofnats".

Doesn't sound amazingly anything. Sounds confused. There is absolutely
no problem with mapping 65 for example to the whole set. I think, as
always, you are using your (powerful) intuition to leap to some
conclusion based on your vast experience with finite sets, finite
combinatorics, von Neumann computing, and so on. Trouble is, your
intuition is useless, utterly useless on infinite sets.

Brian Chandler
http://imaginatorium.org

.

Relevant Pages

  • Re: Orlow cardinality question
    ... >> Tony Orlow (aeo6) wrote: ... and yet you claim it is infinite. ... Simply that the unbounded set of pofnats, ...
    (sci.math)
  • Re: Well Ordering the Reals
    ... Conway knows how to do mathematics. ... > It works if you allow the infinite naturals that are required for the set to be ... >>> Does a well ordering need to have a successor to every element? ... The L set can include all pofnats up to n, some pofnat n, in which case ...
    (sci.math)
  • Re: Calculus XOR Probability
    ... lartgest finite with an infinite successor. ... implies that there is some lartgest finite with an infinite successor." ... that not mean that there is no point at which the set of finite naturals is ... it is the case that the set of pofnats up to ...
    (sci.math)
  • Calculus XOR Probability
    ... lartgest finite with an infinite successor. ... implies that there is some lartgest finite with an infinite successor." ... I suppose what it boils down to here is infinite induction. ... it is the case that the set of pofnats up to ...
    (sci.math)
  • Re: Orlow cardinality question
    ... the ELEMENTS have no upper ... >>> Why do you feel that being infinite is incompatible ... Simply that the unbounded set of pofnats, ... strings constructed from a set of 10 digits, ...
    (sci.math)