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From: stephen on 20 Jul 2005 13:31
In sci.math Han de Bruijn <Han.deBruijn(a)dto.tudelft.nl> wrote:
> stephen(a)nomail.com wrote:

>> It seems like a lot of the "anti-Cantorians" and other
>> mathematical nay-sayers tend to start with their conclusions,
>> and then try to work backward. The idea of starting
>> with a fixed set of well defined axioms and working
>> forward seems totally alien to them.

> Nah, nah. First comes the theorem. And then comes the proof.

> Han de Bruijn

You cannot prove something without axioms. So if you
start with your conclusions, you have to create some
axioms before you can find a proof. Unfortunately they
never seem to actually define any axioms. Look at Tony Orlow
in this thread for a fine example of this sort of behavior.

Stephen
From: Virgil on 20 Jul 2005 13:35
In article <MPG.1d4816be96c6f78d989f36(a)newsstand.cit.cornell.edu>,
Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote:
> Chris Menzel said:

> > Aside from your tenuous grasp of basic transfinite arithmetic, the
> > critical terms in your argument -- "digital number", "length",
> > "exponentially longer" (cardinal or ordinal exponentiation?), "width",
> > "diagonal traversal", "below the line of diagonal traversal" -- are much
> > too vague for your argument to be evaluated. For all we know, you might
> > have some genuine insights. But currently, your argument is smoke and
> > mirrors; it hasn't been expressed as mathematics.


> I think you know exactly what i mean by each of those terms.

In mathematics "think you know" is not good enough. When asked for
precise and comprehensive definitions in mathematics, it is the duty of
the user of any terms to give such precise and comprehensive definitions.

> They are all
> widely understood.

Not by me of by Chis. I can think of at least two antithetical meanings
for each of To's vague terms, so To piles ambiguity on ambiguity.

> It is typical for Cantorians to resort to claims of vagueness on the
> part of their opponents, while presenting such vague proofs as the
> diagonal argument, without defining anything themselves, and assuming
> unfounded postulates such as all infinities are the same, except when
> they're not.

Every bit of "Cantorianism" has been well enough defined for the
understanding of thousands upon thousands of people. That TO fails where
so many have succeeded says more about TO than about the adequacy of
"Cantorianism's" explanations.
From: Virgil on 20 Jul 2005 13:45
In article <MPG.1d4816be96c6f78d989f36(a)newsstand.cit.cornell.edu>,
Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote:

> > > Cantorians seem to think infinity is simply infinity,
> >
> > Hard to fault them on that score. The fact that you have doubts
> > about whether infinity is infinity (what does "simply" add here?)
> > says quite a lot -- though *perhaps* if you were to try to
> > distinguish different senses of "infinite" with any sort of
> > mathematical precision there might be a point there.

> That was a dishonest snip, really. Notice the comma? That means the
> sentence was not finished. The rest of it said that you do this, even
> in the midst of proving that it is not the case.

There is a difference between even naturals and odd naturals, but there
are also differences between one even number and another.

So to say that there are differences between infinite sets is equivalent
to saying that there are different even naturals.


> The diagonal argument assumes that the list is essentially square,

Geometry does not enter into it. Such a "list" assumes that for every
natural there is a real. Decimal reprsentation assumes that for every
real and every natural, n, there is an nth decimal digit to the right of
the the decimal point for that real.

> and can be
> traversed diagonally, but then proves that this is not the case, and
> yet, Cantorians still seem to insist that the diagonalization proves
> the reals cannot be listed, when what it proves is that there are
> more reals than naturals.

The whole point is to prove that there are more reals than naturals. If
it does that, nothing else is needed.
From: Tony Orlow on 20 Jul 2005 13:48
David Kastrup said:
> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes:
>
> > David Kastrup said:
> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes:
> >>
> >> > David Kastrup said:
> >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes:
> >> >>
> >> >> > David Kastrup said:
> >> >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes:
> >> >> >>
> >> >> >> > Even though every subset of the natural numbers can be
> >> >> >> > represented by a binary number where the first bit denotes
> >> >> >> > membership of the first element, the second bit denotes
> >> >> >> > membership of the second element, etc?
> >> >> >>
> >> >> >> Well, what number will then represent the set of numbers dividable by
> >> >> >> three?
> >> >> > 100100...100100100100
> >> >>
> >> >> > Of course, you will argue that this infinite value is not a
> >> >> > natural number, since all naturals are finite, but that is
> >> >> > clearly incorrect, as it is impossible to have an infinite set of
> >> >> > values all differing by a constant finite amount from their
> >> >> > neighbors, and not have an overall infinite difference between
> >> >> > some pair of them, indicating that at least one of them is
> >> >> > infinite.
> >> >>
> >> >> You have not shown such a thing, and of course it would be
> >> >> inconsistent with the Peano axioms defining the naturals.
> >> >
> >> > That is simply not true.
> >>
> >> Sulking won't help.
> >
> > Don't be a jerk.
>
> That's your job description.
I didn't accuse anyone of sulking, especially when they were doing anything but
that. Typical Cantorian ad hominem.
>
> >> > There is nothing in Peano's axioms that states explicitly that
> >> > all natural numbers are finite.
> >>
> >> It is an immediate consequence.
> > Read on.
> >>
> >> > The fifth axiom, defining inductive proof, is used to prove this
> >> > theorem, but it is a misapplication of the method.
> >>
> >> An axiom is not a "misapplication".
> > No, the proof is a misapplication of the axiom. The inductive proof
> > method works well for constant equalities,
>
> What is supposed to be a "constant equality"?
An equality that holds true for n=1, and for n=n+1 given true for n. An
equality, as opposed to an inequality, or a vague property such as "finite",
which is really an inequality that is constantly decreasing.
>
> > but in the case of proving that all naturals are finite, you are
> > incrementing the value at each of an infinite number of steps
>
> Nonsense. Complete bullshit. I do nothing at all like that. There
> are no steps whatsoever involved here, and certainly not an infinite
> number of them. n is finite for n=0, and it is shown that for any
> finite n, n+1 is finite.
So, inductive proof does not rely on proving for n+1 based on n? The infinite
number of successive naturals for which you prove your property do not
constitute an infinite number of implied steps in inductive proof? Because you
only require three lines for the construction of the proof, you think there are
only three steps implied? Talk about nonsense, and complete bullshit. You hit
the nail on the head with that reply.
>
> That is not an infinite number of steps, but at best 2 steps.
>
> > and maintaining that the value remains finite, because you are
> > looking only at individual steps.
>
> That's what induction is about. Showing f(0), and proving
> f(n)->f(n+1). Two steps, and this covers the naturals. If you want
> to crosscheck more than that, feel free to do so, but the axioms don't
> require doing that. That's their whole point: not having to check
> things piece by piece.
The truth for n+1 depends on the truth for n, for all n in N, from 1 to oo, an
infinite number of times, proving the property true for an infinite number of
numbers. If the inductive portion of the proof, where truth is derived for n+1
from truth for n, increments a value, then over N it increments it an infinite
number of times. To deny this is to feign ignorance of the subject matter. That
is not a valid response.
>
> > When you maintain something is "finite", that is to say it is less
> > than any infinite number, but if it grows by a constant finite
> > amount at each of an infinite number of steps, then it has grown by
> > an infinite amount.
>
> It is irrelevant. The axioms don't talk about steps, and certainly
> not an infinite number of them. You are babbling what you consider
> intuitive, but it bears no relation to the math.
the first step is 1, and subsequent steps are denoted by n, and there are an
infinite number, as many as you have n in N.
>
> > Basically, adding X repeatedly Y times is the same as adding Y
> > repeatedly X times, so adding 1 and inifnite number of times is the
> > same as adding infinity once.
>
> Derive that from the axioms. If you can't, it is irrelevant.
So you contend that X*Y<>Y*X? Interesting. I am sure you must have an axiom in
your bag for the commutative property of multiplication? Check the side
pocket....
>
> > Now, I have heard the argument that inductive proof does not prove
> > things for an infinite number of steps, but only a finite number,
> > but if this is the case, then it does not prove anything for an
> > entire infinite set of natural numbers. Either you agree that there
> > are an infinite number of steps involved, or that the set of
> > naturals is finite, or that inductive proof does not prove a
> > property true for all natural numbers as Peano stated.
>
> There is a strictly limited number of steps involved in an induction
> proof, and it provides proof for all natural numbers, which happen to
> form an infinite set.
And I guess only three of them are involved in the proof? How do you prove it
for number x? Does that not depend on x-1? The number of inductive steps
implied in proving a property true for all n in N is infinite.
>
> >> > I offered, and you saw, a deductive proof that proves that the
> >> > largest natural in a set must be at least as large as the set
> >> > size.
> >>
> >> But the set of naturals does not have a largest element. So you
> >> can prove any property you want about it, like it having webbed
> >> feet and dancing swing polka with a vengeance. Doesn't make a
> >> difference, since no such beast exists anyway.
>
> > At each step in the proof, the set has a largest element which is
> > precisely the same as the size of the set.
>
> Fine, and so you prove something for a set that has a largest
> element. The set of natural numbers is no such set, and so your proof
> does not hold for it.
What the proof shows is that, no matter how large the set gets, it never has
more members than the number represented by its largest member. Therefore, if
the elements are all finite, then the set size cannot be infinite, since that
would be larger than the value of every member in the set. I am sure I can work
out a proof without any reference to largest element. I'll put it n my web
page. I am so sick of this lame "largest element" excuse. It's nonsense.
>
> > This constant equality holds for all such sets defined by ANY
> > natural number in the set.
>
> Certainly. Unfortunately, the set of natural numbers is no such set
> (since there is always a larger number than any given number in it),
> and so your proof does not apply to it.
It applies to ALL n in N. Sorry. Take your complaint to Peano.
>
> > Most generally, it is impossible to have ANY set of natural numbers,
> > finite or infinite, which has a number of elements that is greater
> > than its largest element value.
>
> If it has a largest element value. The set of natural numbers has no
> largest element, and so your proof does not apply to it.
Yes, it does. No set of naturals can contain a greater number of elements than
all element numbers in the set.
>
> > This should be clear to anyone who thinks about it. It is impossible
> > to have an infinite set of strings, of which digital numbers are a
> > type, without having either an infinite base, or an infinite number
> > of digits.
>
> Oh, there is an infinite number of digits, but every single string
> only occupies a finite number of them.
You mean it only has non-zero values in a finite number of digits? That's the
same as only having a finite number of digits, in which case one can only have
a finite number of unique digital numbers.
>
> >> > So, which inductive proof do you believe? You cannot add 1 an
> >> > infinite number of times to your maximal element,
> >>
> >> There is no maximal element, and the Peano axioms don't define
> >> "infinite number of times" or similar processes.
>
> > If Peano's fifth is correct, and inductive proof applies to the
> > entire infinite set in a stepwise manner,
>
> There is no "stepwise manner" in the axioms.
true for n+1, given true for n, for each n in N. steps, an infinite number of
them.
>
> > then there are indeed an infinite number of steps implied. It's an
> > "immediate consequence", as you said above.
>
> It is an immediate consequence of the axioms that the described set is
> infinite, since the successor relation gives a bijection to a proper
> subset. But there are no steps involved at all. You just make them
> up for the sake of your personal intuition. They are a personal
> crutch of your own, and cause you to make mistakes that are not
> inherent in the axioms.
Uh huh. And the inductive construction of the naturals using the successor
operator at each step also involves no steps. Maybe you could use a stepwise
crutch so you don't have to crawl on your belly. It is amazing that you cannot
see that incrementing an infinite number of times to generate an infinite set
of naturals does not result in infinite values. Poincare was right.
>
> >> > I will have my web pages published before too long, so I am not
> >> > getting into a mosh pit with you again right now. Just be aware
> >> > that anti-Cantorians are sick of being called crackpots, and the
> >> > day will soon come when the crankiest Cantorians will eat their
> >> > words, and this rot will be extricated from mathematics.
> >>
> >> Oh good grief. Successor in interest to JSH, are we?
>
> > I don't know really what James Harris' point was. I think you
> > fellows ran him out of here before I had a chance to see his
> > points. Maybe they were junk. I don't know. But, I am tired of being
> > told things about infinity and infinite sets that make no sense,
>
> Then get a fscking set theory book and learn. That's what everybody
> else does.
>
> > and are proven using assumptions that are unfounded,
>
> Axioms are not "unfounded", and not assumptions.
Excuse me, but axioms are just that, for the most part. They are statements
assumed to be true for the sake of argument and proof. Often they ARE well-
founded, and justified outside of the axiom set of which they are a part, but I
have heard repeated assertions in sci.math that axioms are "true by
definition", unquestionable atoms of fact, with no need for justification, and
no need to be consistent with anything outside their own set of axioms. This
highly partitioned conception of math is a real problem for me. I would like to
see axioms made to be as universal as possible. Was that not part of Hilbert's
goal, to develop a universal set of axioms for all mathematics? That is still a
worthy goal, but to no one's surprise, a goal detested by Cantorians, whose
"math" is simply not compatible with most other math.
>
> > and told that the Banach-Tarski result is a "paradox" and not a
> > proof by contradiction. There really seems to be a bad influence
> > going on that allows people to think they have an infinite language,
> > when they only allow finite strings, or that the number of paths in
> > a binary tree, which is always half of the number of branches, is
> > suddenly infinitely larger than the number of branches, when those
> > numbers become infinite.
>
> Then look up and understand the definition of an infinite set. It is
> _the_ decisive mark of an infinite set that it does no longer obey the
> pigeon hole principle.
That's one conception, and one I reject. 1 is 1, whether it's in N, or in
{1,2,3}.
>
> > I simply can't understand how mathematics has come to accept such
> > illogical results as "counterintuitive" rather than incorrect.
>
> Then get yourself books and learn.
Yes, spend years trying to qwrap my head around concepts that are clearly
wrong. What a grand use of my time. I am afraid I WILL have to tear it apart
bit by bit to convince anyone, but even the most obvious objections to your set
of naturals are rejected with insults, and no real refutation except your own
established definitions.
> >>
> >> >> >> Let's take the number representing the set of numbers dividable
> >> >> >> by three. Is this number dividable by three?
> >> >> >
> >> >> > Why does it have to be?
> >> >>
> >> >> It does not have to be. But if it is a natural number, it either
> >> >> is dividable by three, or it isn't. You claim that it is a natural
> >> >> number. So what is it? Is it dividable by three, or isn't it?
> >> >>
> >> >> It must be one, mustn't it?
> >> >
> >> > The number is 100100...00100100. It's certainly even, and a multiple
> >> > of 4. Is it divisible by 3? That can only be determined in a binary
> >> > system with a finite number of digits, as far as I can
> >> > tell.
> >>
> >> Oh, certainly not. Natural numbers are defined by the Peano axioms.
> >> Now let us define the set of all numbers with a well-defined remainder
> >> from division by 3:
> >>
> >> a) 0 has a well-defined remainder of 0
> >>
> >> b) If n has a well-defined remainder of 0,1,2 respectively,
> >> S(n) has a well-defined remainder of 1,2,0 respectively.
> >>
> >> c) different numbers n with well-defined remainder have different
> >> successors S(n) with well-defined remainder
> >>
> >> d) 0 is not the successor of any natural with well-defined remainder
> >>
> >> e) if a set contains 0, and for each of its elements x contains S(x),
> >> then this set contains all numbers with well-defined remainders.
> >>
> >> Shiver me timbers, looks just like we have the Peano axioms here. So
> >> all natural numbers have a well-defined remainder from division by 3.
> >> Looks like 100100...00100100 is not a natural number.
> >>
> >> If it is, just point out which of the above laws is wrong.
> > Well, it would appear that, despite the insistence of Cantorians that all we
> > know about finite sets goes out the window for infinite sets, even such basic
> > obvious facts as proper subsets having a smaller size, they would like to
> > insist that infinite numbers behave precisely the same way as finite
> > numbers.
>
> You got it all confused again. There are no infinite natural numbers,
> so the "Cantorians" don't insist on anything for those non-existing
> entities.
You are doing exactly that when I suggest the inclusion of infinite naturals.
You are saying they don't behave the same way, so they are different, and not
in the set. I might as well say that infinite sets are not sets at all, because
my concept of a set is that every set is finite. Don't throw your definitions
at me as if they are proofs. They're not.
>
> > Does this seem a little unfair to you? Why do you insist that
> > eveything one can do with a finite number should work exactly the
> > same for an infinite number?
>
> There is no infinite natural number. The finite natural numbers are
> defined by the Peano axioms, and yes, everything that works with
> finite naturals works with all of them. That is the whole point of
> defining them with those axioms: to have a body of numbers governed by
> the same rules.
And why don't you want all sets to have the same rules? if infinite sets can
have slightly different rules, why can't infinite numbers?
>
> If you don't like it, you can invent your own axioms and numbers, but
> then you can't cry foul that your laws don't hold with the numbers the
> others are talking about.
Sure, except my numbers are a proper superset of your, so if anything holds for
all of them, then it holds for yours too.
>
> > The determination of divisibility by a number which is prime
> > relative to the number base depends on a termination to the process
> > of division.
>
> Good. And since all natural numbers are finite, every division among
> those naturals is guaranteed to finish.
For finite naturals, yes.
>
> >> > Infinite whole numbers aren't always as convenient as finite
> >> > ones, but they still must exist for the set to be infinite,
> >>
> >> Says you.
> > Says the nature of digital numbers, and says the Peano axioms, which
> > define an infinite number of successors,
>
> No. They don't. They only define the existence of a single successor
> for every number. The infiniteness of the set is a _consequence_ of
> the axioms, but it is not defined in them.
It's an "immediate consequence".
>
> > each a constant finite quantity greater than the last. It's an
> > "immediate consequence". The sum of an infinity of 1's is 1
> > infinity.
>
> The Peano axioms are not concerned about sums, neither with a finite
> nor an infinite number of terms. They only have a nonspecified
> "successor" relation without inherent arithmetic properties.
True, the successor operator does not explicitly define quantity, but order.
Quantity is introduced when we define a unit measure between each element and
its successor, which is what we are doing when we define the first element to
be the number 1, and successive elemenets to be defined by incrementing the
value. That's the other side to creating the natural numbers, besides simple
succession. In this sense, Peano does not really portray the full reality of
the numbers, but only the construction of the set.
>
> >> > and they still can be used to represent infinite subsets of the
> >> > naturals.
> >>
> >> Unfortunately not subsets in- or excluding themselves at will.
>
> > Whether the number that is represented by a binary string which
> > represents a subset is a member of that subset is irrelevant.
>
> Not if you are establishing a bijection.
What? Now it's my turn to say "bullshit". There is no requirement that the
subset of N associated with the number x must contain x as an element. I am not
drawing a bijection between the number of a subset and the numbers within it.
The numbers in the subset as a whole determine the number corresponding to the
set, and the bijection is between the numbers representing the subsets and the
natural numbers. Hey, according to Cantorian thought, if I can draw such a
bijection, and it goes on forever, then the sets are equivalent. So what's your
issue here? You don't like the theory contradicting itself? hmmmm.....
>
> >> > If you divide this number by 3 (11) you find it is divisible or not,
> >> > depending on whether you have an odd or even number of 100's in your
> >> > infinite string. Of course, this question is not really answerable,
> >> > so I don't have an answer for you. What do you think? What is
> >> > aleph_0 mod 3?
> >>
> >> Oh, I never claimed that aleph_0 was a member of the natural
> >> numbers, so I don't need to make claims about aleph_0 mod 3.
>
> > But the answer to this question depends on the number of digits in
> > the infinite number.
>
> Since there are no infinite numbers, I need not answer the question.
> You claim that there are, so it is up to you to provide an analysis of
> your claims.
My claim is not that all arithmetic holds for infinite whole numbers, but that
infinite whole numbers are required in any infinite set of whole numbers. The
arithmetic is a sidebar, and if you want to claim that I have to work all that
out to assert their existence, you're full of ptooey. That is an entirely
diferent question.
>
> > If that number is a multiple of 6, then the number representing the
> > set is divisible by 3, but that isn't really decideable, is it?
> > Perhaps not everything that can be determined regarding a finite
> > number can be determined for an infinite number.
>
> Which is why infinite numbers have not been admitted into the
> naturals. It would be impractical to have a set of numbers that are
> not governed by the same laws.
They certainly need to be handled differently, but that's what Cantorians say,
unapologetically, about infinite sets. When asked why a certain property is not
true of infinite sets, the typical answer is, "because not everything that
holds for finite sets must hold for infinite sets", which doesn't answer the
question at all. So, you get no apologies from me for including infinite values
in the set of whole numbers (okay not naturals, see?), in order for that set to
be infinite. By your definition of naturals as all finite, and the basic math
governing quantities and strings, I can only conclude that you have a finite
set. If you want an infinite set, you need to drop that restriction.
>
> > If you can allow discrepancies between you finite and infinite sets,
> > I am not going to apologize for not being able to perform a mod
> > operation on an infinite number. It is still a fact that they must
> > exist for the set to be infinite, says me.
>
> Sulking won't help.
Being a jerk again, are we? Where do you see me "sulking"? Typical.......
Don't get your panties in a twist, David.
>
>

--
Smiles,

Tony
From: malbrain on 20 Jul 2005 13:53


Daryl McCullough wrote:
> Tony Orlow writes:
> >
> >Dik T. Winter said:
>
> >> Back on your horse again. Tell me about the binary numbers (extended to the
> >> left with 0's) where the leftmost 1 is in a finite position. Are all those
> >> numbers finite? Are there only finitely many of them?
> >
> >yes and yes
>
> What definition of "finite" are you using?

Main Entry: fi·nite
Pronunciation: 'fI-"nIt
Function: adjective
Etymology: Middle English finit, from Latin finitus, past participle of
finire
1 a : having definite or definable limits <finite number of
possibilities> b : having a limited nature or existence <finite beings>

This definition from webster should suffice. Binary numbers with ones
in finite positions have a limited number of possibilities.

karl m

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