Re: Cantor and the binary tree

Matt Gutting said:
> Tony Orlow (aeo6) wrote:
> > David Kastrup said:
> >
> >>Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:
> >>
> >>
> >>>David Kastrup said:
> >>>
> >>>>Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:
> >>>>
> >>>>
> >>>>>imaginatorium@xxxxxxxxxxxxx said:
> >>>>>
> >>>>>>
> >>>>>>Tony Orlow (aeo6) wrote:
> >>>>>>
> >>>>>>
> >>>>>>>Well, the diagonal proof is extremely flawed, as I've pointed
> >>>>>>>out. It proves that the infinity of real numbers is greater than
> >>>>>>>the infinity of the digits they have, but it does not mean we
> >>>>>>>cannot list them, conceptually at least.
> >>>>>>
> >>>>>>Ah! "Conceptually"! The proof is a proof that the real numbers
> >>>>>>cannot be arranged in a list. It show that if you assume that the
> >>>>>>real numbers can be arranged in a list, a contradiction
> >>>>>>results. Therefore, mathematicians conclude that it is not possible
> >>>>>>to arrange the reals in a list. (Just as they conclude that the
> >>>>>>sequence of pofnats, 1, 2, 3, ... never ends, and does not have an
> >>>>>>end, even "conceptually".)
> >>>>>>
> >>>>>>But in Conceptual mathematics, a proof you can't do something
> >>>>>>doesn't mean you can't do it "conceptually". Well, will I ever get
> >>>>>>the hang of this stuff, I wonder...
> >>>>>>
> >>>>>>Brian Chandler
> >>>>>>http://imaginatorium.org
> >>>>>>
> >>>>>
> >>>>>Brian you chopped the meat of the argument. The proof is flawed
> >>>>>because it assumes a square grid, which is the only way a diagonal
> >>>>>traversal can possibly traverse the grid of digits.
> >>>>
> >>>>It assumes no such thing. It assumes that the list is indexed by
> >>>>natural numbers, and that the digits can be indexed by natural
> >>>>numbers.
> >>>
> >>>By the full entire set of natural numbers, both vrtically and
> >>>horizontally? the it is square.
> >>
> >>Last time I looked, a square has four corners, not one.
> >
> > Yeah, like any grid. You see one corner, and continuations to infinity
> > horizontally and vertically, and can't imagine an infinite square? Must you
> > look on the finite local level and pretend there is no structure to this list,
> > after having asserted that there is? Do you still think the Earth is flat? This
> > list is the limit as n->oo of any list of all digital numbers of length n. Any
> > such list is exponentially longer than it is wide in digits. This is simple
> > stuff.
> >
> >>>Otherwise, a diagonal traversal on an elongated rectagular grid will
> >>>obviously miss many lines of the grid.
> >>
> >>It is nonsensical to talk about "elongated rectangular" for a quarter
> >>plane. A rectangle has four corners, not one.
> >
> > It is a Nx2^N, or Nx10^N, rectangle. Get used to it. You wanna pretend it's a
> > list of real numbers? Do it right.
>
> Umm, N is not a number; it's the name of a set. You could talk about aleph-null
> rather than N, but then you run into problems since aleph-null is not a number
> but a cardinality.
Set sizes ARE numbers, and if you had been paying attention, I am not fond of
alephs, since that is part of a system that is rife with flaws and
contradictions. In Bigulosity, N is the name of the set, the size of the set,
and its maximal element, as determined by context. This distinction between set
sizes and numbers is entirely artificial, and only serves to confound the whole
issue. The point remains that a list of all binary numbers of ANY given length,
finite or infinite, is exponentially longer in numbers than it is wide in
digits.
>
> >
> >>>>That the digits can be indexed by natural numbers is obvious since
> >>>>they obey the laws:
> >>>>
> >>>> For every digit with significance 10^-n, there is a following digit
> >>>> with significance 10^-{n+1}.
> >>>>
> >>>> Digits with different significances have different significances of
> >>>> the following digit.
> >>>>
> >>>> There is a digit with significane 10^-1.
> >>>>
> >>>> The digit with significance 10^-1 has no preceding digit.
> >>>>
> >>>> Any set including the digit with significance 10^-1, and for each of
> >>>> its digits also containing the following digit contains all digits of
> >>>> the number in question.
> >>>>
> >>>>And that is in obvious correspondance with the Peano laws for the list
> >>>>indices. So it is easy to establish a bijection between list indices
> >>>>and digits.
> >>
> >>>yes, but if it is a digital number, then there are S^L of them. If L
> >>>is the same as |N|, and S is 2 (binary, you have a grid N wide, but
> >>>2^N long. In decimal it's worse, being 10^N long. In any case, the
> >>>list is much longer than it is wide, and connot be fully traversed
> >>>diagonally.
> >>
> >>Nonsense. It is equally wide as long since the sequence of digit
> >>positions and the sequence of list positions obeys exactly the same
> >>laws.
> >
> > That is absolutely incorrect. Here is a list of all 3-digit binary numbers
> > 000
> > 001
> > 010
> > 011
> > 100
> > 101
> > 110
> > 111
> >
> > Are there three of them? No, there are eight. Why? Because 2^3=8. Wake up.
> > If you allow N bits, do you get N binary numbers? No. You get 2^N.
> >
> >>The list cannot be "fully" traversed either across, diagonally, or
> >>downward anyway, since it has no end in either direction.
> >
> > The the proof doesn't prove anything at all, does it? What do you think the
> > logical argument IS in that proof? What exactly does it demonstrate, in your
> > mind?
> >
>
> It demonstrates what it is intended to demonstrate. The proof has nothing to do
> with "fully traversing" a list. It is looking at elements of the list, and
> entities constructed from those elements - not one at a time by concatenation,
> but all at once, by definition.
Excuse me, but nowhere here did you actually answer the question, so this
comment is useless, as well as incorrect.
>
> >>>>"Squares" do not come into play here at all, neither do
> >>>>"diagonals". Only some convenient partial visualization shows
> >>>>something akin to a "diagonal", but it is not relevant to the
> >>>>proof.
> >>
> >>>It is absolutely crucial. The conclusion of the proof is that the
> >>>numbers can't be listed, because a diagonal traversal can be used to
> >>>form a number, the antidiagonal, which not on the line of traversal,
> >>>and therefore not on this list.
> >>
> >>The proof does not depend on the name "diagonal", nor is there
> >>anything leading from one corner to another corner. The proof relies
> >>on a 1:1 correspondence/bijection of digits and list places. Both
> >>obey they laws of natural numbers, and so can be put into
> >>correspondence.
> >
> > That is so far off base I can't even imagine what you're thinking. it assumes
> > something that is simply not true, and jumps to conclusions about the
> > significance of a diagonal traversal and inversion of bits, which you think has
> > nothing to do with the diagonal. Maybe you NEED to smoke something.
> >
>
> I'm not sure what assumption you're seeing that is not true.
I repsonded in another post. It assumes the diagonal has covered all numbers in
the list, and that you have generated a number NOT on the list, when in fact,
all you have done is generate a number ON the list, but not covered by the
traversal. Try it with a finite list, as I showed above, and you'll see what I
mean. As the number of digits increases, the elongation only becomes more
pronounced, not less.
>
> David is right in saying that the proof does not depend on a diagonal;
> it can be rephrased without even relying on any geometric analogy, and
> still holds true.
Give it a shot. I'd love to hear that form of convoluted nonsense. I have never
heard any other real version of this "proof". Enlighten me.
>
> >>>This rests on the assumption that the diagonal covers every digital
> >>>number on the list, which is clearly not the case, since the grid is
> >>>longer than it is wide.
> >>
> >>It does not end horizontally, and it does not end vertically. So you
> >>can walk on in diagonal direction (or any other direction) as long as
> >>you want to without exhausting either dimension.
> >
> > But, what does the diagonal traversal and the generated antidiagonal signify to
> > you, if you don't believe the diagonal has exhausted all the numbers in the
> > set, and yet still has some left? That is the whole point of the proof. As far
> > as I can see, you don't know WHAT it's supposed to mean.
> >
>
> I'm not understanding this clause: "if you don't believe the diagonal has
> exhausted all the numbers in the set, and yet still has some left".
> The clause does not appear to be grammatical - what does "some" refer to?
> In what sense is the diagonal supposed to "exhaust" the numbers in the set?
> As I said above, the diagonal, or antidiagonal (the constructed number which
> is nowhere in the N-indexed list) is not created by "traversing" the list.
> It is created by stating a definition of it, which allows one to define any
> given digit on the list. In any order.
If you have generated a number that is already on the list, but has not been
included in the line of traversal from which the antidiagonal has been
generated, so as to be different from all numbers traversed, what does this
prove? The number IS on the list, below the diagonal. So, you have not
generated a number NOT on the list. Get it?
>
> >>>>>Since the list is a list of digital numbers, the length of the list
> >>>>>is ALWAYS greater than the width in digits.
> >>>>
> >>>>Nonsense. The list and the digits are governed by the same laws and
> >>>>can be mapped 1:1.
> >>
> >>>Absolutely incorrect. In binary, there are 2^n strings of length n,
> >>>and in decimal, 10^n. These numbers are always larger than any
> >>>positive n.
> >>
> >>So what? There is no finite n that would limit either length or
> >>breadth. So there is no 2^n to compare with a 10^n.
> >
> > Uh, if you want to claim there are N digits, then you just gave me an n, and
> > there are going to be 2^N strings on the list. Get it? Consider the list of
> > binary strings to be the powerset of the set of naturals, with each bit
> > representing set membership of the natural number denoting the position of that
> > bit in the string. You have the set of all sets of natural numbers. There are 2
> > ^N of them.
> >
> Again, N is not a number, and David is not saying there are N digits. The "n"
> to which he refers is a finite natural number. You don't appear to have a finite
> natural number to substitute into your exponentials; yet these are precisely
> what you must have if your "2^n" and "10^n" are to make sense *as numbers*.
That is simply not true. I was told that the grid extends to infinity in either
direction, equivalent to the natural numbers, so it is basically square for all
intents and purposes, which is false. Do you contend now that Cantor's list
does NOT extend to infinity in both directions? Maybe you and David should work
that out and get back to me, since you seem to disagree.
>
> >>>>You are again putting your "intuition" about the result into the
> >>>>proof as a prerequisite.
> >>>
> >>>No I am applying mathematics, whether you care to admit it or not.
> >>
> >>You are again talking rubbish. For every m there exists an n such
> >>that 2^n is greater than 10^m, and vice versa.
> >
> > Yeah, so? No kidding. What's that got to do with anything? If you are using
> > decimal, then you have 10^N strings, and if binary, 2^N. This is why digital
> > systems are not a reliable way to measure the complexity of the continuum.
> > There is no consistent result.
>
> Once more, N is *not a number*, but the name of a set. Thus, neither "10^N" nor
> "2^N" are numbers - they are undefined expressions, at least until you provide
> a definition for them.
N IS a number, in Bigulosity. I am not playing by your confused rules, but by
rules that make sense, so drop cardinality while discussing this with me.
That's not the topic.

N is the number of natural numbers in the set, N. Any set size IS a number,
finite, infinite, or "transfinite". Why don't you try forming a precise
distinction between transifnite "numbers" and numbers in general. Did Cantor
err by calling the alephs numbers?
>
> >
> >>>>That is simply wishful thinking.
> >>>
> >>>No, it is correct thinking. How many different values can you
> >>>represent with three decimal digits? Three? A thousand? Who's
> >>>blowing smoke now?
> >>
> >>You are. Because "three" is a fixed finite number, and n can take on
> >>any value.
> >
> > For any n-length decimal number, there are 10^n possible values. Three is an
> > example of n. Take any n>0, and try it out. See whether the realtionship
> > changes as you have bigger sets. Ask yourself why this constant equality would
> > suddenly disappear at infinity, and if you don't have a good explanation, then
> > dump that notion.
>
> Of course there's a good explanation. Infinity is not a natural number. If it
> were, then we would have to dump the Axiom of Choice. (There is a very simple
> proof of this, which may or may not have been provided. I can easily provide
> a recapitulation if desired.)
Good. Dump it. It's not necessary. I don't need Cantorian arguments to see
that.
>
> But we've specifically been assuming that we're all talking about standard
> math, which to my knowledge includes the Axiom of Choice.
No, we are discussing the contradictions in what you consider standard math,
which implies that we don't all hold it to be correct.
>
> >
> > At each step in Cantor's proof (binary version), for each n digits traversed,
> > there are 2^n-n binary values of length n that have NOT been traversed. The
> > antidiagonal is simply one of those.
> >
>
> What is this about "traversing" the diagonal? The diagonal need not be moved
> through sequentially to construct the antidiagonal. Indeed, as I stated above,
> the whole proof can be given algebraically, without any reference to sides,
> lengths, diagonals, or other geometric analogies. If you like, I can easily
> provide such a proof.
It still assumes that a digital number has as many possible values as its
length, which is patently false.
>
> >>>>>The proof only shows that this is true, but doesn't prove that no
> >>>>>list can exist in concept, though of course it is infinitely longer
> >>>>>than it is wide. The assumptions used in this "proof" are typical of
> >>>>>proofs in this area.
> >>>>
> >>>>You are talking bullshit. The strawman you try to discredit is
> >>>>entirely of your own invention.
> >>
> >>>Bullshit, as you have just had your nose rubbed in. What kind of
> >>>mathematician contends that three digits can only represent htree
> >>>values in a digital number? Sounds like chicken-scratch math to me.
> >>
> >>Your problem is still that you are confusing "three" with "cardinality
> >>of the naturals". They don't obey the same laws.
> >
> > Pay attention. I used 3 as an example of a number of bits, to show you
> > concretely that three digits does NOT mean three possible strings. If that
> > example wasn't concrete enough for you then you're hopeless.
> >
>
> This will hold for every natural number. However, every natural number is
> finite, whereas the number of items in the list is not finite. Hence, the number
> of items in the list is not a natural number, and there is no reason to believe
> that what is true of natural numbers is necessarily true of other entities.

Yes there is. Why don't you try explaing to me how lim 1->oo (S^n/n)=1 (ratio
of number of strings to number of digits) for nonzero S? I am sorry, but that
limit is not 1, but infinite, meaning there are infinitely many more strings
than digits. Is aleph_1 infinitely bigger in your book than aleph_0? Isn't this
what the proof is supposed to demonstrate? And yet, the proof ignores this fact
while purporting to prove it. The thing it proves is the thing that falsifies
the proof. No wonder Cantor ended up institutionalized, proving things that
invalidate that same proof.
>
> Matt
>

--
Smiles,

Tony
.

Relevant Pages

  • Re: Cantor and the binary tree
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  • Re: abundance of irrationals!)
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  • Re: Cantor and the binary tree
    ... It proves that the infinity of real numbers is greater than the infinity of the digits they have, but it does not mean we cannot list them, conceptually at least. ... As I said above, the diagonal, or antidiagonal is not created by "traversing" the list. ... Consider the list of binary strings to be the powerset of the set of naturals, with each bit representing set membership of the natural number denoting the position of that bit in the string. ... Indeed, as I stated above, the whole proof can be given algebraically, without any reference to sides, lengths, diagonals, or other geometric analogies. ...
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  • Re: abundance of irrationals!)
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