Cantor Confusion
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From: Virgil on 17 Dec 2006 16:57 In article <1166360707.217376.57470(a)l12g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > Consider a binary tree which has (no finite paths but only) infinite > > > paths representing the real numbers between 0 and 1 as binary strings. > > > > So the paths are not "those which can be obtained by repeated > > multiplication by 2"; it was /given/ by you at the beginning of your > > argument that a path exists for each real number in [0,1]. > > Yes. And obviously the cardinality of the set of these numbers can be > obtained by repeated multiplication by 2, namely 2^aleph0. This is a > repetition, repeated as often as there are natural numbers. As no finite number is then number of natural numbers, that means more than any finite number of times. But in set theoretical notation A^B mean the set of all functions from set B to set A, and card(A)^card(B) = card(A^B). > > > Note that you have already given that there are as many paths in the > > tree as there are real numbers in [0,1]. We are investigating how we > > can construct a surjection of the naturals onto the reals. > > But we do not presuppose that the reals are uncountable. At least we do > not presuppose that ZFC is free of contradictions which may result in > two valid proofs with different results. But WM presupposes things that do contradict ZFC. > > > > Your edges are not > > being broken into 2 "shares" each; they are being divided into an > > infinite number of shares, and that is very different from 2 shares! > > For this sake we have mathematics, I mean really correct mathematics, > not based on dubious infinite sets, which states that 1 + 1/2 + 1/4 + > ... = 2. That is not an equality but a hidden limit statement: Lim_{n --> oo} Sum[k=0..n: 1/2^k] = 2 > > > > The notion "rational relation" is a further development of the notion > > > "relation". While a relation connects elements of a domain to elements > > > of a range, the rational relation connects shares of the elements of > > > the domain to (shares of the) elements of the range. A simple example: > > > > > > > > > A) Consider the relation: > > > > > > > > > 1 ---> a > > > 2 ---> b > > > > > > > Is this drawing supposed to represent a bijection {1,2} -> {a,b}? > > Of course. Then it presumes a bijection and is of no relevance unless some bijection is already presumed. But that is precisely the issue, whether such a bijection exists. So that once more WM begs the question. > > Wouldn't it be simpler to just give a clear /definition/ of what you > > mean by your "rational relation", instead of endless examples? > > > I gave a clear definition at the biginning of this discussion. I did > not give it in the language of ZFC because it seems not possible to do > so. That might be a clue that it does not hold in ZFC. > > > > > I am only interested in your "rational relation" argument. > > > > > > Fine. Why are you interested only in that argument. And why are you > > > interested in it at all? > > > > I am interested in that argument because it has a surface plausibility > > of "mathiness" while being obviously wrong. > > The reason for the differet result is easy to see. In the tree a real > number = path must really exist in order to be accepted. In set theory > the real numbers do not exist. They do in some set theories, such as ZFC and NBG, at least there are models having all the properties of a complete Archimedean ordered field, and there is, valid in both ZFC and NBG, a theorem that all such complete Archimedean ordered fields are isomorphic as complete Archimedean ordered fields. > It is enough to have a sequence of > digits which can be distingusihed from any given sequence. The problem > is exactly the same as in my discussion with William Hughes. He must > insist that in the Equilateral Infinite Triangle > > 1 > 12 > 123 > ... > > the diagonal must be longer than any line, which is obviously contrary > to the definition of a diagonal but required to save the real numbers > in ZFC. It is not at all obvious to anyone who does not assume that the set of all naturals is finite. And such an assumption directly contradicts ZFC and NBG and most other set theories. > > In the tree we require "intuitively" that every path is separarted from > another one not other than by an edge. Every two paths have a non-empty set of nodes in common, and separate at the edges branching from the "last" node (furthest from the root) in that common set. Those common nodes and the edges connecting consecutive pairs of them form an "initial finite subpath","initial" meaning starting at the root node and "finite" meaning that it ends at some terminal node within the tree. In fact each node is the end node of an initial finite subpath.
From: Virgil on 17 Dec 2006 17:01 In article <1166360896.635013.20520(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166184731.253435.242720(a)l12g2000cwl.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > In article <1166092755.336596.309060(a)l12g2000cwl.googlegroups.com>, > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > Virgil schrieb: > > > > > > > > > > > WM deceives himself over the number of lines versus the number of > > > > > > elements in any one line. The number of lines is not finite but the > > > > > > number of elements in any one line is finite. > > > > > > > > > > Each line differs by 1 element from the preceding line. If the number > > > > > of lines is actually infinite then the number of differences must be > > > > > actually infinite too. > > > > > > > > So far so good. > > > > > > > Their sum is an infinite number. If all elements are there, then also > > > all sums are there. All of the /finite/ sums of an infinite series are always there , but the infinite sums are only "there" when the series converges. That does not prevent an infinite series from having infinitely many finite terms and infinitely many finite partial sums, even when it does not converge. > > > > > > > > > If not all sums are there, not all elements are > > > there. > > > > There are infinite sequences that do not converge. > > > > And 1 + 2 + 3 + ... is one of them. > > And 1 + 1 + 1 + ... also is one of them and, in addition, it is an > element of the infinite set. Therefore, in an actually infinte set N, > there must be an actually infinite number. Then 1 + 1/2 + 1/4 + 1/8 + ... must be equally infinite.
From: Virgil on 17 Dec 2006 17:23 In article <1166361248.951652.84540(a)l12g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166185295.562736.84050(a)j72g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > > > > > > No I simply state that the diagonal proof fails in case of the tree. > > > > > > > > AS the diagonal proof is not allied to trees, so what? > > > > > > > Its contradiction is. > > > > Show me! > > If you believe in a diagonal longer than any line of the matrix and in > paths which have no beginning, then I can't show you anything. You have things backwards, as usual. The reason that you cannot show me things is that you have not presented any axiom system on which to base them. When you want to prove something, you merely pull assumptions out of the air to serve as justifications. You cannot show me things within in ZFC or NBG that require you to assume things that contradict the XFC or NBG axioms because in an axim system you are not allowed to make assumptions that contradict the axioms already assumed. > A minimum of logic is required for any arguing. And WM does not have that minumum. > > > > Every finite beginning of a set of paths, however long, is the beginning > > > > of as many paths as the set of all paths. > > > > > > > > So there are as many paths following any such beginning as there are > > > > paths altogether. > > > > > > And there are as many beginnings following such a beginning. > > > > if a beginning is either a node or an edge, there are fewer beginnings > > than unending paths from such a beginning in an infinite binary tree. > > So, most of the paths have no beginning? Uncountably many paths have the same beginning, at least for any finite beginning to any paths. >Or where, do you think, they > begin as separated entities? The single root node is the beginning of all paths. The finite sub-path from the root node to any other node is the beginning of just as many paths. > > > > > > > > > Therefore the beginnings of separated parts of the paths are > > > > > countable. > > > > > > > > > > > > It is the endings that are not countable. > > > > > > It is the endings, which do not exist. > > > > Each path has a head node and a tail consisting of everything else. > > > > The number of heads in an infinite binary tree is countable, the number > > of endless tails from any node is equal to the number of endless binary > > strings which is not countable. > > So how many tails has one head? Uncountably many tails to each head, where a head is the finite sub-path from the root node to any node. > And where begin these mysterious uncountably many tails? At the end node of their head. > > For you it is not a problem that there are more tails than heads and > simultaneously that every head has one and only one tail? Who said a head could only have one tail? By my definitions, a "head" is a sequence of nodes and branches from the root node to any end node and a "tail" is the endless subpath starting at the end node of some head. Thus every path is the tail for the root node as head. For every head, there is a natural bijection of its tails onto the set of all paths. So that the set of tails for any head is always as uncountable as the set of all paths.
From: Virgil on 17 Dec 2006 17:26 In article <1166361677.475397.105840(a)l12g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > The notion "rational relation" is a further development of the notion > > > "relation". While a relation connects elements of a domain to elements > > > of a range, the rational relation connects shares of the elements of > > > the domain to (shares of the) elements of the range. A simple example: > > > > Again, what are "shares"? > > An edge is considered to be a set of shares. Since as many paths pass through each edge as there are paths altogether, every edge must then be shared by infinitely many paths.
From: cbrown on 17 Dec 2006 19:00
mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > Consider a binary tree which has (no finite paths but only) infinite > > > paths representing the real numbers between 0 and 1 as binary strings. > > > > So the paths are not "those which can be obtained by repeated > > multiplication by 2"; it was /given/ by you at the beginning of your > > argument that a path exists for each real number in [0,1]. > > Yes. And obviously the cardinality of the set of these numbers can be > obtained by repeated multiplication by 2, namely 2^aleph0. This is a > repetition, repeated as often as there are natural numbers. You are wandering off topic again. Your proof is supposed to be a /proof/ that the theorem in ZFC that the reals are uncountable /does/ constitute a contradiction in ZFC; becuase your proof constructs a bijection between N <-> R. It is pointless to start by asserting "obviously, the reals are countable" until you have actually /proven/ that the reals are countable. You have claimed that you can use the binary tree and your "rational relation" to prove this. This proof should be independent of any /other/ proof that the reals are countable. > > Note that you have already given that there are as many paths in the > > tree as there are real numbers in [0,1]. We are investigating how we > > can construct a surjection of the naturals onto the reals. > > But we do not presuppose that the reals are uncountable. But if you instead presuppose that the reals /are/ countable, then your rational relation argument is pointless. We start by assuming /only/ that the edges are in bijection with the naturals; and that paths are in bijection with the reals in [0,1]; and that we /know of/ no bijection N <-> R. You are to prove that therefore there is a bijection between the naturals and the reals. > > > > Your edges are not > > being broken into 2 "shares" each; they are being divided into an > > infinite number of shares, and that is very different from 2 shares! > > For this sake we have mathematics, I mean really correct mathematics, > not based on dubious infinite sets, which states that 1 + 1/2 + 1/4 + > ... = 2. > "Really correct mathematics" doesn't simply /state/ that "1 + 1/2 + 1/4 + ... = 2". Mathematics /defines/ what the series of symbols "1 + 1/2 + 1/4 + ... = 2" means, and then /proves/ that it is true from these definitions. Similarly, I ask that you don't simply /state/ "there is a rational relation which is a surjection of edges onto paths". I ask that you /define/ what that statement means; and then /prove/ that that statement is true from those definitions. > > Is this an example of a rational relation? > > > > (C) > > > > 1#1 -> b > > 1#2 -> b > > 1#3 -> b > > > > 2#1 -> b > > 2#2 -> a > > 2#3 -> b > > 2#4 -> a > > Sure. It is an example of a rational relation. In order to prove > surjectivity of f: {numbers} --> {letters}, we have to show that 1#1 U > 1#2 U 1#3 U 2#1 U 2#3 = one element of the set of numbers and 2#2 U 2#4 > is also one element of the set of numbers. The latter is false unless > 2+1 and 2#3 are empty and no furter shares 2#5 etc exist. > Ah! So shares can be "empty"; or presumbaly any non-negative real number size, under the constraints that sum (1#x) = 1, sum(2#x) = 1. And we have a "surjection" if also sum(sizes of shares mapping to b) = 1 and sum(sizes of shares mapping to b) = 1? Is that so hard to say? So the above "drawing" is /not/ sufficient to say "(C) proves a surjection"; we must /also/ know the /sizes/ of the shares, as well as "what they are mapped to", before we can say "we have proven a surjection". > > Does this (C) prove the surjection {1,2} -> {a,b} "is possible, even if > > it cannot be found"? If instead 1#3 -> a, then does it prove it? Could > > you explain why? > > If 1#3 is a share which contains half of 1, and if all the shares of 2 > are of equal size, then we have a surjection. So 1#3 can "contain 'half of' 1". Are shares sets then? Can 1#3 also contain two thirds of 1 at the same time, but not one quarter? > > > Wouldn't it be simpler to just give a clear /definition/ of what you > > mean by your "rational relation", instead of endless examples? > > > I gave a clear definition at the biginning of this discussion. I did > not give it in the language of ZFC because it seems not possible to do > so. > Feel free to point me to such a "clear definition", in ZFC or any other mathematical system. For example, when first you stated: > > > 1#1 ---> a > > > 1#2 ---> a you did not also state "and 1#1 has size 1/2, as does 1#2"; and that lead to some confusion. I'm not asking you to use exclusively the language of ZFC; but it is reasonable to ask you to include /all/ the properties of "shares" in your definition at /one time and one place/, and not scatter random aspects of the definition until your are forced to reveal them through repeated questioning. Clearly, we cannot understand a "rational relation" until we can first understand what "shares" are. Here's what I can deduce from your examples and occasional fragments of statements so far: Given an element x of a set X, we call a set x' "a subdivision of x into m shares" if: (1) m is a natural number. (2) if y is an element of x', it is called "a share of x". (3) there are m distinct shares of x in x'; i.e., card(x') = m. (4) each y in x' is associated with a real number shareSize(y) >= 0. (5) sum (y in x') shareSize(y) = 1. Is that your definition, when m is a natural number? Are there any other properties of "shares" that we need to know about, or does this cover everything you want to say about "shares" (we will address "rational relations" when our knowledge of "shares" is complete)? If so, then what is your definition when m is /not/ finite, e.g., when there is one distinct share of x for each natural number, or when there are "as many shares as there are real numbers"? How should we determine "sum (y in x') shareSize(y) = 1" in these cases? I know how to determine it in the case when m is finite, and I can /guess/ what it means when m =|N|. But what does it mean when m = |R|? How do you define it, without first /assuming/ that |N| = |R|? <snip> > > I am only interested in that argument, because to pursue your other > > arguments regarding |N| = |R| would be a distraction from the argument > > in question. > > Ok. Then let us stay with the rational relation. > > Regards, WM > > PS: I saw that my URL das been posted incomplete again. So I will give > a shorter address for a sample. But the book is written in German. Sadly, I only read English and French (a little). Cheers - Chas |