Cantor Confusion
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From: mueckenh on 30 Nov 2006 04:03 Virgil schrieb: > In article <1164818051.564389.180950(a)14g2000cws.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > MoeBlee schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > > > > > > > > > > > > If both, W and N, are countable, then renaming the elements of > > > > > > > > one of > > > > > > > > them leads to an identity map. > > > > > > > > > > > > > > You are confusing bijections within the model to bijections > > > > > > > outside of the model. > > > > > > > > > > > > And who told you that you were outside? > > > > > > > > And who told you that you were outside? > > > > > > > > > You are noting that it is not possible to construct a bijection > > > > > in a nonstandard model and that it is > > > > > possible to construct a bijection in the standard model. This > > > > > is true, however, it is not a contradiction. > > > > > > > > In particular because there is neither a stadard model nor a > > > > non-standard model of ZFC. > > > > > > You haven't proven that there is no model of ZFC. > > > > Mathematics exists nowhere except in the minds of mathematicians. At > > present there is no mind with such a model. So, where should it exists? > > That such a model does not exist in WM's mind is not evidence that it > does not exist in minds competent to comprehend it. Do you know of such a mind? Why can't the model be written down? > > > > > > > And in particular because the expression "contradiction" is not pat of > > > > ZFC. > > > > > > In Z set theory, we can formulate definitions of 'S is a contradiction' > > > and 'T is inconsistent'. > > > > One could, but one wouldn't. Never! > > One has. But defining something does not instanciate it. > > One can define square circles and 4 sided triangles, too. Wrong definitions alltogether, the four-sided triangle as well as the possibility of set theory being proven inconsistent by one of its advocates. Regards, WM
From: mueckenh on 30 Nov 2006 04:05 William Hughes schrieb: > > > So what we have to do to prove that all sets of natural numbers > > > are finite is to assume that there is no set of natural numbers > > > without a largest element. > > > > No. We have to accept thart there are sets which are capable of > > growing, as Fraenkel et al. expess it. Then we have finite sets without > > a largest element. Then we describe reality correctly. > > (Leave Fraenkel et al. out of it. Finite sets without a largest > element are your idea). I did not ascribe those sets to Fraenkel et al. I said, "growing, as Fraenkel et al. express it." And, in fact, they used this expression. > > O.K. Let A be a finite set without a largest element. > What is the cardinal number of A. Clearly it cannot > be any finite cardinal number? It cannot have a cardinal number at all. We have no cardinals in potential infinity. Cantor knew that. > > Recall, you want to do more than just state that A does > not have a cardinal number. You want to show that assuming > that A has a cardinal number leads to a contradiction. Please do not mix up potential and actual infinity. Actual infinity is required, e.g., for the real numbers. They are not potentially infinite series, but actually infinite series. Consider just the paths in the infinite binary tree. In this case I have shown a contradiction. Here it is: The binary tree Consider a binary tree which has (no finite paths but only) infinite paths representing the real numbers between 0 and 1 as binary strings. The edges (like a, b, and c below) connect the nodes, i.e., the binary digits 0 or 1. 0. /a \ 0 1 /b \c / \ 0 1 0 1 .......................... The set of edges is countable, because we can enumerate them. Now we set up a relation between paths and edges. Relate edge a to all paths which begin with 0.0. Relate edge b to all paths which begin with 0.00 and relate edge c to all paths which begin with 0.01. Half of edge a is inherited by all paths which begin with 0.00, the other half of edge a is inherited by all paths which begin with 0.01. Continuing in this manner in infinity, we see by the infinite recursion f(n+1) = 1 + f(n)/2 with f(1) = 1 that for n --> oo 1 + 1/2 + 1/ 4 + ... = 2 edges are related to every single infinite path which are not related to any other path. (By the way, the recursion would yield the limit value 2 for any starting value f(1).) The load of 2 edges is only related to infinite paths because any finite segment of a path with n edges will carry a load of (1 - 1/2^n)/(1 - 1/2) < 2 edges. The set of paths is uncountable, but as we have seen, it contains less elements than the set of edges. Cantor's diagonal argument does not apply in this case, because the tree contains all binary representations of real numbers within [0, 1], some of them even twice, like 1.000... and 0.111... . Therefore we have a contradiction: |IR| > |IN| || || |{paths}| => |{edges}| Regards, WM
From: mueckenh on 30 Nov 2006 04:10 Virgil schrieb: > In article <1164834278.356038.117340(a)80g2000cwy.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > In your terminology every digit 3 in 0.333... represents a number. The > > first 3 represents the number 0.3, the second 3 represents the number > > 0.33 > > In my number system it represents 0.03. > > > > > and the 10th 3 represents the number 0.3333333333. > > In my number system it represents 0.0000000003. Read what Dik has written. > > > What does this > > interpretation have to do with the existence of the decimal > > representations of the number 1/3, namely 0.333... in a list of real > > numbers? > > The way you have stated it, nothing. > > > > > As there is no digit "3" at an infinite distance from the "0." in > > 0.333..., there is no infinite decimal expansion. > > There is that same confusion surfacing again. > > An endless sequence of single decimal digits creates an "infinitely > long" expansion with no digit more than finitely far from the decimal > point. That is my point. (I use this argument, although it is easily to recognize as matheological nonsense.) > > Until WM learns this, his attempts to work with set theory will continue > to fail. > It was Dik who assumed an infinite distance. Please try at least to understand what you read before you answer. Regards, WM
From: mueckenh on 30 Nov 2006 04:21 Dik T. Winter schrieb: > In article <1164816790.379338.139370(a)h54g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > MoeBlee schrieb: > > > > I think if one swells to explosion about his knowledge of set theory, > > > > he should at least know the very foundation. But I know, that you do > > > > not even understand the simple texts of Fraenkel et al. > > > > > > No, YOU radically MISunderstand what Fraenkel, Bar-Hillel, and Levy > > > wrote. > > > > How can you judge about that without the slightest idea of what they > > wrote? Only for the lurkers: Fraenkel et al. write: "Platonistic point > > of view is to look at the universe of all sets not as a fixed entity > > but as an entity capable of "growing", i.e., we are able to "produce" > > bigger and bigger sets." So a set (like the set of all sets) is not a > > fixed entity. > > There is nothing in that that shows that a set is not a fixed entity. The universe of all sets can grow. Define: "The universe of all sets is called the set of all sets", and you see it. > You are able to produce bigger and bigger sets, but they are all > different. That is a matter of definition. If you consider a fixed set then it is fixed. Small wonder. If you consider a variable set then it is variable and perhaps changes its cardinal number. An easy example which should not escape you: The set of states of the EC has been growing and probably will continue to grow. > When the universe has grown it allows bigger sets than > where originally allowed, but the sets originally allowed are still > sets and still the same and did not grow. How you conclude from the > above statement that sets themselves are growing escapes me. It is simply a matter of definition. Regards, WM
From: mueckenh on 30 Nov 2006 04:38
Dik T. Winter schrieb: > > > > In my tree there are no finite (terminating) paths! > > Sorry, as far as I see your tree has also finite paths. I see paths > from the root to the first nodes below the roots. That is an edge. But no path does stop there. Every path is continued, infinitely. > > What is the tree? What are the nodes? You again switch to something > different. I only want to make clear that decimal representations of real numbers and binary representations of real numbers (like the paths of the tree) are not different. > But in a decimal tree, where each node is connected by > edges to ten subnodes and where each of those subnodes is assigned a > decimal digit, it can be shown that also each node represents a > finite decimal number. And in that tree 1/3 is not represented by > any node. But 0.333... is not a path in that tree, just as I wrote > earlier. And, indeed, 0.333... does not represent a number until > that notation has been defined. And I use exactly the same definition, adapted for the case of binary representations. Instead of writing a number like 0.11000... I write this same number 0. 1 1 0 0 0 .... And that is all the difference! I cannot understand how someone familiar with decimal representations of real numbers should have any difficulties with the binary tree. > > > > In your infinite tree, all edges come fome a node > > > at finite distance, and so all edges together only represent the numbers > > > with a finite binary expansion. > > > > So all decimal representations represent finite decimal expansions? Or > > what is the difference? > > The difference is that infinite decimal representations are not defined > as given, there is a concept behind it: limit. Take simply the same concept for my tree: The limit. But whether there is a concept behind or not: Your mentioning of an infinite distance from the root to a node fails completely. In any case there is no digit of a real infinitely far from the decimal point. Why? Its position could not be indexed by a natural number. Its position, therefore, would be undefined. The same is true for the nodes of the binary tree. Its paths are infinitely long but no node has an infinite distance from the root. Do you see your error? > > > > As there is no digit "3" at an infinite distance from the "0." in > > 0.333..., there is no infinite decimal expansion. > > What is the difference to a representations by paths? > > Limits. No. Paths are only another notation for the reals in usual representation and usual definition. Regards, WM |