Cantor Confusion
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From: mueckenh on 1 Dec 2006 06:54 cbrown(a)cbrownsystems.com schrieb: > > 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. > > So your relation is supposed to be a function mapping edges -> sets of > paths, correct? Yes. But the mapping is not the usual one (one edge --> one path). The edges are subdivided in shares. > > > 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. > > Why is only 1/2 "inherited"? Why not 1, or 1/3, or any other number? Because each of the two sets of paths (such paths which begin with 0.00 and paths which begin with 0.01) shall inherit the same share. This is the most simple way to execute justice. > Isn't "inheriting" a different function than the original relation > (which was presumably a function from edges to sets of paths)? Yes, it is a new idea not yet entertained. I know this because this idea leads to a contradiction in ZF and hence would have become popular already. > How is > it that now we have a function mapping (paths X edges) -> Q? I am not sure what you mean here. We have a function f mapping the paths onto the real numbers of he interval [0,1]. The domain of f is the set of all infinite paths. The function is not injective because some real numbers have two representations. But the function is surjective. > > > 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. > > Your notation is confusing. As you have described it, edges are not > related to paths; they are related to /sets/ of paths. The best I can > understand by "edge e is related to path p" is that p is in the set of > paths related to e; but clearly for any path p of length greater than > 2, there are more than 2 edges related to p by this interpretation. > > To clarify, can you indicate which 2 edges are "related" to the path > which represents the real number 1/3, which are not related to any > other path? > The binary representation of 1/3 is 0.010101... So the edges related to this set are the limit of the series which has the following partial sums of edges (compare the sketch below): a a/2 + b (a/2 + b)/2 + c ((a/2 + b)/2 + c)/2 + d .... Since every edge counts one unit, the limit is 2. 0. /a \ 0 1 / \b / \ 0 1 0 1 /c 0 \d .......................... Regards, WM
From: mueckenh on 1 Dec 2006 07:09 Dik T. Winter schrieb: > In article <1164878462.838895.276420(a)n67g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > 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. > > And that is the confusion. If it is a set it cannot grow, but as Fraenkel > et al. do not define it as a set it is allowed to grow. They talk about the set of all sets, but cannot use it for the well known reasons. > They do not state > that a set can grow because they do not state that the universe is a set. Here is a book on ZF set theory: Karel Hrbacek and Thomas Jech: "Introduction to Set Theory" Marcel Dekker Inc., New York, 1984, 2nd edition. they write: The letters X and Y in these expressions are variables; they stand for (denote) unspecified, arbitrary sets. Therefore we can denote a set by X and we can say that the set X grows. That is nothing else than to say that the number of states in the EC grows. Of course the number 6 has not gown to 25. But it is simply a matter of definition, how one interprets "to grow" and "number". > > > > 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. > > That is not "the set of states". You can talk about "the current set of > states" or about "the set of states in 1957" or whatever. At least > mathematically. In mathematics, by definition, a set can not grow. Wrong. Read my explanation above. > You are, of course, entitled to use another definition, but that will > not clarify the discussion at all (and you are not using standard set > theory). Hrbacek and Jech teach standard set theory including the fact that in ZF everything is a set. > > > > 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. > > Yes, with your definition a set can grow, but you put yourself outside > set theory, and you must at first consider all results from set theory > unproven theorems in your theorem, and you need to prove them (if > possible). Hrbacek and Jech teach standard set theory including the fact that in ZF everything is a set. Regards, WM
From: mueckenh on 1 Dec 2006 07:18 Dik T. Winter schrieb: > I see paths from the root to the nodes below the nodes below the tree. There is no "below the tree", since there is no "after the last digit" in decimal or binay representations of real numbers. > > 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. > > I have difficulty with the tree because your explanations are confused > and sometimes contradictionary. Which one? > > > > 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? > > What has this to do with everything else? You stated that there must be an infinite distance between the root and some node in an infinite tree. That is as wrong. I simply explained why. > > > > > 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. > > You state that you are using limits with your infinite paths? Of course. The paths are nothing else but another way of denoting a real number in binary representation. Regards, WM
From: mueckenh on 1 Dec 2006 07:21 Dik T. Winter schrieb: > In article <456EEA86.20001(a)et.uni-magdeburg.de> Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> writes: > > On 11/30/2006 3:19 AM, Dik T. Winter wrote: > ... > > > There is nothing in that that shows that a set is not a fixed entity. > > > You are able to produce bigger and bigger sets, but they are all > > > different. > > > > So far I recall "the set IN of naturals is countable". > > I understood set theory means a hypothetical (alias fictitious) set off > > all natural numbers. > > Oh, for once, try to talk mathematics. By the axiom of infinity the > set of all naturals is neither hypothetical nor fictitious. You mean that by stating the axiom this fictitious set gains some reality? It is assumed to exist. That's all. "Ficititious" is an adjective describing its state extremely well. Regards, WM
From: mueckenh on 1 Dec 2006 07:26
stephen(a)nomail.com schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. > > So you do not know what a path is, or a tree apparently. First you should know: There is no final judgement what a path is or a tree. > In a tree, there exists exactly one path between any two nodes. That may be your definition and it may be useful for some purposes. But here it would confuse the matter. In the binary tree as I use it there is no finite path and there is no path which connects nodes in horizontal direction. Regards, WM |