Cantor Confusion
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From: Dik T. Winter on 29 Dec 2006 20:22 In article <1167392633.967611.252860(a)n51g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Apparently your limits work only for real numbers in lists but no for > > > real numbers in trees. > > > > Oh, they do. If the trees are properly defined. > > Imagine a tree which contains only paths of rational numbers in > infinite representation, i.e., ending with a period of 000... or > 111.... > > The first rational tree contains only the paths 0.000... and 0.111... > > 0. > / \ > 0 1 > / \ > 0 1 > ............. > > The second tree contains the paths 0.000..., 0.0111..., 0.1000..., and > 0.111... > > and so on until all rational numbers are represented. At what point is 1/3 represented? > Take the union of all these rational trees. It contains all paths > representing rational numbers. No. It contains all paths representing rational numbers for which the denominator is a power of two. > What distinguishes this union of rational trees from the complete tree? That some rational numbers are missing. If the union contains all rational numbers, there must be one of the rational trees that contains (for instance) 1/3. There is none. > There is no node and no edge of the complete tree which is not > in the union of all rational trees. As a path is an ordered set of > edegs (and nodes), two paths are not different unless they differ by an > edge (or node). Hence, there is no path in the complete tree which is > not in the union of all rational trees. Where is 1/3? At what point does it get into the union of rational trees? > That is the reason why the cardinal number of the paths in both trees > cannot be different. That is the reason why the cardinal number of the paths in both trees *is* different. > But if we subtract the rational paths from the complete tree, then > nothing remains but the empty set which obviously is the set of all > paths representig irrational numbers. > > This shows that irrational numbers are nothing (but a chimera). And apparently a lot of rational numbers are nothing? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 29 Dec 2006 20:46 In article <1167394001.555223.74290(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1167256678.283824.142350(a)42g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > > Yes. So it is essentially different from a finite path. The same > > > > with your trees, Each finite tree has a natural number as maximum > > > > level and has no infinite paths. The completed infinite tree has > > > > no maximum level and has infinite paths. So any conclusion you > > > > can make from the finite trees does not necessarily hold for the > > > > infinite tree. > > > > > > We conclude only that one can follow a path infinitely, and that always > > > there is an edge which continues it. There is no continuation without > > > an edge. ok? That is unavoidable and it is enough to prove that there > > > are not less edges than paths. > > > > Well, in that case, pray, show a proof. > > Follow a path. Look whether it splits somewhere without needing an > edge. That is not a proof of your statement "there are not less edges than paths". I ask for a proof. > > > Yes. That shows that there are no infinite sequences of edges (or > > > digits) which individually represent nunmbers like 1/3 or ideas like > > > irratinal numbers. > > > > Why? > > Because non-distinguishable sets are not different sets. You just > proved that there is no element of a path which distinguished it from > every other path. Right. But that does *not* mean that it can not be distinguished from every other path. Quantifier dyslexia again. > > > Not at all. Forget about all finite trees. In the infinite tree there > > > is always an edge which continues a path. There is no continuation > > > without an edge. That is unavoidable and it is enough to prove that > > > there are not less edges than paths. > > > > In that case: prove it. As I have already shown, all edges can be made > > to represent the rational numbers where the denominator is a power of 2. > > That is because all edges are at a finite distance from the root. > > What distinguishes this union of rational trees from the complete tree? I answered that already in another article. Posting it twice does not make it more correct. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 29 Dec 2006 20:40 In article <1167393772.865867.58640(a)73g2000cwn.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Yes. And they say: As long as 10^-k is not zero, the sum {n=1 to k} a_n > > > + 10^-n is not an irrational number. > > > > Right. And as 10^-k is never zero, that sum is never an irrational number. > > You need limits to get irrational numbers. > > The infinite binary tree contains this limit. That's why he tree is > called infinite. Yes, so what? > > > > > There is no edge in the *finite* domain which is "passed in > > > > > full". In the infinite series there is an edge passed in full. > > > > > > > > Which edge is passed in full to the path that goes at each step to > > > > the left? > > > > > > I don't claim the existence of this representation of 0. > > > > Do you claim that is an answer to my question? > > Yes. There is no acually infinite sequence of digits (or anything else) > which can be distinguished from any other infinite sequence. Strange. You state that in the infinite tree for each path has an edge that is passed in full. When I ask you what particular edge is passed in full by a particular path you claim something about not claiming something about a representation. I ask you to substantiate your claim that for each path there is an edge that is passed in full. Especially for the path that goes to the left at each step. I am not talking about representations. > > Again, re-read my question. > > I am not talking about representations, I am talking about paths. You were > > talking about paths. Infinite paths. Hence my question. > > A path is a representaton as well as any decimal or n-ary number. Does not matter. Your claim was about edges and paths, my question was about a specific path. But apparently you decline to answer the question. > > Your statement was that if there is no edge in the infinite series that is > > passed in full (meaning that the edge does belong only to that path), the > > path does not exist. That is what you have to prove. You again fail to > > do so. > > A path is an (ordered) set. If there is no element distinguishing it > from every other set, then it is not different from every other set. Each path can be distinguished from each other path by some edge. There is *no* edge that distinguishes a particular path from all other paths. Back to that again. Quantifier dyslexia. Similar to: each natural number written in decimal can be distinguished from each other natural number by some (actually in most cases many) decimal digit. For each natural number there is *no* decimal digit that distinguishes it from all other decimal digits. > > > > You should know that you can *not* reverse absolutely converging > > > > series. > > > > > > Proof? > > > > The infinite series has no last term, so the reversal does not have a first > > term, and so is not an infinite series. > > But it has the same value. Why? Proof? > > > No reordering of an absolutely converging series changes its sum. > > > > But reversal makes it something quite different. What is the sum of the > > first 10 terms of the reversal of the above series? What is the sum > > of the first 20 terms? > > Not defined. What is defined is the sum of all terms. That's enough. How do you define that? With your own particular definition? Is that definition consistent? > > > There is no largest element, but the infinite path is nothing than the > > > union of all finite paths. Therefore we can calculate the union of all > > > finite sequences 1 + 1/2 + 1/4 + ...+1/2^n. > > > > Interesing. How do you calculate the union of sequences? > > By taking the limit n --> oo. (The union of sequences is only a > *slightly* sloppy expression.) But that union of sequences does actually not cover the path that leads to 1/3. > > But again my > > question: care to give a "proof" that the tree is not complete if there > > is no edge that is completely assigned to a path? > > The tree is as complete as a tree an be. The lacking personal edge > shows the non-existence of irrational numbers. And a host of rational numbers. But even of the tree of rational numbers ending in either 111... or 000... there is no personal edge for any of those numbers. A lack of personal edges abounds in your tree. > > > > No. I never did state that. > > > > > > So at least one path has its own edge? > > > > No, I never did state that. And it is false. > > Correct. Even in an infinite tree this is wrong, because there are no > irrational numbers. Irrational. In its second meaning we see: Nomen est > omen. And a host of rational numbers also do not exist. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 29 Dec 2006 21:00 In article <1167394330.729252.306460(a)k21g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > If the number > > > of all paths is a meaningful notion, then half of it is a meaningful > > > notion too. > > > > Oh. How do you *define* aleph_0 / 2? > > I do not define aleph_0. No, but that is the number of all paths in one of the trees. Apparently you think it is not meaningful if one half is meaningless. That depends on you interpretation of the word "meaningful". > > > > The point is that you think that also in the infinite tree you can > > > > assign shares of edges as 1 + 1/2 + 1/4 + ... to a path. But that > > > > is false. > > > > There is *no* edge that is shared by finitely many paths. > > > > > > So there is no irational number? > > > > Can you give your reasoning behind this statement? > > See the union of all rational trees jus posted. It implies that the > union of all rational numbers contains all irraional numbers. By the reasoning in that article also numbers like 1/3 do not exist. Apparently you have no idea how an infinite union is defined. > > > But even without terminal nodes, the paths consist of edges. Where no > > > edges are, there is no path. Even "in the infinite" that is true. The > > > number of paths cannot "overtake" the number of edges "in the > > > infinite". > > > > Why not? > > It is the same as with the set {2,4,6,..,2n} the cardinality of which > can never overtake all numbers contained in it. If you don't see it, > there is no help. Again assuming that what holds for the finite case also holds for the infinite case. You apparently fail to see that that is not necessarily true. > > > > Note that you conclude things about the infinite here based on > > > > the finite. > > > > > > We cannot conclude anything other than based on the finite. > > > > So you think. > > and Leibniz, for instance. Eh? You mean the person who invented infinitesimals but was not able to define them so that he only gave the rules using them and did show that they worked, without any justification? The man who invented the word "transcendental"? Well, I understand why he is your favourite. You also almost never come with proper definitions, only examples, just like Leibniz. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 29 Dec 2006 21:05
In article <1167429289.476619.27000(a)48g2000cwx.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > Further, the algorithm for 0.010101... works only as long as you can > determine whether the number n of the level is even or odd. For certain > natural numbers like the first 10^100 digits of pi this is impossible. Again, a finitistic view. There are algorithms that do calculate that number. That they can not be implemented in a finite world is of no concern to the mathematician. On the other hand, there *is* a natural number that is equal to that value. But apparently you also eschew the proof that li(x) and pi(x) cross each other infinitely often, because it is impossible to even calculate the first cross-over. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |