Cantor Confusion
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From: Virgil on 29 Dec 2006 04:11 In article <1167372153.122093.156470(a)n51g2000cwc.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > Virgil wrote: > > In article <1167338300.513980.203010(a)a3g2000cwd.googlegroups.com>, > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > Virgil wrote: > > > > In article <1167316205.832225.238540(a)48g2000cwx.googlegroups.com>, > > > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > > > > For NO "levels of an infinite tree" is it true, but it is true for > > > > > > finite trees. > > > > > > > > > > It is true for ALL nodes, blue, green, red, finite, infinite, all > > > > > nodes > > > > > of the infinite path, simply all. > > > > > > > > Then prove it. > > > > > > Proof: > > > 1) lim{n-->oo} (2*2^n - 2)/2^n = 2 > > > 2) for n = 1 we get (2*2^1 -2)/2^1 = 1 > > > 3) (2*2^n - 2)/2^n is a monotonic function of n > > > Hence An(1 <= (2*2^n - 2)/2^n <= 2), that is for all n 1 <= > > > #edges/#paths <= 2 > > > All n means ALL levels and ALL edges in any infinite paths. > > > > > > Actually, "For all n in N, #edges(n)/#paths(n) >= 2", not "<= n". > > > > "For all n" means for all n in some set. and in this case the set is > > the set of finite naturals, N. > > Yes, but there is one-to-one mapping between the naturals and the edges > in each infinite path. Which rebuts nothing. > > > > > But For n not in N, #edges(n)/#paths(n) is not defined, > > And #edges(N)/#edges(N) = 0, if it has any definable value at all. > > > > > > > > > > > > > > I can prove that in an infinite tree there are a countably infinity of > > > > nodes or edges but an uncountable infinite of paths. > > > > > > > > Since every edge terminates in a unique node, there is at least one > > > > node > > > > for each edge. > > > > > > > > One can biject the nodes with the naturals by: > > > > root node <--> 1 > > > > 2nd level nodes <--> 2,3 > > > > 3rd level nodes <--> 4...7 > > > > 4th level nodes <--> 8...15 > > > > ... > > > > nth level nodes <--> 2^n ...(2^(n+1)-1) > > > > ... > > > > > > > > One can biject the the set of infinite binary sequences with the set of > > > > infinite paths by matching each 0 in a sequence with a left branching > > > > edge of a path and 1 with a right branching edge. > > > > > > > > And while there are injections from the set of naturals to the set of > > > > binary sequences, a simple diagonal argument shows that there are no > > > > surjections from the set of naturals to the set of binary sequences. > > > > > > > > Thus there are "more" paths than nodes (or edges). > > > > Proof not refuted!
From: mueckenh on 29 Dec 2006 06:38 Virgil schrieb: > In article <1167255246.090261.62550(a)h40g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > > I did not calculate |N|/|R| but edges per path. This calculation yields > > a real number for any finite path - and the infinite path is nothing > > else but the union of all finite paths. > > But what happens for every finite path need not be true for an infinite > path. For example, every finite path has a root node at one end and a > leaf node at the other, but this is not true for the union of all of > them. > > > What is your definition of an infinite path? You feel that it is not > > the union of all finite paths but something much much greater and > > longer, more exciting and most mysterious? Do you have a definition how > > or any reason why the infinite path is not the union of all finite > > paths? > > One can, indeed, view an infinite path as the union of a nested sequence > of finite paths, just as one views the first limit ordinal as the union > of all finite ordinals, but one must be careful about presuming that > properties of the finite sets carry over to the infinite union so > obtained, as not all properties carry over. One need no other observation at all but this: Every bunch of paths is distinguished from another bunch of paths by a special edge. If individual numbers with infinitely many digits in fact exist, then they are represented by bunches with only one element. If such bunches with only one element do not exist, even in an infinite tree, then there are no numbers with infiniely many digits which can be distinguished from others. In short: Irrational numbers do not exist, but only approximations of such "numbers" (which better would be called ideas in order to express the impossibility to apply the diagonal argument to them). You would avoid inconsistencies like those with the edges and paths in the tree. If the tree is the union of all finite trees, is there a path in it representing an irrational number? If so, does this path differ from any path representing a rational number? Does the unions of all rational numbers include all irrational numbers? Notice: there is no irrational number SUM{n= 1 to k} a_n * 10^-n for any k with 10^-k > 0. This shows that Cantor's diagonal agument fails. Regards, WM
From: mueckenh on 29 Dec 2006 06:43 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. Take the union of all these rational trees. It contains all paths representing rational numbers. What distinguishes this union of rational trees from the complete tree? Nothing. 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. Therefore the union of all rational paths contains all irrational paths. That is the reason why the cardinal number of the paths in both trees cannot be 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). Regards, WM
From: mueckenh on 29 Dec 2006 06:48 Dik T. Winter schrieb: > Where is the contradiction with what I wrote? ZFC is not sufficient > to give a definable well-ordering, but a definable well-ordering is not > inconsistent with ZFC. Why don't you simply give the formula for the definable well ordering? The formula could be evaluated, and, step by step, we would get a list of all reals. A list of all reals would not be inconsistent with ZFC??? > > > > it follows from ZFC+V=L that a particular formula > > > well-orders the reals, or indeed any set." > > > > Instead of casting assertions: Why don't you simply give the formula > > for the definable well ordering? The formula could be evaluated, and, > > step by step, we would get a list of all reals. > > You are too lazy to look it up? If I understand it well enough, V=L > immediately gives a well-ordering of the reals. > > But whatever, you stated that a definable well-ordering of the reals would > lead to a contradiction in ZFC. But you have nowhere shown any proof of > that, and it is false. It has been proven that a definable well-ordering > of the reals is consistent with ZFC. Why don't you simply give the formula for the definable well ordering? The formula could be evaluated, and, step by step, we would get a list of all reals. Regards, WM
From: mueckenh on 29 Dec 2006 07:02
Dik T. Winter schrieb: > > > It has been > > > proven that *every* Cauchy sequence yields a real number. And it is > > > easy to see that *every* decimal expansion is in fact a Cauchy sequence. > > > 10^-n is *never* zero. You need limits. > > > > 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. > > > > > 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. > 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. > > > > > Otherwise the path > > > > does not exist at all. > > > > > > That is what you are trying to prove. But you can not just state it. You > > > hav to *prove* it. > > > > I need not prove that two entities which have no different property, > > cannot be distinguished. > > 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. > > > > > The infinite sum 1 + 1/2 + 1/4 + ... yields 2, although there is no > > > > smallest term 2^-n. > > > > > > Indeed. > > > > > > > It can be applied to calculate the edges per path. You should know that > > > > absolutely converging series can be reordered. > > > > > > 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. > > > > And that is what you are doing. Because that would result in a series > > > without start. > > > > 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. > > > > > > But in the complete tree > > > > >the parts of edges that are assigned to an infinite path are not 1, > > > > >1/2, 1/4 etc. There is *no* edge that is completely assigned to a > > > > > path, > > > > > > > > Then the tree is incomplete. You fail to understand infiniteness > > > > already on this low level? > > > > > > Circular reasoning abounds here. Care to give a *proof* that the tree > > > is not complete in that case? This statement is completely similar to > > > the statement that the completed set of natural numbers should have a > > > last element. > > > > 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 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. > > > > > > Note that in each finite tree, the complete edge assigned to a path is > > > > > the last edge on the path. As there is no last edge in an infinite > > > > > path, this does not hold in the infinite tree. > > > > > > > > So there are different paths without different edges? > > > > > > 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. Regards, WM |