Cantor Confusion
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From: mueckenh on 29 Dec 2006 07:06 Dik T. Winter schrieb: > In article <1167256678.283824.142350(a)42g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > 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. > > > > For each two paths you can identify an edge where they separate. But there > > > is *no* edge that separates (for instance) the path leading to 1/3 from > > > all other paths. And through each and every edge run infinitely many > > > paths. > > > > 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. > > > > You are extending conclusions valid for finite trees to the infinite tree > > > without proof, it is just your intuition. > > > > 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? 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 07:12 Dik T. Winter schrieb: > In article <1167256922.435883.99380(a)a3g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > What is an infinite path? > > > > > > Why do you not answer my question? (Note: rood -> root.) > > > > The second left branch is assigned to half of all paths. Error (slight): The second left branch is assigned to quarter of all paths. > 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. > > > > 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. > > > > In the finite case all paths end at a terminal node. So when you compare > > > number of edges to number of paths at finite levels you are comparing > > > edges to paths with terminal nodes. > > > > 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. > > > > 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. Regards, WM
From: Virgil on 29 Dec 2006 15:10 In article <1167392332.648616.111760(a)s34g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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, Every infinite path determines an endless binary numeral between 0 and 1 by identifying its branchings in one direction with 0 digits and branchings in the other direction with 1's. > then they > are represented by bunches with only one element. Is WM claiming that through a given edge passes only one path? If so, he is even further from reality than I thought. > 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. WM is claiming that if more than one path passes through a given edge, then no two distinguishable infinite paths can exist. WM has become completely out of touch with the reality of infinite binary trees. > > If the tree is the union of all finite trees, is there a path in it > representing an irrational number? Yes! > If so, does this path differ from any path representing a rational > number? It differs from all of them. > Does the unions of all rational numbers include all irrational > numbers? What does WM mean by the "union" of two or more numbers, rational or otherwise? > > 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. Consider the decimal analog of the infinite binary tree in which there are 10 child nodes for each parent node. Every infinite path in such a tree represents a real number between 0 and 1, inclusive, including all the irrationals once and some of the rationals twice.
From: Virgil on 29 Dec 2006 15:24 In article <1167392633.967611.252860(a)n51g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Except that it is easy to construct a path in the complete tree not in any member of the union, and therefore not in the union itself. x = 0.101001000100001000001.... cannot be in any member of that union. Alternately, x = sum{n=1..oo} 1/(2 ^ ((n^2+n)/1) > There is no node and no edge of the complete tree which is not > in the union of all rational trees. But there are lots of paths which are not in that union. E.g., given any f:N -> N which is a strictly increasing polynomial function of degree greater than 1, then y = sum 1/2^f(n) is not in your union. > 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. WRONG! See above for specific examples of transcendental numbers corresponding to paths not in your tree. > Therefore the union of all rational paths contains all irrational > paths. WRONG! See above. > That is the reason why the cardinal number of the paths in both trees > cannot be different. WRONG! See above. > > 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. WRONG! See above. > > This shows that irrational numbers are nothing (but a chimera). WRONG! See above.
From: Virgil on 29 Dec 2006 15:47
In article <1167393772.865867.58640(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > There is no acually infinite sequence of digits (or anything else) > which can be distinguished from any other infinite sequence. As any two actually infinite sequences f binary digits, if they differ at al, will differ at a finite "first" digit position, in any comparison of two values one is only dealing with finite digit positions. That WM claims HE is incapable of dealing with that is his problem, not that of mathematics. > > 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 only size comparison (order relation) definable on naturals, rationals, reals, or paths compares them one pair at a time. One can distinguish one path in any tree from ANY other path in that tree by finding an edge contained in one but not both. Such an edge, if it exists, will always be only finitely distant from the root node. > > > > 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. Not so, as the reversal is not defined as an infinite series, there is no definition of "convergence", thus no 'value" > > Not defined. What is defined is the sum of all terms. Not so. Limits of series are determined by the finite sums. No finite sums, no limit. > The tree is as complete as a tree an be. The lacking personal edge > shows the non-existence of irrational numbers. WMs "union" does not contain the path which branches right at the n^2 level node and left otherwise. So that, by specific example, WM's tree is proved incomplete. |