Cantor Confusion
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From: hagman on 30 Dec 2006 10:42 mueckenh(a)rz.fh-augsburg.de schrieb: > cbrown(a)cbrownsystems.com schrieb: > > > Have you given up on your "rational relation" proof that |N| = |R|? > > No, why should I? It is correct. (A series with a first but no last > term can be reversed to have a last but no first term - without oosing > its value.) > > But it is easier to see, and should be visible even for such as you, > that the tree built from the union of all finite trees (with > representations of rational paths) is the same as the complete infinite > tree. > > Regards, WM and while a finite binary tree with n levels has 2^n-1 vertices 2^n-2 edges 2^(n-1) leaves 2^(n-1) paths starting at the root and ending in a leaf, the union of these finite trees is an infinite tree with |N| vertices |N| edges 0 leaves |R| paths starting at the root and dashing off to nirwana. What shall this demonstrate? a) Because #edges >= #paths for all finite subtrees, the same relation holds for the union (i.e. |N|>=|R|)? b) Because #paths=#leaves for all finite subtrees, the same relation holds for the union (i.e. |R|=0)? c) Because #vertices<2*#leaves for all finite subtrees, the same relation holds for the union (i.e. |N|=0)?
From: Lester Zick on 30 Dec 2006 12:37 On Sat, 30 Dec 2006 02:00:09 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >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. And what exactly makes your definitions quite so proper? The fact that you can't even claim they're true? What is this, a joke? You complain about Leibniz? Talk about the mouse that roared! ~v~~
From: Virgil on 30 Dec 2006 15:14 In article <1167492220.138771.111750(a)h40g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > Have you given up on your "rational relation" proof that |N| = |R|? > > No, why should I? It is correct. (A series with a first but no last > term can be reversed to have a last but no first term - without oosing > its value.) How does one "sum" a series with no first term? There is no place to start! > > But it is easier to see, and should be visible even for such as you, > that the tree built from the union of all finite trees (with > representations of rational paths) is the same as the complete infinite > tree. In the union of all finite trees, there are no infinite paths at all, since to be represented in the union, a path would have to be in one of the finite trees, ergo a finite path. In the same way, the union of any family of sets of finite naturals does not contain anything but finite naturals.
From: Virgil on 30 Dec 2006 15:28 In article <1167492502.295005.312080(a)i12g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > Dik T. Winter wrote: > > > In article <1167393772.865867.58640(a)73g2000cwn.googlegroups.com> > > > mueckenh(a)rz.fh-augsburg.de writes: > > > > > > 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. > > > > Even more trivially: let A = {a,b}, B = {b,c} and C = {c,a}. For each > > set, there is no element that distingushes it from the other sets; yet > > for each set can be distinguished from the others. > > If you want to talk to the topic presently discussed, then you should > recognize the elementary constraints: Paths are linearly ordered sets. > Your example misses the clue. If one is talking of the internal ordering, let C = {a,c} with the usual alphabetical ordering on {a,b,c}, and A, B and C are linearly ordered in the same sense as paths. If one is talking about an order relation of the set of all paths, one can define it by P1 < P2 if and only if at the first node from which they differ, P1 branches left and P2 branches right. But in this case, there is no single node or branching which distinguishes any path from /all/ others, at least in an infinite tree. In fact, there is no /finite/ set of branchings that distinguishes one path from /all/ others in an infinite tree. WM's inability to grasp the nature of infinite trees is because of his inability to grasp the nature of infiniteness at all.
From: Virgil on 30 Dec 2006 15:44
In article <1167493176.710056.286330(a)s34g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > 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? > > In the binary tree it is not represented, but it could be represented > by using a ternary tree. It is not represented in any finite binary tree, but it is trivial to represent it in an infinite binary tree. > But this is the same with decimal numbers. There is no representation > of 1/3, unless it were in the infinite. But if you insist on an > existing decimal representation of 1/3 in the infinite, then also the > union of all finite binary trees contains such an existing binary > representation of 1/3. A union of finite sets of finite naturals does not contain as a member anything but finite naturals. In the same way, a union of finite trees, each of which contains only finite paths, does not contain any but finite paths. WM just doesn't understand infiniteness at all. > > > > > 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. > > Every decimal number gives a power of 10. We are discussing binary trees, not decimal trees. > > > > > 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. > > My question is only the following: Do you think that the union of all > rational binary trees is the complete infinite binary tree or not? I am not sure what WM means by a "rational binary tree". The union of all finite binary trees does not include any infinite binary tree. Since every path in every finite binary tree is finite, the union of all such trees can contain only such finite paths. A union cannot contain anything that is not a member of at least one member of that union. That is true in every version of set theory. > > > > > 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? > > > My question is only the following: Do you think that the union of all > rational binary trees is the complete infinite binary tree or not? Clearly not. Finite binary trees contain only finite paths. So unions of finite binary trees can contain only finite paths. At least in a sane world. > > > This shows that irrational numbers are nothing (but a chimera). > > > > And apparently a lot of rational numbers are nothing? > > A lot of binary representations, yes. But every rational number has > some representation - in contrast to irrational numbers. In a binary tree, sqrt(2) is as representable as 1/3. > > Regards, WM |