Cantor Confusion
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From: imaginatorium on 10 Dec 2006 16:26 David Marcus wrote: > imaginatorium(a)despammed.com wrote: > > David Marcus wrote: > > > Tony Orlow wrote: > > > > In any case, I think his objections are specific enough to warrant > > > > some attention. "Calculus XOR Probability" was about the loss of > > > > additive probability measure, when you have an infinite set of equally > > > > likely possibilities, > > > > > > Can you give me a procedure for generating a natural number at random? I > > > can give you a simple procedure using a coin for generating a real > > > number in the interval [0,1] at random. (By "at random", I mean each > > > number is equally likely.) > > > > Me, me, me!!! I can do this one. > > > > Choose a real number in the interval [0,1] at random (you say you can > > do this): now multiply it by Big'un, and Bob's your uncle. (Not you, > > Bob) > > So, if I pick 0.5, then multiply it by Big'un, I get my aunt? Which > natural number does she correspond to? Oh, wait a minute. I don't have > any aunts. Huh? You really are having difficulty with this, aren't you? And it's so utterly elementary. Look, Big'un times 0.5 is obviously equal to Big'un/2. That's it. Halfway from zero to your declared unit infinity. (You didn't declare one? Perhaps not, but Tony has done that for us. See the cranks have so much more flexibility in the withoneboundhewasfreeness dimension.) Brian Chandler http://imaginatorium.org
From: cbrown on 10 Dec 2006 18:32 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > Because your argument is of the form: "There exists a bijection f > > between the naturals and the edges. > > This is evident... I'm glad you agree. > ... and has nothing to do with shares. On the contrary; we will use f to prove that there is a bijection between the shares composing any given path p and N: each path p is composed of a countable number of shares of edges. > > > There exists a bijection h between > > the paths and the reals. > > Again tat is evident ... I'm glad you agree that h exists. .... and has nothing to do with shares. On the contrary; we will use h to prove that there is a bijection between shares of an edge and R: each edge is composed of an uncountable number of shares of edges. > > > There exists a surjection T between edges and > > paths. Therefore, the composition h o T o f is a surjection of the > > naturals onto the reals; contradiction." > > > > I assume you know what "a surjection T from edges onto paths" is, yes? > > It doesn't map 1/2 of an edge to 1/4 of a path! It is a function that > > maps each edge /in the original diagram/ to a path; so that every path > > is in the image of the function. > > Of course. But what I do is somewhat more sophisticated. It requires to > kow that 1/2 + 1/2 = 1 and some stuff that like. "... 1/2 + 1/2 = 1 and some stuff that like" is hardly a mathematical definition, sophisticated or otherwise. Either you have a surjection from edges in the original diagram to paths, or you don't. If you do, you have proven your claim that paths are both countable and uncountable. If you don't, your claim that edges are equinumerous with paths must be proved some other way. <snip> > Every edge is divided into shares, namely in as many shares as there > are paths to which this edge belongs. Sweet Mary! Finally, something resembling a definition! Let B(e) be the set of paths to which the edge e "belongs". Then the set of shares of an edge e is a set of elements called "shares", such that there exists a bijection from these shares to B(e). I suppose that if a share s is a share of edge a and s is also a share of edge b, then edge a = edge b; is that correct? I.e., a "share of an edge" is always "marked" as a share of some /specific/ edge e in the original diagram, yes? If that is the case, I would suggest that we define "shares of edges" as members of the set (paths X edges). We then define "the set of shares of the edge e in the original diagram" as: S(e) = {(p,e) : p in B(e)}. There is an obvious bijection from S(e) <-> B(e); so as required every edge is "divided" into "as many shares as there are paths" to which e "belongs". And if s is a share of edge a, it is not a share of edge b, unless a = b; so a "share of an edge" is uniquely associated with an edge in the original diagram. By these definitions, if edge e belongs to path p, then the share of edge e that belongs to path p is (p,e). Every share of an edge e (p,e) belongs to some path p in B(e). Every path p is "composed" of shares (p,e) where e belongs to p. To anticipate, let us also define "the composition of path p" as C(p) = {(p,e) : p in B(e)} So C(p) contains a unique "share of an edge" for each edge e that "belongs to" the path p. Do you agree with the above definitions of B, shares, S and C? Now, B(e) = {p : e "belongs to" p}. We haven't defined "belongs to" explicitly, but if you mean the "obvious" thing, then the set of paths to which edge e belongs can be bijected with the set of all paths (informally, by appending some unique finite length prefix associated with the edge e to all paths). Next, recall that we have assumed a bijection h : paths <-> R. Therefore, by composition, we have a bijection B(e) <-> paths <-> R for every edge e; and thus a bijection S(e) <-> R for every edge e; in other words, every edge is "divided" into an uncountable number of "shares". Clearly, the cardinality of C(p) is the same as the cardinality of some subset of the edges in the original diagram; and so card(C(p)) <= card(edges). Equally clearly, the cardinality of C(p) is at least countable ("every path is infinite"). Therefore, by the bijection f, it follows that there is a bijection C(p) <-> N; in other words, every path is composed of a countable number of shares of edges. <snip> > > But it is no necessary to divide an edge at all, because ... Let's stick to one argument at a time, shall we? Or have you given up on trying to prove your "rational relation" argument? You now have the following functions whose existence we have established: (1) A bijection f between the naturals N and the edges. (2) A bijection h between the paths and the reals, R. (3) A function B : edges /in the original diagram/ -> sets of paths, where B(e) = { p : edge e "belongs to" path p}, and card(B(e)) = card(paths) = card(R). (4) The set of all "shares" is a subset of (paths X edges); i.e., "s is a share" implies s is of the form (p,e) for some path p and edge e in the original diagram, and where e belongs to p. (5) A function S: edges /in the original diagram/ -> sets of shares, where S(e) = {(p,e) : p in B(e)}; and card(S(e)) = card(R) for all edges e. (6) A function C: paths -> sets of share, where C(p) = {(p,e) : p in B(e)}; and card(C(p)) = card(N). (7) A function g : (paths X edges X N) -> R, with the property that lim n->oo sum(over all edges) g(p,e,n) = 2. In English, these 7 statements correspond to: (1) Edges are countable. (2) Paths can be bijected with the reals. (3) B(e) is the (uncountable) set of all paths to which edge e belongs. (4) Shares (of edges in the original diagram) are denoted (p,e) for some path p and some edge e where e belongs to p. (5) S(e) is the (uncountable) set of all shares of edge e. (6) C(p) is the (countable) set of all shares which belong to path p. (7) 1 + 1/2 + 1/4 + ... +1/2^n + ... = 2 Now, how do we use the above functions to produce a surjection T : edges /in the original diagram/ -> paths? You have previously mentioned "full edges" as distinct from "edges in the original diagram". How do they fit into the picture? What is the role of the function g in your argument? Cheers - Chas
From: Virgil on 10 Dec 2006 18:44 In article <1165778176.600438.234910(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Eckard Blumschein schrieb: > > > > > Then why do you disagree with Cantor's results? > > > > This is a good question. > > I disagree with those of his results which are in error. Then you agree with everything he said about Cardinals. > Cantor knew > how to distinguish potential from actual infinity, that is more than > most discussers here can achieve, but he was not infallible. Neither are you! And Cantor made fewer goofs than you make. > > Regards, WM
From: Virgil on 10 Dec 2006 18:51 In article <1165778477.143980.296870(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > So why did you state that I erronously believed that nodes represent > > numbers? > > Because you erroneously did. (The real numbers in the tree are all > represented by infinite paths like > 3.1000... > 3.14000... > etc.) Then every node represents an infinite path cut off at that node, just as an in finite decimal can be cut off after finitely many digits to give a number. > > > It is easy enough to construct a surjection from the edges to the > > binary rational numbers with terminating expansion. But indeed, WM > > does not show a surjection at all. > > Indeed you are unable to understand fractions? As WM's "construction" requires a unique infinite sequence of nodes or branches for each separate path, he is matching sequences with paths, not individual branches or nodes. And there are uncountably many such sequences.
From: Virgil on 10 Dec 2006 19:00
In article <457c6023(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > There is no number aleph_0 or > > omega counting all natural numbers. > > ========================= > > > > I agree. There is no smallest infinity, and no discernible end to R. That makes 2 in favor and millions against. > > >> Repeating a falsehood does not make it any less false. > > > > Why then did you repeat always that here are no balls in the vase at > > time 0? Why do you repeat without reasonable arguments that there are > > more paths than edges in the binary tree, although no path springs off > > without its own egde? > > ========================= > > > > Again I agree. There are 9 n balls after n iterations, and there are > half as many paths as edges or nodes in the tree, despite supposed > bijection hat tricks. Banach-Tarski is ridiculous, and omega is a phantom. > > >> Why he maintains this this somehow bijects the individual branches with > >> paths, is not clear. > > > > Because every branch consists of two edges, one for the each of the > > resulting paths. > > > > ========================= > > > > To put it another way, each right branch may be considered a > continuation of the same original infinite path, like adding a 0 to the > right of the decimal point - it doesn't change the value. The left > branch from every node produces a new path, creating a new value with > the concatenation of a 1 to the string. Therefore, for every two > branches, there is one additional path. From every branch point (node) there spring exactly as many paths as from the root node, and if WM, or anyone, asserts that the set of all of them is countable, that person owes us a proof, which so far has not been forthcoming. > > Is the diagonal longer than any line? Nope. Which line(s) is it not longer than? > Sure, it's a fun game, for the mystically minded. > > >> I did not claim to have personally shaken hands with each of them, as > >> Ramanujan was rumored to have done, but I know where they live. > > > > You think you know where some of them live. And you know them as well > > as you know the integer [pi*10^10^100] Some of them much better. > > > > ============================= > > > >>> You cannot imagine the integer [pi*10^10^100]. > > > >> Why not, you just did! > > > > No. I did not. Well perhaps WM did not, but he provided a name for it, which is good enough for me. |