Cantor Confusion
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From: Han de Bruijn on 21 Dec 2006 04:06 William Hughes wrote: > The statements are > > A "potentially infinite set" is a function on sets. > > A potentially infinite set takes on the values true and false. > > Any claim that the putative domain and range are not clear > is playing word games. On the contrary. The domain "on sets" of your function is not defined. > If you have a substantive objection (e.g. one cannot define > such a function because you can only define functions on sets, > and the set of all sets of natural numbers does not exist) > then the answer is "No". > > If, however, you wan't to answer this question No, but refuse > to do so because you feel that answering it No would make > you look stupid, I do not feel for you. > > Does a potentially infinite set exist? I have another definition that I can handle, but I can not handle yours. Han de Bruijn
From: William Hughes on 21 Dec 2006 06:09 Han de Bruijn wrote: > William Hughes wrote: > > > The statements are > > > > A "potentially infinite set" is a function on sets. > > > > A potentially infinite set takes on the values true and false. > > > > Any claim that the putative domain and range are not clear > > is playing word games. > > On the contrary. The domain "on sets" of your function is not defined. Word games. Clearly you had no difficulty determining what was intended. > > > If you have a substantive objection (e.g. one cannot define > > such a function because you can only define functions on sets, > > and the set of all sets of natural numbers does not exist) > > then the answer is "No". > > > > If, however, you wan't to answer this question No, but refuse > > to do so because you feel that answering it No would make > > you look stupid, I do not feel for you. > > > > Does a potentially infinite set exist? > > I have another definition that I can handle, but I can not handle yours. > More word games. For some reason you do not wish to give an actual reply like: No, this would require defining a function on all sets which cannot be done. My preferred definition is... - William Hughes
From: Han de Bruijn on 21 Dec 2006 06:45 William Hughes wrote: > Han de Bruijn wrote: > >>William Hughes wrote: >> >>>The statements are >>> >>> A "potentially infinite set" is a function on sets. >>> >>> A potentially infinite set takes on the values true and false. >>> >>>Any claim that the putative domain and range are not clear >>>is playing word games. >> >>On the contrary. The domain "on sets" of your function is not defined. > > Word games. Clearly you had no difficulty determining what > was intended. > >>>If you have a substantive objection (e.g. one cannot define >>>such a function because you can only define functions on sets, >>>and the set of all sets of natural numbers does not exist) >>>then the answer is "No". >>> >>>If, however, you wan't to answer this question No, but refuse >>>to do so because you feel that answering it No would make >>>you look stupid, I do not feel for you. >>> >>>Does a potentially infinite set exist? >> >>I have another definition that I can handle, but I can not handle yours. > > More word games. For some reason you do not > wish to give an actual reply like: > > No, this would require defining a function on all sets which cannot > be done. My preferred definition is... the Naturals Construction Set, as I've indicated before: http://groups.google.nl/group/sci.math/msg/13795822737a77ca?hl=en& BTW, you seems to know very well what my objections would be. Han de Bruijn
From: Albrecht on 21 Dec 2006 06:59 On 20 Dez., 14:55, Franziska Neugebauer <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > Albrecht wrote: > > There is not set N. Should say: There is no set N. (Is your English as weak as you are unable to interpret the wrong spelling?) > Adieu! > > F. N. > -- > xyz
From: mueckenh on 21 Dec 2006 14:07
Virgil schrieb: > > So there are not less real numbers in the interval [0, 1] > > than paths in the binary tree. > WM's sloppy logic again fails him! > What WM has argued justifies "there are not less paths than > reals", which is not what he is claiming, but not "there are not > less reals than paths", which is what he is claiming. Yes, I made an error, the "not" is wrong in this text. However, why did you snip the next line? You would have seen the correct statement in mathematical form: aleph_0 = |{edges}| >= |{paths}| >= |{reals of [0,1]}| = 2^aleph_0 Or did you see it and tried some faking? Perhaps you think that some faking might be necessary in order save your position? Perhaps you are even right? > > remember that N contains only finite numbers. > But infinitely many of them. And even more paths in the tree. We know your creed. Nevertheless it is wrong. > Consider the model given within any infinite tree by any of > its (infinite) paths. > In this model, every natural appears in some line. > In this model, there is no line in which every natural appears. > This model disproves WM's contentions about his EIT. This model has been devised in order to show that, *given, aleph0 exists*, 2^aleph0 > aleph0 can be contradicted. > > Show me two elements wich cannot belong to the same line. > How is that in any way relevant? What I have said allows that for > each pair there is SOME line which they do not both belong to. How is that in any way relevant? What I have said allows that for each pair there is SOME line which they do both belong to. Are you really thinking that your statement is in any way stronger than my? Do you really thinking that the possibility of a bijection is in any way a stonger stateent than the possibility of a non-bijection? >> Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n >> different paths and E(n) = 2*2^n - 2 different edges. So we find >> lim{n-->oo} E(n)/P(n) = 2, or, if we are unable to determine limits, >> E(n)/P(n) >= 1. Nevertheless "E(n)/P(n) < 1 in the actual infinity". > Why not? There is no reason for "continuity" to be required here. But in the bijection 1, 2, 3, ... 2, 4, 6, ... there is some reason for continuity and some proof of same cardinality? > By the same limit argument as WM has just claimed, there must be as > many leaf nodes in the infinite case as there are paths, since the > ratio of number of leaf nodes to number of paths is identically 1 for > all finite trees. No, there are no leaf nodes in my tree. But for any path which separates itself from another path there is an edge where this happens. This cannot be circumvented by the magic of infinity. Therefore there are not more paths than edges possible. >> We are talking about shares, but we do not need more than 3/2 per >> path. > How are the edges at the root node to be divided up into shares, and to > which paths are those shared to be assigned? > Then the same questions need answering about the next level of edges, > and then the next, and so on ad infinitum. Didn't you say that there is no continuity in infinity? >> If there are infinitely many different equations of form >> "1+1+1+...+1 = n, >> then there must be infinitely many different sums of ones. Actual >> infinity of the umber of equations implies actual infinity o at least >> one sum of ones. > That is only true in WM's mythical axiom system, not in any actaul axiom > system. If infinity is to be only potential, then we may speak of infinitely many finite numbers, because potentially infinite sets are always finite. If infinity is to be actual, then we have a number greater than any natural number for the lines and simultaneously for the columns of the EIT. >>You cannot have an actually infinite finite number. > WM may not be able to, but his inadequacies are personal, not universal. So you are able to have it? You accept a diagonal of a matrix longer than any line. You accept more separated paths than splitting positions (= origins of separated paths). You accept that 1 + 1/2 + 1/4 + ... > 10 is possible. And at last you accept the existence of an actually infinite finite number. Sorry, but then we do no longer have a common basis of discussion any longer. Have a merry Christmas and a happy New Year. Regards, WM |