Cantor Confusion
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From: Virgil on 14 Dec 2006 18:39 In article <1166093270.978792.13680(a)l12g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > (It is contained in the union of all lines, but the > > > > union of all lines is not a line) > > > > > > That is a void assertion unless you can prove it by showing that > > > element by which the union differes from all the lines. > > > > Not quite. In order to achieve that the diagoal is not in any linem all > > that is required is: > > Given any line there is an element of the diagonal not in THAT line. > > It is not requires that: > > There is an element of the diagonal that is not in any line. > > > For linear sets you cannot help yourself by stating that the diagonal > differs form line A by element b and from line B by element a, but a is > in A and b is in B. This outcome is wrong. As this has not been claimed by anyone, except possibly WM himself, its rightness or wrongness is irrelevant. > > Therefore your reasoning "there is an element of the diagonal not in > THAT line. It is not required that: There is an element of the diagonal > that is not in any line." is inapplicable for linear sets. It is quite applicable provided the linear sets do not have a lst or largest element, which is the case with every inductive set, which must exist in ZFC and NBG. Until WM presents his own axiom system for inspection, we have no alternative but to use one which already exists. > > > > For S = the set of all naturals, or for S = the set of all reals, or for > > S = the set of all rationals between 0 and 1, and for lots of other > > ordered sets, it is true that > > (Ax in S) (Ey in S) (y > x) > > but it is false that > > (Ey in S) (Ax in S) (y > x) > > That looks innocent and can be satsfied by a potentially infinity set > (except that "A" (forall) is not applicable there but only "Awhich > exist" or so.) Until WM presents us with is own axiom set, we will continue to use ZFC or NBG, whichever seems most appropriate. > > But it is false that such a set can by an actually infinite set of > finite elements only. In what axiom system is it false? With no axioms, nothing is true, or false, so WM must be making unvoiced assumptions to support any conclusions at all. What are these assumptions? Until WM states them, he cannot conclude anything.
From: Virgil on 14 Dec 2006 18:49 In article <1166093469.219744.116050(a)16g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1165923504.410525.226600(a)79g2000cws.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Dik T. Winter schrieb: > > > > > > > > > > Please give me all the bits of 1/3. Then I will show the > > > > > > > bijection. > > > > > > > > > > > > I can't so you can not show a bijection. > > > > > > > > > > Representations (= paths) which do not exist cannot be part of a > > > > > bijection. > > > > > > > > So you can not show a bijection (in your opinion), but nevertheless you > > > > state that you have given a surjection. Do you not think you are > > > > contradicting yourself a bit? > > > > > > I give a surjection on all existing paths. > > > > > > > > But what has been surjected by WM were not the individual nodes or > > branches, as required, but /infinite sequences/ of nodes or /infinite > > sequences/ of branches. Which is not at all the same thing. > > OK. If you agree that this is a surjection, then you must only > recognize that the series 1 + 1/2 + 1/4 + ... > consists of countably many terms. And the set of all countably many > series I used consists of countably many terms. Therefore I gave a > surjection from a countable set onto the set of paths. But the set of all sequences of edges is NOT countable, so what restrictions on sequences does WM have to limit then to only a countable subset of all such seqeunces? > > Further, the tree is continuous in that at every level n there arrive > 2^n edges, that means there arrive 2^n separated paths. WRONG! there "arrive" 2^n separated subsets of paths with each subset as numerous as the whole set. > If in the whole > tree 2^aleph0 paths have been separated, then there must be a level up > to which (or within which) less than 2^aleph0 paths have been > separated One is not separating individual paths but sets of paths with each set uncountable as the set of all paths. The set of all paths is clearly the same as the set of all endless sequences of left-right branchings. Cutting off the first n edges from each of these still leaves a set of endless sequences of left-right branchings no smaller than before.
From: Virgil on 14 Dec 2006 18:51 In article <1166093612.342256.123940(a)16g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <457ece72(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > > Dik T. Winter wrote: > > > > In article <457d8cc0$1(a)news2.lightlink.com> Tony Orlow > > > > <tony(a)lightlink.com> > > > > writes: > > > > > Dik T. Winter wrote: > > > > > > In article <1165761763.908889.34550(a)80g2000cwy.googlegroups.com> > > > > > > Han.deBruijn(a)DTO.TUDelft.NL writes: > > > > > > ... > > > > > > > Let P(a) be the probability that an arbitrary natural is > > > > > > > divisible > > > > > > > by > > > > > > > a fixed natural a. Then P(a) = 1/a . Forbidden by set theory. > > > > > > > > > > > > No. Not specifically forbidden by set theory. Forbidden because > > > > > > there > > > > > > are > > > > > > no appropriate definitions for the words you are using (they are > > > > > > not > > > > > > used > > > > > > conforming to standard definitions, so you better supply > > > > > > definitions). > > > > > > In probability theory (as is commonly use) you have to define how > > > > > > you > > > > > > *select* your arbitrary natural. You have not done so, so > > > > > > probability > > > > > > theory does not have an answer. > > > > > > > > > > Why does that matter? > > > > > > > > It does matter because if you do not properly define your problem, > > > > mathematics is not able to give an answer. > > > > > > It's sufficiently defined if one assumes that there is a uniform > > > probability distribution. > > > > From which assumption, added to the others, one can deduce that 0 = 1, > > and all sorts of peculiar things. > > > > > > > > > > > > This is the same thing as your stupid ball and > > > > > vase trick. Why do you need to label anything, or know what you're > > > > > choosing from the infinite set? > > > > > > > > Because that is part of the problem setting. Giving that setten will > > > > allow mathematics to model the question and give an answer. > > > > > > > > > > That problem has a clear answer with or without the labels: the sum > > > diverges as f(n)=9n. The labels are confounding, not clarifying. > > > > > > What is confounding to TO is clairifying to anyone with the wits to > > understand it. The result depends on the labeling. Eliminating the > > labeling makes the result impossible to determine. > > > > For any clear mind the result is completely independent of labelling. WM mistranslates. What he means is "For any blank mind the result is completely independent of labelling."
From: Virgil on 14 Dec 2006 18:53 In article <1166093706.459154.39710(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <457ecfca(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > > > > If the number of levels in the tree is countable, > > > then every path in the tree is finite, and marked by a specific edge > > > where the value arises. Is that not true? > > > > If TO means by "level" of a node in a binary tree the number of branches > > between it and the root node, then to have every path finite requires > > that every path end in a leaf node ( a ode with no child nodes ). > > > > That requires the number of "levels" be finite, not merely countable. > > > > > In order to allow infinitely > > > long strings, you must have an uncountable number of levels in the tree, > > > in which case 1/3 will exist with a specific edge, infinitely far from > > > the root node. > > > > There are no such things as trees with uncountably many "levels", at > > least according to any standard definition of trees. > > And there are no such things as trees with uncountably many paths. There are in ZFC and NBG. What axiom system does WM refer to in which there are not? > > > > there is a unique root node at level 0, and each level thereafter is the > > successor to a previous level through a connected chain of levels down > > to the root. > > Therefore we have a continuous growth of patghs separations from 1 to > 2^aleph0. Therefore the function 2^n must be continuous Non sequitur. > > Regards, WM
From: Virgil on 14 Dec 2006 18:55
In article <1166094152.380162.322300(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166007320.816114.309670(a)79g2000cws.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > Franziska Neugebauer schrieb: > > > > [...] > > > > >> 1. You do not present a convincing definition of "number". (Most > > > > >> likely you have none). > > > > > > > > > > Definitions are abbreviations like the following: > > > > > > > > [too long, too old, > > > > > > Impossible. Its from my new book to appear within few days. Can future > > > be too old? > > > > A new book with nothing in it but old ideas is not part of the future > > but merely an echo of the past. > > This book combines old knowledge and new insigts in a brilliant way. Any author would, of course, think so, but what do the reviewers say? |