Cantor Confusion
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From: mueckenh on 14 Dec 2006 05:47 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. 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. You see it best if you try to give an example using a finite element a or b. You will say there is no finite example a or b. That is correct. And from the obseration that there are no other than finite elements you could then conclude that your reasoning is wrong, if ... > > 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.) But it is false that such a set can by an actually infinite set of finite elements only. To see that compare the IET and your statement that a diagonal can be longer than every line. Regards, WM
From: mueckenh on 14 Dec 2006 05:51 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. Further, the tree is continuous in that at every level n there arrive 2^n edges, that means there arrive 2^n separated paths. 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 but more than any finite number of paths. Regards, WM
From: mueckenh on 14 Dec 2006 05:53 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. Regards, WM
From: mueckenh on 14 Dec 2006 05:55 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 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 Regards, WM
From: mueckenh on 14 Dec 2006 06:02
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. > > > > > too German; > > > > it is impossible to be too German. > > Only in the minds of Germans. > > > > > no definition at all] > > > > It is clear that you have not understood. > > On the contrary, it is likely that he understood too well. Well, meanwhile even you should have understood that he is a she. Regards, WM |