Cantor Confusion
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From: Virgil on 20 Dec 2006 16:15 In article <1166635862.474008.237960(a)t46g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166545363.371163.143890(a)i12g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > There are only finite numbers in the diagonal. QED. > > > > WM again loses track of the point at issue. > > > > No one wants anything but finite numbers IN the diagonal, but we all > > want infinitely many of them. > > But there are not actually infinitely many. Maybe not in your world, but you do not rule in ZFC and NBG. > > > > And since WM has used infinite binary trees as justifying his arguments, > > he must allow that a complete path in such a tree is like a diagonal of > > WM's EIT with its edges being the last members of the lines of that EIT. > > > > Which gives every line being finite but the diagonal being infinite. > > Every edge is in some line but no line contains all edges. > > The tree proves that, given aleph_0 exists, aleph_0 is not less than > 2^aleph_0. The EIT proves that aleph_0 does not exist. > > Regards, WM
From: Virgil on 20 Dec 2006 16:19 In article <1166635911.407021.21870(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166529653.157491.94810(a)48g2000cwx.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > So you consider it nonsense that mathematics should consist of clear > > > > definitions and proofs of theorems. > > > > > > No. But I need no proof in order to see that > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > > > > Absent some formal definition, "1 + 1/2 + 1/4 + ...+ 1/2^n + ..." is > > without any mathematical meaning whatsoever. > > No. It is less than 10. It was so before your kind of formal definition > existed and it will remain so after your kind of formal definiton will > long have ceased to be remembered by mankind. When interpreted, symbols can have meaning. Absent interpretation they are no more meaningful than chicken scratches in the dust. And much of WM's scratchings are about on that level even with interpretation
From: Virgil on 20 Dec 2006 16:24 In article <1166635980.571001.240220(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > In the infinite tree there are no finite paths. > > But if we calculate the limit of a finite tree, the result is the same. According to that rule every path must have a germinal (leaf) node, even in an infinite tree. The ratio of terminal nodes to paths is exactly 1 for every finite tree, so in the limit as the lengths are allowed to increase unimitedly that ratio must remain 1, at least according to WM's logic. So in an infinite tree, WM argues that every infinite path must have a last node.
From: William Hughes on 20 Dec 2006 16:46 Han de Bruijn wrote: > William Hughes wrote: > > > Han de Bruijn wrote: > > > >>William Hughes wrote: > >> > >>>Han.deBruijn(a)DTO.TUDelft.NL wrote: > >>> > >>>>William Hughes schreef: > >>>> > >>>>>Does the "potentailly infinite set" of natural numbers > >>>>>exist? A potentially infinite set is a function on sets > >>>>>that takes on the values true and false. The > >>>>>potentially infinite set of natural numbers > >>>>>takes on the value true for a set containing only > >>>>>natural numbers, and false for any other set. > >>>>>(i.e. is there a way of recognizing a set of > >>>>>natural numbers?) > >>>> > >>>>I don't understand this question. > >>> > >>>Stop me at the first line you don't understand. > >>> > >>>[Note if you do not like the term "potentially infinite > >>>set" substitue "recognizer".] > >>> > >>>A "potentially infinite set" is a function on sets. > >> > >>It's already here: "function" and "set". > > > > You answered the question > > > > "Is there an unbounded set of natural numbers?" > > > > No. > > > > Now you claim you don't know what I mean > > by "set". . > > > > Clearly you cannot answer simple questions. > > Whatever. It's the combination that confuses me: what are the domain and > the range of your "potential infinite set" function? 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. 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? - William Hughes
From: mueckenh on 20 Dec 2006 16:47
William Hughes schrieb: > > Please give a counter example for a finite number n. > > No such counterexample exists. The claim is that despite > the fact that given any n we can find a single line, L(n), that works > for this n, there is no single line, L_D, that will work for all n. This claim is obviously false. At least my counter claim cannot be disproved by an example.> > > > Remember, even if there is not a last line, every n eps N is finite. > > Which means that for any n we can find a single line L(n) that works > for this n. It does not mean that that we can find a single > line L_D, that works for all n. That is impossible, beause there is no "all n". If "all n" were somewhere, then we had no problem to put them in one single line. Due to the finiteness of every n, every position of the line was finite. And due to the fact that a number with only finite indexes is a natural number, the whole number was a natural number. Recently we had the following claim (by Virgil) The number of equations 1+1+1+...+1 = n with finitely many 1 is infinite. Considering the fact that there are exactly as many 1 as different numbers, it is impossible that the number of numbers can be infinite while the sme number of 1 is finite. That is obviously our old EIT. 1 11 111 .... The number of 1 is not *actually* infinite without one line being actually infinite. > > Therefore your claim is void, unless you can find a linear set of > > finite elements which forbids exchange of quantifiers. > > > Consider the (potentially infinite) set of natural numbers. > This is linearly ordered. Every element is finite. > > For any natural number n, there exists a natural number m(n), > such that m>n > > However it is not true that: > > There exists a natural number m, such that for any natural > number n, m>n. It is true for any set of natural numbers which can be written down. You cannot give a counter example. The only things you can do is either to present a line which does not contain all numbers, or to assume that a line contains all numbers and then to show that this is not the case. What a surprise! Or you can try to talk of dancers each of which has a women partner but not all have the same. ==================== > B: There exists a single line L_D, such that for every natural > number n, n is in L_D. > B does not follow from A, and there is a simple proof that if the > diagonal is unbounded, then B is false. There exists a simple proof that all elements of the diagonal are in the lines. And by the linearity and finity of the lines we find that all elements which exist in some lines exist in one line. This is a proof by induction covering all finite natural numbers, i.e., all elements of the diagonal. Regards, WM |