Cantor Confusion
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From: mueckenh on 14 Nov 2006 08:18 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> mueckenh(a)rz.fh-augsburg.de wrote: > >> > Franziska Neugebauer schrieb: > >> >> mueckenh(a)rz.fh-augsburg.de wrote: > >> >> > Franziska Neugebauer schrieb: > >> >> >> mueckenh(a)rz.fh-augsburg.de wrote: > >> >> >> > Franziska Neugebauer schrieb: > >> >> >> [...] > >> >> >> >> Are there really three vertices in WM's "triangle"? > >> >> >> > If finished infinities [...] > >> >> >> Verbiage. > >> >> > Yes. But, sorry to see, it is the fundament of modern > >> >> > mathematics. > >> >> "Finished infinities" is your wording. > >> > Precisely describing the fundament of modern mathematics. > >> Cbjre Bs Oryvrs. > > Always insisting on having the last word? > > There is no last word. Amen. Excuse me: omega. Regards, WM
From: mueckenh on 14 Nov 2006 08:25 Dik T. Winter schrieb: > In article <1163431235.417059.113430(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > Well, my only advise is, read it. > > > > > > > > If he says so, then it wil not be a good idea to waste my time with it. > > > > > > Do you really think the node 1/3 is finitely far from the root in the tree? > > > > Dik, are you joking? There is no node yielding any number like 1/3. > > Every node represents one bit of the binary representation of many real > > numbers. > > You switch so many times from what represents what, that I lost track. > But with this notation: there is no node in the tree where 1/3 is > completed. There is no binary digit where 0.010101... is completed. >If there were such a node, it would be infinitely far away. It would be the last one, number omega, but we agree that it does not exist. > But, whatever. You can just as well state that each node carries also > the decimals above it, the distinction is extremely small. I call this collection of decimals "a path". > > > The first two levels of the tree contain the following initial > > segments: > > 0.00... > > 0.01... > > 0.10... > > 0.11... > > Note that at every level n of your tree the numbers represented are > all of the form a/2^n, with 0 <= a < 2^n. So neither 1/3 not 1/5 > are in any of the finite subtrees. They have no finite binary representation. But if they can appear in a list, then they can appear in a tree. > > > > Your node 1/3 is infinitely far from the root because there is no finite > > > (natural) number that can state the distance. > > > > There is no node 1/3. There is the number 1/3 consisting of infinitely > > many nodes 0.010101... None of them has infinite distance from the > > root. > > But that one is not in the tree. If (as I state above) we consider that > each node also carries the decimals above it (which is equivalent to > your statement), each rational number in [0, 1) with a power of 2 in > the denominator is in the completed tree, but no other numbers. Even of [0, 1] because 0.111... = 1. If you have this opinion, I will happily agree, but then you must also apply it to every binary representation of the reals. The tree is nothing other than such a representation, a special one. You just stated that there are no infinite strings. A agree. Regards, WM
From: William Hughes on 14 Nov 2006 08:27 mueck...(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > If we add one element to each of the columns then > > the supremum of the lengths of the initial segments of the columns > > changes to omega +1. If we add one elelment to each of the lines > > the supremum of the lengths of the lines does not change, it > > remains omega. > > Fine. This new matrix has a diagonal, if the old matrix had one. Let he original matrix be A. Let B be the matrix we get by adding one element to each of the columns of A. This matrix does not have a diagonal. Let C be the matrix we get by adding one element to each of the lines of A. This matrix has a diagonal. Let D be the matrix we get by adding one element to each of the lines of B. This matrix does not have a diagonal. Only for matrixes B and D is there a last line. So if there is a diagonal, there is no last line. >It > bijects the columns to the lines. The element d_nn maps the n-th column > to the n-th line. It should do so, at least. > > > > And it does. > > Could you construct this bijection? It is easy to begin: > > Column 1 <--> Line 1 by the diagonal element d_11 > Column 2 <--> Line 2 by the diagonal element d_22 > Column 3 <--> Line 3 by the diagonal element d_33 > ... > Column ? <--> Line omega+1 by the diagonal element d_?? > > But t is difficult to end, although there is a last element in the set > of lines No. If there is a diagonal there is no last line. .. - William Hughes
From: Franziska Neugebauer on 14 Nov 2006 08:29 William Hughes wrote: > If we add one element to each of the columns then > the supremum of the lengths of the initial segments of the columns > changes to omega +1. If we add one elelment to each of the lines > the supremum of the lengths of the lines does not change, it > remains omega. Could you explain in a formalized way what "add one element to each of the columns" means? F. N. -- xyz
From: Franziska Neugebauer on 14 Nov 2006 08:36
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > {1,2,3} is the collection of, and a convenient expression to write >> > {that we are talking about, the numbers 1 ,2, and 3. >> >> What are we talking about, when we write >> >> { } >> >> ? > > The same as when we write > > > > > > > > > Regards, WM I would like to suggest the name "finished emptiness" for that. F. N. -- xyz |