Cantor Confusion
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From: David Marcus on 3 Nov 2006 12:47 mueckenh(a)rz.fh-augsburg.de wrote: > > David Marcus schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > In article <1162405520.008395.100850(a)e64g2000cwd.googlegroups.com> mu= > eckenh(a)rz.fh-augsburg.de writes: > > > > > Dik T. Winter schrieb: > > > > > > > > Where in the above quote is 1/oo defined? > > > > > > > > > > It is defined that omega (which Cantor used later instead of oo) i= > s a > > > > > number larger than any natural number n. Omega is the limit ordinal > > > > > number. Therefore 1/omega must be a number smaller than every frac= > tion > > > > > 1/n. > > > > > > > > Why? As long as 1/omega is not defined you can not talk about it. Y= > ou > > > > simply assume that 1/omega is a number. But that is not the case, it > > > > is not defined. > > > > > > If omega is a number > n, then 1/omega is a number < 1/n. For all n e > > > N > > > > What do you mean by "number"? Normally, "omega" denotes a certain > > ordinal. Division is not normally defined for ordinals. If you want it > > to be defined, then you have to define it. So, using a plausible > > interpretation of your words "number" and "omega", your statement is > > false because 1/omega is not an ordinal (since it is not defined). > > By "number" I understand those mathematical objects which can be > compared by size or magnitude, i.e., which observe trichotomy. I know > that division is not normally defined for transfinite ordinals, but > with omega > n we can define 1/omega < 1/n for n e N as some kind of > abbreviation. You can define anything you want. But, if it is not a standard definition that people already know, you need to give the definition. You say that you are defining "1/omega < 1/n for n in N" as an abbreviation. An abbreviation for what? For "omega > n for n in N"? The abbreviation seems to be longer than the statement it is abbreviating. > Don't forget, most things come into being by intuition, not by > formalization. > > It is by logic we prove, it is by intuition that we invent. (Henry > Poincar=E9) > > When Cantor introduced omega, there was no formal definition. > > Das Zeichen oo, welches ich in Nr. 2 dieses Aufsatzes gebraucht habe, > ersetze ich von nun an durch omega, weil das Zeichen oo schon vielfach > zur Bezeichnung von unbestimmten [d. h. potentiellen] Unendlichkeiten > verwandt wird. [G. Cantor, Gesammelte Anhandlungen, p. 195] -- David Marcus
From: David Marcus on 3 Nov 2006 13:04 mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > What do you mean by "understood"? You concede that your logician agrees > > that the argument is not valid in ZFC. > > He does not know how to translate it. So, until proven otherwise, we will assume it is not valid in ZFC. > > So, what logical system does your > > logician say it is valid in? > > Do you really believe that ZFC is the only and ultimate system of > thinking? I asked a question and you replied with a question that is asking me if I believe something that I never said I believed. Let's try again. Your logician can't show your argument works ZFC. However, you say he agrees with the argument. So, in what system does he (or you) say the argument can be given in? > > > The proof of uncountability of R therefore shows ZFC is false. > > > > The statement "ZFC is false" is meaningless. All you are saying is that > > your proof cannot be given in ZFC. > > The proof has not yet been given. That is a bit different from cannot > be given. Certainly. But, you can't assume it can be given until you give it. > > > A supremum is a least upper bound of a set (like the set of columns of > > > a list). If there is no maximum, then the supremum is not taken, i.e., > > > it does not belong to the set. > > > > It is true that if there is no maximum, but there is a sup, then the sup > > is not an element of the set. So what? That doesn't prove anything about > > the diagonal. > > For the diagonal (d_kk) of the list > > 0 > 1 2 > 3 4 5 > 6 7 8 9 > ... > > we have the following mappings: > d_kk --> d_mk and d_kk --> d_km with k, m eps N. Normally, the word "mapping" means function. If you are trying to define functions, it isn't clear to me what you mean since m appears on the right, but not the left. Is d_mk the kth element in the mth row? > It is curious that the set of terms of (d_km) has omega as a maximum > for every fixed m eps N while the set of terms of (d_mk) has omega not > as a maximum for every fixed m eps N. For m in N, you appear to be considering two sets: A = {d_km | k in N}, B = {d_mk | k in N}. If we take m = 0, then A = {d_00, d_10, d_20, ...} = {0,1,3,6,...}, B = {d_00, d_01, d_12, ...} = {0, undefined, undefined, ...}. Regardless, no set of natural numbers has omega as a maximum. I've no idea why you think your list shows that omega is the maximum of a set of natural numbers. -- David Marcus
From: David Marcus on 3 Nov 2006 13:08 mueckenh(a)rz.fh-augsburg.de wrote: > > David Marcus schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > Any "diagonal" for this function will have to be of length greater than > > > every n in N. > > > > > > This shows the inconsistency of ZF and NBG. > > > > It is amazing that you keep repeating claims after you yourself admit > > that you have not shown the claim. Is this intentional or don't you > > realize that you are contradicting yourself? > > It is amazing that you don't understand the meaning of my proof. > I have shown that the set of paths of the binary tree has a cardinality > not less than R. > I have shown that the set of edges has the cardinality of N. > And finally I have shown that the cardinality of the set of paths is > not larger than the cardinality of the set of edges. But, you don't use ZFC for your proof. > In ZFC Cantor's proof is valid, according to which the cardinality of R > is larger than that of N. > > Therefore there is a false result obtained from ZFC. By "there is a false result obtained from ZFC", you mean that your system proves some statement P and ZFC proves not P. Since your whole point seems to be that the two systems are different, this is not surprising. However, the statement "ZFC is inconsistent" does not mean this. The meaning of the word "inconsistent" in logic is that the system proves both P and not P. You could say, "ZFC and my system are not consistent with each other." -- David Marcus
From: David Marcus on 3 Nov 2006 13:14 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > For reference, here is my proof: > > > > A sequence is a function such that the domain of the function is an > > ordinal. A finite sequence has a natural number as its domain. A > > denumerable (countably infinite) sequence has omega as its domain. An > > uncountable sequence has an uncountable ordinal as its domain. > > > > The entries in a sequence are members of the range of the sequence. > > Each entry is indexed by a member of the domain. The elements of the > > sequence are ordered pairs of the form <x y> where x is a member of the > > > > domain (which is an ordinal) and y is a member of the range of the > > sequence (so the y's are the entries). > > > > dom(S) = the domain of S. > > > > length(S) = dom(S). > > > > In my example S is the denumerable sequence recursively defined as > > follows: > > > > S(0) = {<0 0>} > > > > S(n+1) = the unique finite sequence f such that length(f) = > > length(S(n))+1 and such that f(0)= S(n)(max(dom(S(n)))) + 1 > > > > So S is a denumerable sequence such that each entry of S is a finite > > sequence. > > > > The diagonal of S = the unique denumerable sequence D such that, for > > all n in omega, D(n) = S(n)(n). > > > > The diagonal of S is a denumerable sequence. > > > > Therefore, there exists a denumerable sequence S of finite sequences > > such that the diagonal of S is denumerable. > > Now map it on a line. It is longer than any line entry. But all line > entries which can exist are already there. Hence the mapping of the > diagonal cannot exist. Moe gives his proof. Then you say, "map it on a line", and a couple of sentences later say that the mapping doesn't exist. So, you can't have started off with "map it on a line". So, what you wrote appears to have no relevance to Moe's proof. -- David Marcus
From: David Marcus on 3 Nov 2006 13:18
mueckenh(a)rz.fh-augsburg.de wrote: > > Virgil schrieb: > > > In article <1162470874.593282.36250(a)b28g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > WM merely repeated his automatic error several more times here. > > > > WM claims that a list in which the nth listed element is a string of > > length at least n characters cannot produce a diagonal of length > > greater that any finite number of characters. > > > > The diagonal needs an element from every line, the n-th element from > the n-th line. Therefore it cannot be longer than every line. By "it cannot be longer than every line", do you mean its length can't be greater than the sup of the lengths of the lines? > Map the diagonal (d_kk) on some line (d_nk). You should see it. -- David Marcus |