Cantor Confusion
in [Math]
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From: Dik T. Winter on 15 May 2007 22:33 In article <1178815921.746264.51330(a)y80g2000hsf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 10 Mai, 04:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > Obviously not. For each node there is a co-path p' which has been with > > > p all the way long. > > > > Where is the contradiction? > > If you say > 1) for every path p', there is a node where p' leaves p > and > 2) there is no node where every path p' has left p Indeed. > then you must be able to name a node K(n) such that for nodes K(m < n) > at least two paths p' and p'' are required to accompany p. Well, this holds for *all* natural n and nodes K(n) of p. For each n there are uncountably many different paths that accompany p to that node. > Because, if for all nodes which you can enumerate, one path p' is > sufficient to accompany p, why should anybody share your belieflthat > for all paths you can enumerate, not one path p' is sufficient to > accompany p? And this makes no sense. > > > How do you address, represent or use it in any other form "in > > > mathematics"? > > > > By having a set of it. Why is representation needed? > > Because you cannot use a number without address (= representation). > You can only use the set. But why do you need each number? The mere existence is sufficient. The only numbers I really do use (apart from the integers and rationals) are numbers like sqrt(5), (99831352 + 2146257.sqrt(-311))^(1/19), the largest real root of x^5 - 8822 x + 1, and so on. O, and pi and e on occasion. > > > That's your (and others') error. > > > > Did you ever correctly read the proof? > > More than once.And I wonder, why the case 1 = 0.999... must be > excluded, if only finite positions played a role. Because 1.000... and 0.999... differ in all finite positions (there are no infinite positions). > > > The proof in the tree is, that for > > > every path p there is another path *existing in the tree* which is not > > > different from p. > > > > Nonsense, if two paths are different there is a node where they differ. > > Do you believe that there is a real number between 0 and 1, which is > not represented in the tree? No, and I wonder why you ask. I already stated such many times. > > > Why should this be different in Cantor's list? > > > > Because it is a list. A tree is not a list. > > It is possible to change a digit of every number, because the nodes of > the tree are a countable set. You can even change as many nodes as you > like (also in a decimal tree) and construct a number from the changed > tree. That number which you construct has already been in the tree. Yes, so what? The nodes form a countable set, and you can re-order them, giving a different infinite tree. I do not understand. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 May 2007 22:42 In article <1178817355.571399.114020(a)y80g2000hsf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 10 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1178723452.432944.32...(a)e51g2000hsg.googlegroups.com> WM <mue= > ck...(a)rz.fh-augsburg.de> writes: > > > On 9 Mai, 15:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > But that is not the definition, their formal definition is: > > > > "A binary relation F is called a function (or mapping, corresponden= > ce), if > > > > aFb_1 and aFb_2 imply b_1 =3D b_2 for any a, b_1, and b_2." > > > > > > And how do you think the a's and b's are selected? And from what sets > > > do you think are they selected? > > > > The first thing you should do is find how they define "binary relation". > > But if you skip introductory material you can be lead to errors. > > Be sure, I studied every page very carefully, so carefully that I > typed every letter (some time ago). Some results are available here. Again leading to a soup of (to me) not interpretable nonsense. > A binary relation is, therefore, determined by giving all ordered > pairs of objects in that relation; it dos not matter by what property > the set of these ordered pairs is described. > > !!!!!!!!!!!! But it obviously does matter *that* it is > described. !!!!!!!!!!!!!!! And I think that this is additional material by you. The point is that what may describe the property depends on the model of set theory, and what properties are allowed. So in mathematics the desription itself is left outside the discussion. You will find that in the axiom of subset (or how it calls itself currently) the subset is described with an arbitrary property 'phi'. Whether 'phi' is a formula, or something else, depends on the model you are working in. For set theory the underlying possibilities of the properties are not interesting. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 16 May 2007 15:55 On 14 Mai, 22:09, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 14, 3:17 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > Which of the following statements is not true > > Every element of R is different from p. > > Every node of p is in some element of R > Back to the facts: 1) No element r =/= p of R is required to cover any node of p. (I can show of every such element r =/= p which you may name that it is not required.) 2) No element of R is sufficient to cove all nodes of p. Hence your set R is not suitable to demonstrate your claims. Regards, WM
From: WM on 16 May 2007 16:08 On 16 Mai, 03:55, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178795907.957410.94...(a)o5g2000hsb.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 10 Mai, 03:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > So in your opinion there are only finitely many paths in the > > > > > infinite tree. > > > > > > > > I did not say that. > > > > > > Quote: "Only from those few you can ask for". I can ask only for finitely > > > many paths. > > > > That is correct. But that does not mean that I said that there are > > only finitely many paths in the tree, as you implied. > > Why not? I can ask only for finitely many paths, and (if I interprete your > thinking correctly) path which I can not ask for do not exist. That MatheRealism. > > > > > In the same way: "what is sum{i = 1..n} i" a single > > > question. You want to split the first question into infinitely many > > > questions about all paths. In that case you should also split the second > > > question into infinitely many questions about all n. Using your logic > > > the split questions can only be answered for finitely many cases, so the > > > formula "sum{i = 1..n} i = (n + 1) * n / 2" is only valid for those cases > > > that you did ask. > > > > It is valid for those cases I did prove. I did not prove it (and one > > cannot prove it) for "all natural numbers which obey this formula", > > because all natural numbers which obey this formula do not obey this > > formula. > > This sentence makes no sense. I do not ask it for "all natural numbers which > obey this formula", I ask it for "all natural numbers". And what natural > numbers that obey that formula do not obey that formula? Up to each one they obey it. *All* natural numbers obviously do not obey the formula. > > > > > > > > There is a subtle difference. With the diagonal proof we always > > > > > > > remain in the finite. > > > > > > > > > > > > That's why it fails for numbers with infinitely many digits. > > > > > > > > > > No. > > > > > > > > Assertion, no argument. > > > > > > Yours was also an assertion without argument. > > > > My argument is: Why do you have to exclude the case 0.999... = > > 1.000... if you remain always in the finite? For every finite index > > both numbers are different. > > But when you use all finite indices they are the same. But the indices > are still finite. And by the axiom of infinity we can use the set of > all finite indices. But not in sum{i = 1..n} i = (n + 1) * n / 2 ?> > Here is a part of the introduction, but I have not bothered to > > translate the Greek and other symbols: > > Sorry, that makes it unreadable to me. But from what I can decipher is > that they define the infinite sum of cardinal numbers as the cardinality > of the infinite union of disjoint sets. In common set theory the infinite > union of sets is defined, and so this makes sense. But this has nothing > to do with > sum{i = 1..n} i = (n + 1) * n /2 > where the sum is constrained to natural numbers n. Analysis and algebra > are about real numbers, and in those fields infinite sums are not defined, > except as a limit. Moreover, using their definition of limits of > sequences of ordinal numbers we would get omega. It can make sense to > make > sum{i = 1..oo} i = aleph_0 > but it does not follow in any way from the usual theorem, and requires > some additional definitions. No. Then sum{i = 1 to oo} i could also yield another result. But that is certaily false. It can neither be finite nor can it be uncountable. The same holds for sum{i = 1 to oo} 1 = aleph_0. Regards, WM
From: WM on 16 May 2007 16:12
On 16 Mai, 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178805746.803872.67...(a)n59g2000hsh.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 10 Mai, 03:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1178716585.114658.276...(a)h2g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > ... > > > > > > > There are subsets of a finitely definable set which are not > > > > > > > finitely definitable. > > > ... > > > > There does not exist anything in mathematics, unless it is finitely > > > > definable. In particular, any infinite definition is not a definition, > > > > because there is a definition of "definition" which says: Definition: > > > > A definition has an end, i.e., a last word and a point. > > > > > > Try to find a finite definition of the set of Ulam's lucky numbers. > > > > Here it is. Use the prime sieve of Eratosthenes but don't count the > > numbers already eliminated. > > What is the problem? > > I would expect a finite definition of the sieve. If there was not such a finite definition, then we would not yet know what Eratosthenes had meant. > > > The set is finitely defined. Not all lucky numbers can get finitely > > defined. > > In general a recursive definition is not considered a finite definition. Every definition which ends after finitely many words is a finite definition. Regards, WM |