Cantor Confusion
in [Math]
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From: WM on 10 May 2007 10:14 On 10 Mai, 03:59, Virgil <vir...(a)comcast.net> wrote: > In article <1178738274.909825.208...(a)e65g2000hsc.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 9 Mai, 20:16, Virgil <vir...(a)comcast.net> wrote: > > > In article <1178707943.860551.269...(a)h2g2000hsg.googlegroups.com>, > > > > > But the formal definition does not specify how a and b are related > > > > other than by mentioning F. This F however is undefined unless you > > > > know from the first paragraph that it is a procedure or rule. > > > > If it is a set, it is defined as sets are defined, which does not > > > require anything as undefined as "procedure" or "rule". > > > How do you define a set, unless you cannot specify the elements > > belonging to it? > > I am reasonably certain that that question does not mean in standard > English, what you intended to ask. Should have been: How do you define a set, unless you can specify the elements belonging to it? > > The axioms of ZF tell how to specify sets in ZF. Yes, how to specify, but they do not specify sets except few. > > > > > > > Further definition 3.1 contains: It is customary to use the notations / > > > > F: A -> B/. Why do you think A and B were not two sets and F was not a > > > > formula > > > > Define "formula". Until it is defined, it has no mathematical > > > significance, and it hasn't yet been defined. > > > A formula is a rule or prescription specifying or determining which > > element of the domain is mapped on which element of the range. The > > formula in its widest sense can even consist of several expressions. > > In case every element of the domain has to be mapped by its own > > expression, the formula can even be a catalogue or set. > > So a set can be defined by being a set! How marvelously circular! A set can be specified by specifying its elements. This can be done very well, in particular for finite sets, by a list. Yes, the rule or formula can be a list listing the elements of a set. > > Do you know what a relation is? Do you know that there is also the > > domain and the range required? > > There are a domain and range to a relation, but they are consequences of > its definition, not antecedents to it. Answer only by yes or no: Do you know that there is also the domain and the range required in a relation and all the more in a function? Can you imagine a relation without having a domain and range? Regards, WM
From: WM on 10 May 2007 10:47 On 10 Mai, 03:59, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > There is a slight difference. Finite definitions do not imply finitely > definable. Consider the computable numbers as they are presented by halting > Turing machines. The definition is indeed finite. The definition of the set is finite, not of the numbers. > Nevertheless they are > considered not to be finitely defined because it is impossible to find out > whether a particular Turing machine indeed *does* halt. Therefore only few of the numbers are finitely definable. > > > > But that a set of entities is finitely definable does *not* > > > mean that each element of the set is finitely definable. Consider Cauchy > > > sequences. By (a finite) definition a Cauchy sequence denotes a real > > > number. This is a *finite* definition of the set of real numbers. > > > > Yes, of the set. But it is only a definition of those real numbers for > > which Cauchy sequences can be finitely defined. > > No. It is a definition of the set. It is not a definition of the real > numbers at all. A well defined Cauchy sequence, for pi, for instance, makes pi well defined - not as a number in the sense of MatheRealism, but as a number in the sense of mathematics. > > > > This does *not* > > > mean that each real number is finitely definable. > > > > And it does not mean that every Cauchy sequence has a finite > > definition. > > Yes, no problem with that. > > > I gave a bijection between the set of nodes and the set of branching- > > offs of paths bunches. > > You did not. What node is bijected with 0.1010101010...? Please read: I gave a bijection between the set of nodes and the set of branching- offs of paths bunches. > > > As there can be not more results of branching- > > offs than branching-offs, the task is done. > > As in each node the number of path bunches decreases I wonder about this > statement. In each node the number of path bunches increases. If we add one bunch going into the first node, we an say: In all nodes one bunch goes in and two come out. There is no way to circumvent the fact that bunches - nodes = 2-1-1+2-1-1+2-1-1+- < 3. It is remarkable how this simple fact could have been overlooked for such a long time. It is similar to 1+2+3+...+n = n(n+1)/2 This equation is not satisfied by all natural numbers which are asserted to exist. Regards, WM
From: WM on 10 May 2007 10:54 On 10 Mai, 04:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178722003.108344.38...(a)y80g2000hsf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 9 Mai, 03:52, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Sometimes it is preferable to have a calm atmosphere. > > > > > > I have first to sign up, next I get probably an interface that can only be > > > handled with a mouse. No, thanks, I have already too many problems with > > > mouse use. > > > > I have got the same interface as here. > > I have not. I do not use a web-browser for discussions. But you apparently > do not understand the principles of Usenet newsgroups. To a certain extent you are right. I am not an expert in IT. But, nevertheless, it was claimed that I had influenced some remote computer! (I am glad if my own does what I want it to do.) > > > I will answer there. > > I will not see it, so you clearly cut short the discussion. Please try, we could reduce the noise significantly and could kill any polemics and insults. If you are not able to see it, I will repeat it here. But I wonder why Google introduced this technique if it remains inaccessible to most users and even to experts. http://groups.google.nl/group/AOTI/browse_frm/thread/6386cb0c4c23e619/# Regards, WM
From: WM on 10 May 2007 12:52 On 10 Mai, 04:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178722674.540579.289...(a)e51g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 9 Mai, 04:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Even in the *infinite* tree a path cannot be distinguished from all > > > > other paths. > > > > > > Why not? For each other path there is a finite node where it goes a > > > different way. > > > > 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 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. 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? > > > > That is simply opion. Indeed, only countably many can be described by a > > > finite formula (as is clear from the work of Turing). In mathematics that > > > does *not* mean that the other numbers do not exist. > > > > 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. > > > > > Cantor's diagonal proof fails, because the diagonal number is never > > > > distinguished from all other real numbers (if uncountably many real > > > > numbers exist). > > > > > > But that is not the proof. The proof is that the diagonal number is > > > distinguished from the real numbers in the list (which are countably many). > > > > 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. > > > 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? > > > And if you apply Cantor's proof in the tree, by > > forming a "diagonal" by switching a bit for every path, then it is > > undisputed that the constructed diagonal number is represented by a > > path in the tree. > > I need a list of paths before I can even try to start this. The set of > paths is not a list. So, first, provide me with a list of paths and > I will come up with a path not in the list. > > > 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. Regards, WM
From: WM on 10 May 2007 13:15
On 10 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178723452.432944.32...(a)e51g2000hsg.googlegroups.com> WM <mueck...(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, correspondence), if > > > aFb_1 and aFb_2 imply b_1 = 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. 2. RELATIONS Mathematicians often study relations between mathematical objects. Relations between objects of two sorts occur most frequently; we call them binary relations. .... 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. !!!!!!!!!!!!!!! .... 2.3 Definition Let R be a binary relation. (a) The set of all x which are in relation R with some y is called the domain of R and denoted by dom R = {x | there exists y such that xRy}. dom R is the set of all first coordinates of ordered pairs in R. (b) The set of all y such that, for some x, x is in relation R with y is called the range of R, denoted by ran R. So ran R = {y | there exists x such that xRy} 2.9 Lemma The inverse image of B under R is equal to the image of B under R-1. 2.11 Definition The membership relation on A is defined by ïA = {(a, b) | a ï A, b ï A, and a ï b} The identity relation on A is defined by IdA = {(a, b) | a ï A, b ï A, and a = b} 2.12 Definition The set of all ordered pairs whose first coordinate is from A and whose second coordinate is from B is called the cartesian product of A and B and denoted A ï´ B. In other words, A ï´ B = {(a, b) | a ï A, b ï B}. ???You see a A and B here???? > > > It is ridiculous to > > follow this discussion. I can assure you, if one of my students would > > not know that a function is a formula (or rule or whatever) together > > with a domain where it is defined and a range, then he or she would > > not pass the exame. And this is the same in the better math courses in > > Germany. > > If that is true, it tells us a lot about mathematics education in Germany. > Functions in set theory are *not* the same as functions in analysis. I talked about mathematics, not about set theory. Nevertheless functions *in set theory* consist of a prescription (formula, rule, whatever), domain, and range. Exercises 2.1 Let R be a binary relation; show that dom R c U(UR), ran R c U(UR). Conclude from this that dom R and ran R exist. The last exercise is a good homework for such "mathematicians" who studied in Oxford or Harvard or at home and, therefore, do not know that dom R and ran R do exist for any binary relation, or in other words: without domain and range R would not be a binary relation. Regards, WM |