Cantor Confusion
in [Math]
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From: Carsten Schultz on 10 May 2007 15:03 WM schrieb: > 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). That does not seem to have lead to much understanding. > 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. !!!!!!!!!!!!!!! No, it does not. > ... > 2.3 Definition Let R be a binary relation. Is there not a definition of a binary relation, or have you omitted it? It seems to me that it would have been significant for a discussion on what a relation is. Another sign of dishonesty or just carelessness? > (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} [...] >> 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. Playing semantic games again? Every natural number x has a square, x*x. (At least in mathematics.) There is no natural number which does not have a square. Would you therefore say that a natural number consists of a number and its square? Do I have to give the square of a number when I define a number? But you are good at it. Since you never state /exactly/ what you mean, you can always claim to have meant something else. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: Carsten Schultz on 10 May 2007 15:14 Virgil schrieb: > In article <1178793587.372588.308230(a)q75g2000hsh.googlegroups.com>, > WM <mueckenh(a)rz.fh-augsburg.de> wrote: > >> On 9 Mai, 23:42, Virgil <vir...(a)comcast.net> wrote: >>> I could hardly have learned less. >> Of course, "infinite definitions" are always majoring anything else. >> >>>>> You need it taught by anyone at all with any more mathematical >>>>> competence than you have yourself. Only the ignorant laugh at their own >>>>> ignorance. >>>> Then try to stop it. Do you know the definition of a binary relation >>>> by H&J? >>>> 2.1 Definition A set R is a binary relation if all elements of R >>>> are ordered pairs, i.e., if for any z à R there exist x and y such >>>> that z = (x, y). >>>> 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}. >>>> Can you find the words domain and range in this text? >>> I find 'domain' and 'range' to be defined in terms of a set of ordered >>> pairs, not the reverse, as WM has been trying to imply >> I did not try to imply anything as you try to imply, but I stated >> clearly that domain an range belong to a function. And here you see >> that they belong to a binary relation too. >> >> At Oxford and Harvard, to my knowledge, functions are formulas (or >> rules or whatever) with two sets, just as in Germany. > > At british and American universities, functions are first of all sets of > ordered pairs. While some functions may be defined using 'formulas or > rules or whatever', those 'formulas or rules or whatever' are not the > functions themselves, the functions 'are' the sets of ordered pairs. I can assure you that mathematics in Germany is the same as in the US or Great Britain. Mückenheim is not a mathematician and does not teach mathematicians. Afaik, he teaches mathematics and other things to students of e.g. engineering at what is called a `Fachhochschule'. You would have to find out for yourself what exactly that is, but while I would not generally say anything negative about their importance or quality (Mückenheim definitely is an exception and a disgrace to Fachhochschule Augsburg, even though I still hope that this does not affect his teaching too much, but I cannot tell), they are not what you would probably think of when you think of German universities. Best regards, Carsten (from Germany, obviously) -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: Virgil on 10 May 2007 15:21 In article <1178795907.957410.94020(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. Since your other comments imply that what one cannot ask for does not exist, it is reasonable to conclude you meannt only finitely many. > > > > Not at all. I ask the single question: "are all other paths different from > > a particular path?" > > That is circular, because by "other" you imply that path p can be > distinguished by a node from all the others. Not so. It only implies that any two distinct paths can be distinguished from each other at some node. And they can, for if they are not the same it is only because at some node they branch differently. > But we know, that this > not possible. Is that a royal "WE"? It is certainly not a plural "WE" in this thread. > Hence there is another path with p for all the time. If two paths share all their nodes, how can you tell that there is more than one of them? If > you state > > 1) p is not alone at any node. > 2) There is no single path p' with p at all nodes. > > Then you should show at least two (or more) paths p' which are with p > such that one of them is always with p but not always the same. Prove > that more than one further path p' is required. The finite chain of nodes (partial path) from the root node to any node, p, is the initial chain of infinitely many complete (infinite) chains. If path p' is to be "with" path p at all nodes of p, in what way can they be different paths? Definition of equality of paths in any tree: if every node of path p is a node of path p', and vice versa. then p = p'. > > You always state that this was true, but you cannot prove it. Is that > mathematics? There is nothing to prove once we have the relevant definitions. For every node in a CIBT, there are at least two paths through that node. If p' is "with" p at every node of p, then p' = p. > 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. So that. Apparently, one of WM's axioms is that finite induction does not hold. Since in ZF and NBG it does hold, that means that WM is assuming as an axiom that ZF and NBG are self-contradictory. That leaves us free to conclude that it is WM's system that is self-contradictory. > > > > > > > > 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. We, and Cantor, exclude them by having a diagonal rule that never uses either 0 or 9. > > > Do you think it would be > > > possible to sum an infinite set by (n + 1) * n / 2? > > > > No, as such has not been defined in mathematics. But you reject the > > formula above for any 'n' for which you did not ask. So each time you > > use the formula with a different 'n' you should prove it for that > > particular > > 'n'. > > It is obviously wrong for all n for which it holds, if such set of all > n exists. For which naturals n is WM saying Sum[x = 1..n, x] n*(n+1)//2 is false? He seems to be saying it its false whenever it is true, which h is a bit thick, even for him. > > > > > > > > > "completed" means nothing remains. > > > > > > > > > > > > Perhaps. > > > > > > > > > > Sure. > > > > > > > > Oh. A mathematical proof, please. > > > > > > The set of natural numbers is completed such that there is no natural > > > number outside of that set. Do you disagree? > > > > Ah, now you define completed, I think. > > That meaning should be obvious without definition. In mathematics, the standard is that nothing is so obvious that it does not require definition, except for the primitives of an axiomatic system. > > > But it is not true that nothing > > remains. We have still the rational numbers, the real numbers, and > > whatever. > > Read again what I wrote, please: > The set of natural numbers is completed such that there is no natural > number outside of that set. I think he's got it! That certainly looks like N. > > > > > > > But again, in mathematics it is *not* necessary to test > > > > > > each and every individual. Otherwise you could not even prove > > > > > > that > > > > > > sum{i = 1..n} i = (n + 1) * n / 2 > > > > > > for all n, but only for finitely many n. > > > > > > > > > > Of course this is only true for finitely many n. For infinitely > > > > > many > > > > > natural numbers, we get aleph_0. > > > > > > > > You are wrong. We do not get that. If so, pray show a proof. > > > > > > See, for instance, Hrbacek and Jech, p. 188, if you do not believe me. > > > > Again you misread what I state. I do not state "for finitely many i", I > > state "for finitely many n". The formula is a statement about a single n. > > Only on the right-hand side. The left-hand side formula is a statement > about a set. So what? It is about a finite set as specified by a (finite) natural. > > And here we can see again, that "infinite set of finite natural > numbers" is a selfconradiction. Where is this alleged contradiction? For any n in N, the equations "sum{i = 1..n} i = (n + 1) * n / 2 " is a statement involving only a finite number of well-defined operations to verify.
From: Virgil on 10 May 2007 15:30 In article <1178796129.155631.279870(a)u30g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 03:43, Virgil <vir...(a)comcast.net> wrote: > > In article <1178737478.076726.140...(a)n59g2000hsh.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > I did not say that I could name a path which does never branch off > > > from p. But there must be such a path. Why not call it p''. > > > > Why must there be any different path that is not also a separate path? > > > > > > > > > > <snip> > > > > > > >The proof in the tree is, that for > > > > > > every path p, and level M, there is another path *existing in the > > > > tree* > > > > which is not different from p at a level less than or equal > > > > to M. > > > > > Yes. But why only see the one side of the medal? For every path p > > > there is another path *existing in the tree* which is not different > > > from p at any node, because p is not single at any node. What the above says is that for every path p there is a path p' in the same tree which is the same as p at every node but somehow not as a whole the same as p. But that is false in ZF or NBG trees. > > Given any path p and any path p', if there is no node in one but not > > the other, then p = p'. > > That is completely irrelevant! That is entirely relevant to my understanding of paths in a CIBT. > Give me an example of two nodes, such > that p' is not with p at the first but the second, and p'' is not with > p at the second but the first. Then I will believe you. Otherwise you > are wrong. Whatever have I said that WM claims would be wrong if I do not do as WM asks? I have certainly never said anything liked what he seems to be implying I said.
From: Virgil on 10 May 2007 15:36
In article <1178806498.433405.226930(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. And having those few, with intersections, differences, etc., one can build many without a single "rule" f the sort WM implies are necessary. > > > > > > > > > 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 The questions were irrelevant to the issue of whether a function is, first of all, a set of ordered pairs. |