Cantor Confusion
in [Math]
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From: William Hughes on 10 May 2007 18:25 On May 10, 6:32 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 22:43, William Hughes <wpihug...(a)hotmail.com> wrote: > > > >...there must be such a path. > > >Why not call it p''. > > > because the name p'' has already been used > > for a path that does branch off from p. > > No, it has not. p'' is, by definition, that path which is with p also > at the next node, wherever you rest. > > Funny, I thought it was different. Let's check the definition. Oh, you inadventently snipped it. No problem, let's just restore it. For each other path there is a finite node where it goes a different way. Call this other path, p'', and the node., M. So, at node M, p'' goes a different way from p. So, by definition, p'' does branch off from p. You must have been distracted by the pink elephants. - William Hughes
From: Carsten Schultz on 10 May 2007 19:11 WM schrieb: > > In Germany mathematics belongs to "Wissenschaft" which is usually > translated by science. The English word `science' usually refers to what is called `Naturwissenschaft' in German. -- 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: Ralf Bader on 10 May 2007 19:39 MoeBlee wrote: > On May 10, 1:47 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: >> >> I said that a function consists of formula, domain and range. >> This is true and provable, whatever you try to counter argue. > > Provable? In what system of logic and from what axioms? > > Define, in set theory, 'consists of'. Maybe he is asserting that it is common usage and understanding to define the notion of function in that way. However, in any source I know the notion of function is explained in basically the same way as in Hrbacek-Jech. This is even the case in the textbooks of mathematics for engineers in German "universities" (say: community colleges) of applied sciences (I checked the books of Papula and Stingl). Nobody talks about a "formula" as a necessary ingredient of a function, and there are reasons for this: 1 - "Rules" or "procedures" have a semantic component. A formula is a purely syntactical object and by itself doesn't tell anything about how to obtain a value for an argument; the formula "f(x)=x^2" doesn't tell anything about the meaning of "^2". To explain that meaning, one already needs the notion of function, or something similar. If that notion is available, then one can use formal expressions to denote such functions. So Mückenheim's idiotic (in the classical sense) attempt is kind of impredicative, a crime he abhors if others commit it. 2 - Theorems like the existence theorem of Peano for solutions of differential equations become dubious, if not outright wrong, if a function needs a "formula". I don't see how one could sqeeze out a formula for the solution function out of the proof of this theorem, although it gives a kind of rule for that function. The development of the notion of function, leading to the present and unanimously applied definition, is a prominent chapter in the history of analysis, and there were stages where a function was understood to be given by various kinds of formulas or analytic expressions. But all of these definitions left something to be deserved. However, from Mr. Mückenheims previous utterances there can be no reasonable doubt that he is absolutely incapable of understanding these issues. Ralf
From: WM on 11 May 2007 02:54 On 10 Mai, 21:21, Virgil <vir...(a)comcast.net> wrote: > In article <1178795907.957410.94...(a)o5g2000hsb.googlegroups.com>, > > WM <mueck...(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 meant only finitely many. My personal opinion is not the starting point here. Since mathematical entities exist only in the mind, such entities, which cannot exist in the mind, cannot exist at all. Hence, mathematical entities which cannot be asked for, cannot exist. But this should come to you as an unavoidable conclusion, not as my personal opinion. > > > > > > 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. Then all different paths should be distinguishable. This is contradicted by the fact that every node of p which you can ask for is shared by the same path p'. > > > But we know, that this > > not possible. > > Is that a royal "WE"? No, it includes only people capable of logic thinking and refuting the notion of infinite sets, in particular of infinite definitions. > > 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? That is impossible. Therefore there are not all nodes. > > 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? There is an antinomy. Therefore there are not actually infinitely many nodes. > > 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. Your last two sentences show the antinomy, because for any node we can show that one and the same path ' accompanies p for all nodes investigarted. > > > 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. In mathematics it holds for the potentially infinite set of natural numbers MatheRealism shows that it holds only for few natural numbers. > > > 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. Why is that rule required? > > > > > 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? For the actually infinite set of "all naturals" it is false. All naturals are finite, so we can sum all naturals by this formula, but as the result is 1+2+3+...= aleph_0 there must be an infinite natural, which is a contradiction. > > 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. Every natural is finite. So we can use this formula for every natural (right) and for all naturals up to every natural, i.e., for all naturals. Or is "all up to every" not the same as all? Regards, WM
From: WM on 11 May 2007 04:00
On 10 Mai, 22:10, Virgil <vir...(a)comcast.net> wrote: > In article <1178817355.571399.114...(a)y80g2000hsf.googlegroups.com>, > > 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. !!!!!!!!!!!!!!! > > That depends on what one wants to do with the set. H&J could have written: "it does not matter that the set of these ordered pairs is described by a property" in your opinion? > In dealing with sets > of ordered pairs in general, one need not describe it as anything but a > set of ordered pairs. Did I specify the rule, process, mapping, in short: the formula? When dealing with the general notion of function, i.e., when not dealing with a special function you need not specify the formula. But without a formula or formal description determining by what property the set of ordered pairs is described, there is no set of ordered pairs, i.e., *there is no function*. And without having a domain, the function is not defined at all. Of course you can name every element of the domain separately. But that does not veil the fact, that a domain is required and is existing for any well defined function f. Regards, WM |