Cantor Confusion
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From: mueckenh on 13 Dec 2006 06:00 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> mueckenh(a)rz.fh-augsburg.de wrote: > >> > Dik T. Winter schrieb: > >> [...] > >> >> Again you have provided neither a definition of "number", nor of > >> >> "grow". > >> >> Are you unable to do so? In common parlance, but that is not > >> >> mathematics. In mathematics functions can grow in relation to > >> >> their argument, but not the entities they denote. > >> > > >> > Functions cannot grow, according to modern mathematics. > >> > >> Wrong. I have provided a definition: > >> > >> ,----[ <45742128$0$97220$892e7fe2(a)authen.yellow.readfreenews.net> ] > >> | Definition: A function f: A |-> B grows iff there exist a1 < a2 of > >> | dom(f) and f(a1) < f(a2). We use the abbreviation "f grows" for of > >> | "the function f grows". > >> `---- > > > > The natural number n is a particular set of n elements. > > If the number n can take a value n_1 and can take a value n_2 > > with n_1 =/= n_2 then the number n can vary. > > How is that related to your sentence that "functions cannot grow"? It is the usual set theoretic view and as such in direct opposistion to my view. > > >> > The expression > >> > "variable" is merely a relict from ancient times when people knew > >> > that the objects of mathematics do not exist in some nirvana but > >> > have to be present in a mind where not everything can be present > >> > simultaneously. > >> > >> How do you call "Textbaustein" in English? > > > > Sorry, I did not expect that you read every word of mine addressed to > > other people. > > You may assume that most of the subscribers at least skim over the > postings of the threads of interest. Hence your copy and paste maneuver > will hardly go unnoticed. I am glad to hear that many people read my texts. Repeated reading supports understanding. Repetition is a Baustein of learning. Regards, WM
From: mueckenh on 13 Dec 2006 06:14 William Hughes schrieb: > > > A_1 = {1} > > > A_2 = {1,2} > > > A_3 = {1,2,3} > > > > > > B = {1,2,3} > > > > > > then B is contained in the last A_i. If there is no last A_I, then > > > there is > > > no A_i that contains B > > > > That has nothing to do with "last". > > If A_i contains B, then A_i contains any A_j. > Therefore A_i is "last". > > >It has all to do with "every A is finite". > > No. From the statement "every A is finite" we cannot conclude > that there exists and A_i that contains B. That is correct. But this conclusion is derived from: 1) N is a linear set (every finite initial segment (= line) includes all preceding segments) 2) There is no element of the complete segment (= diagonal) outside of every finite segment (= line). > > > > > > > > Which of the following three do you claim > > > (note that none of them require assuming that > > > the set of natural numbers actually exists). > > > > > > - {1,2,3 ...} is not an initial segment > > > > If no actual existence is assumed, what is the meaning of "..."? > > > Which of the following three do you claim > (note that none of them require assuming that > the set of natural numbers actually exists). > > - The potentially infinite sequence of > natural numers is not an initial segment The potentially infinite sequence of natural numers is an initial segment. (If we refrain from the physical constraints of numbers.) > > - there is a natural number which is > not an element of the potentially infinite > sequence of natural numbers No. Every natural number belongs to the sequence and to at least one finite initial segment. > > - there is a line for which you cannot > find a natural number which is not an > element of the line Not so many nots please. For each line you can find a natural number which is not an element of that line. And as soon as you have found (created) that number, you have determined (created) the first line to which it belongs. There is no line with every natural number. Similarly you cannot find every elment of the diagonal. If you name an element, then the diagonal is constructed up to that element at least. As there is no line with every number there can be no diagonal with every number. Regards, WM
From: mueckenh on 13 Dec 2006 06:52 Tony Orlow schrieb: > That's fine. If you don't like irrationals, ignore them. Oh I would certainly love them but I never met one. I met only names, different lengths in squares and circles, cubes and spheres, Schall und Rauch. >There are an > actually infinite number of rationals in a unit interval. How "are" they? > Physics used to be more continuous, but atoms and quantum effects have > been discovered. Time and space may even be discrete. Mathematics can > reflect that, or treat things as continuous. I don't think we've > determined for sure that nothing is continuous. Do you? What *in principle* can't be measured, is not existing. Now make a photon with all the energy of the universe 5*10^55 g * c^2. There is no distance smaller than the wavelength of this photon, lamba (or lambda / 2 pi). Regards, WM
From: mueckenh on 13 Dec 2006 06:57 Han de Bruijn schrieb: > William Hughes wrote: > > > Tony Orlow wrote: > > > >>Well, the proof is simple. Any finite number of subdivisions of any > >>finite interval will only identify a finite number of real midpoints in > >>that interval, between any two of which will remain more real midpoints. > >>Therefore, there are more than any finite number of real points in the > >>interval. > > > > This just shows that the number of real points is unbounded. > > It does not show it is infinite (unless of course you use the > > fact that any unbounded set of natural numbers is infinite). > > Isn't unbounded the same as infinite, i.e. = not finite = unlimited = > without a limit? Unbounded is potentially infinite but it is not necessarily actually infinite. Regards, WM
From: mueckenh on 13 Dec 2006 07:04
Dik T. Winter schrieb: > In article <1165923504.410525.226600(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > So you can not show a bijection (in your opinion), but nevertheless you > > > state that you have given a surjection. Do you not think you are > > > contradicting yourself a bit? > > > > I give a surjection on all existing paths. > > But not from the edges. But from parts of edges. > > > > > No? Even if there are enough parts to gather more than a whole edge to > > > > be mapped on every path? > > > > > > No. Each part of an edge may map to a single path, that does *not* give > > > a map from each edge to a single path. > > > > Since infinity = infinity became an allowed equation, 1/2 + 1/2 = 1 can > > no longer be true? > > So if the first half of ab edge maps to, say, 1/2 and the other half to, > say, 3/4. To what single path maps the edge itself? Why isn't it sufficient to collect the shares of two edegs for every path? That shows that there are not more paths than edges. Can you explain your objection to factions? > > > > I can. And I see that each edge maps, when you do this, to a > > > plethora of paths, so that is *not* a surjection from edges to paths. > > > The crucial thing in a surjection (and indeed for a mapping) from A to B > > > is that each element of A maps to a *single* element of B. So, let me > > > ask a different question. To which single path does the edge that goes > > > left from the root map. > > > > That one which goes left from the root is not engaged, because we need > > only half of the set of edges. > > So your construction is not a surjection. Of course it is! It is a surjection from the set of edges onto the set of paths. > > > That one which goes right, could be > > mapped on that real number (path) which is asked for most frequently, > > namely 1/3. That edge leaving the node 1 on level 1 left is mapped on > > the next frequently mentioned number, namely pi. You will not be able > > to ask for more numbers than I can name edges. And you will not be able > > to construct a "diagonal" path. > > And you do not know what a mapping is either. It is not even necessary to know what a mapping is in order to see that you cannot construct (or define) a path representing a number of [0,1] which is not in the tree. Regards, WM |