Cantor Confusion
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From: Han de Bruijn on 13 Dec 2006 07:37 mueckenh(a)rz.fh-augsburg.de wrote: > 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. Yep! And since actually infinite does not exist, we are just READY for doing some _real_ mathematics. No? Han de Bruijn
From: Franziska Neugebauer on 13 Dec 2006 07:45 mueckenh(a)rz.fh-augsburg.de wrote: > 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. Can't see any support for your claim, that "functions cannot grow". >> >> > 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. It is you who repeats. Are you still not convinced of your mathematical revisionism? F. N. -- xyz
From: Han de Bruijn on 13 Dec 2006 07:54 William Hughes wrote: > Han de Bruijn wrote: > >>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? > > Keep your cranks straight. > > To TO, the set of finite naturals is unbounded but > not infinite. > > To WM, the set of finite naturals does not actually exist. > It is a potentially infinite set. > > To you, the set of finite integers is bounded > by a largest integer > so there is no unbounded set of naturals. > Any statement made about an unbounded set of > naturals is vacuously true. Huh, no! It is _you_ who "can" distinguish that many kinds of different infinities. There is only _one_ for me. It's exemplified in the theorem that there are infinitely many primes. Suppose there are not, then form the number p1.p2.p3. ... .pN of all primes { p1,p2, ... ,pN } we have so far and add 1 to them. The number you get is not divisible by any of the { p1,p2, ... ,pN } hence our premise is false. Therefore the number of primes is not bounded = unlimited = without a limit = not finite. If you call _this_ infinity, then I suppose we may be on speaking terms. Han de Bruijn
From: Dik T. Winter on 13 Dec 2006 07:59 In article <1166011440.201958.55080(a)73g2000cwn.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1165923504.410525.226600(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > > > 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 have explained. Why do I need to repeat that? You have *not* provided a surjection from the edges to the paths. And I have no idea what you mean by "collect the shares of two edges for every path". > > > 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. Than again. To what number maps the edge that goes left from the root? If it is a surjection, there should be one. > > > 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. You had said that you had constructed a surjection. A surjection is a mapping with specific properties. It is necessary to know what a mapping is in order to construct a surjection, which you did not do. And I have *no* idea what you mean with the second half of your sentence. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: William Hughes on 13 Dec 2006 08:00
Han de Bruijn wrote: > William Hughes wrote: > > > Han de Bruijn wrote: > > > >>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? > > > > Keep your cranks straight. > > > > To TO, the set of finite naturals is unbounded but > > not infinite. > > > > To WM, the set of finite naturals does not actually exist. > > It is a potentially infinite set. > > > > To you, the set of finite integers is bounded > > by a largest integer > > so there is no unbounded set of naturals. > > Any statement made about an unbounded set of > > naturals is vacuously true. > > Huh, no! It is _you_ who "can" distinguish that many kinds of different > infinities. There is only _one_ for me. It's exemplified in the theorem > that there are infinitely many primes. Suppose there are not, then form > the number p1.p2.p3. ... .pN of all primes { p1,p2, ... ,pN } we have > so far and add 1 to them. The number you get is not divisible by any of > the { p1,p2, ... ,pN } hence our premise is false. Therefore the number > of primes is not bounded = unlimited = without a limit = not finite. If > you call _this_ infinity, then I suppose we may be on speaking terms. > Therefore, as there are at least as many digit positions in 0.111... as there are primes, there are an infinite number of digit positions in 0.111... - William Hughes > Han de Bruijn |