Cantor Confusion
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From: Franziska Neugebauer on 13 Dec 2006 13:35 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> >> Mind and brain and representation of (abstract) entities therein >> >> is still off topic in sci.math. >> > >> > This decision is >> > 1) wrong >> > 2) not yours. >> >> Which _mathematical_ institution is doing research in the field of >> "representation of (abstract) entities in mind or brain"? > > New aspects are new because they have not been treated yet. I thank you for confirming that "representation of (abstract) entities in mind or brain" is not a mathematical issue. Though off topic in sci.math I would like to ask you which _neuroscientific_ knowledge you are referring to when you write ,----[ <1165819653.080811.290680(a)l12g2000cwl.googlegroups.com> ] | "Of course [am I discussing issues of neuro sciences --F.N.], because | math requires mind and brain." `---- F. N. -- xyz
From: mueckenh on 13 Dec 2006 13:52 Dik T. Winter schrieb: > > > > 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. > A surjection onto the paths covers all elements of the *range* = every edge. It is not necessary that every edge is mapped on a path, to show that there are not less edges than paths. It is only necessary that there is no path without edge mapped on it. > > > > 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. If someone makes an asserted list of all real numbers and hands it out to Cantor, then Cantor can construct a diagonal number which is not in the list, contradicting the assertion. If someone makes a binary tree of all real numbers and hands it out to Cantor, then Cantor cannot construct a diagonal number which is not in the tree, so no contradiction of completeness is possible. The problem with Cantor's theorem aleph0 < 2^aleph0 is that the function f(x) = 2^x must be discontinuous: For natural x = n we have 2^n < aleph0, for the least infinite x = aleph0 we have 2^aleph > alep0. So there is a gap. The values between aleph0 and 2^aleph0 cannot be taken by the function 2^x. This is highly suspicious, if aleph0 is to be a number which is in trichotomy with the natural numbers, but I can't prove it wrong. By means of the binary tree I can exclude any discontinuity. Therefore the number of branches isolated at a certain level n grows continuosly from 1 to its maximum value. If this was 2^aleph0, then the number of branches could not but call on aleph0 or at least at a value between aleph0 and 2^aleph0. To show that this is impossible is achieved by the big advantage of the tree: its undoubted continuity. Regards, WM
From: mueckenh on 13 Dec 2006 13:56 Han de Bruijn schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Tony Orlow schrieb: > > > >>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. > > Sigh! The common confusion about what continuity means. Okay, I'll never > become tired of repeating this over and over again: CONTINUITY IS IN THE > EYE OF THE BEHOLDER. The _same_ phenomenon in nature may be discrete as > well as continuous. Fluid Flow of water consists of a zillion molecules, > hence it is Discrete. But it is described with the continuum equations > of Fuid Dynamics, hence it is Continuous. A less well-known example is > the Fluid Tube Continuum model, by which I've been able to calculate the > flow and temperature distributions in a shell and tube heat exchanger. > It's found at: > > http://hdebruijn.soo.dto.tudelft.nl/www/programs/pascal.htm#Nerat > Graphically: > http://hdebruijn.soo.dto.tudelft.nl/www/programs/plaatjes/slide01.htm > > It is even possible to calculate critical values for certain variables > in such models. Beyond such values the continuum model breaks down and > the discrete substrate becomes sensible. See: In mathematics it is the same. Say with Robinson: "(i) Infinite totalities do not exist in any sense of the word (i.e., either really or ideally). More precisely, any mention, or purported mention, of infinite totalities is, literally, meaningless. (ii) Nevertheless, we should continue the business of Mathematics 'as usual', i.e., we should act as if infinite totalities really existed." And exchange "infinite" by "continuous". =============== > > 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? Of course, Han, after having grasped that actual infinity does not exist, we can say unbounded = infinite, and that's it. But I am afraid most of the participants here have not yet taken the plunge. They could easily become confused. Regards, WM
From: mueckenh on 13 Dec 2006 14:00 Dik T. Winter schrieb: > In article <457ece72(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > Dik T. Winter wrote: > > > In article <457d8cc0$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > ... > > > > Why does that matter? > > > > > > It does matter because if you do not properly define your problem, > > > mathematics is not able to give an answer. > > > > It's sufficiently defined if one assumes that there is a uniform > > probability distribution. > > You can assume as much as you want, that does not make it a definition. > What *is* a uniform distribution of all natural numbers? > > > > > This is the same thing as your stupid ball and > > > > vase trick. Why do you need to label anything, or know what you're > > > > choosing from the infinite set? > > > > > > Because that is part of the problem setting. Giving that setten will > > > allow mathematics to model the question and give an answer. > > > > That problem has a clear answer with or without the labels: the sum > > diverges as f(n)=9n. The labels are confounding, not clarifying. > > Yes, but that does *not* indicate anything about the limit, as I did show > below: > > > > And it is bad to think that because for a sequence of sets holds that > > > lim{n -> oo} |S_n| = k > > > with some particular value of k, that also > > > | lim{n -> oo} S_n | = k > > > because the latter statement contains something that has not been > > > defined in mathematics. > > > > I'm not sure what that statement is supposed to say. Can uoi give an > > example? > > You have: |S_n| = 9n and you think that > | lim{n -> oo} S_n | = lim{n -> oo} | S_n | = lim{n -> oo} 9n. > but first "lim{n -> oo} S_n" is not defined, and second, when you define > it you have to prove that the first part is equal to the second part. > > > > But even when we define it, it is not certain > > > that it holds. Given the following (I think reasonable) definition: > > > lim{n -> oo} S_n = S > > So, what, S_n is supposed to be an initial segment of the sequence? > > I do not understand. S_n are sets indexed by n and so form a sequence. > > > > if: > > > (1) for every element a in S there is an n0 such that a is in each of > > > the sets S_n with n > n0 > > > (2) for every element a not in S there is an n0 such that a is not in > > > each of the sets S_n with n > n0. > > In (2), it sounds like a would not exist in ANY S_n if it's not in S. > > No. Any S_n with index larger than some n0. > > > > So from some particular point an element either remains in the sets in > > > the sequence or remains out of the sets. > > > > You mean, at some point you can tell whether a given element a is in S, > > because if it were, it would be there by then? > > and if it were not it would not be there by then. (Note that the point > where that is the case can be different for each element.) > > > > With this definition (when we look at the rationals) we have that > > > lim{n -> oo} [0, 1/n] = [0] > > Okay that interval degenerates to 0.... > > > > > and so: > > > lim{n -> oo} | [0, 1/n] | = aleph0 != 1 = | lim{n -> oo} [0, 1/n] | > > > (I am talking standard mathematics here). > > > > Are the |'s supposed to denote set size? If so, how can you claim that > > [0,0] contains aleph_0 elements? > > Where do I claim that? > > > > So taking cardinality and limits can not be interchanged except in some > > > particular cases. But that is not unprecedented in mathematics. > > > limits and integrals can also not be interchanged except in particular > > > cases. And so can the interchange is not in general passoble if one > > > of the things you interchange is a limit. Even interchanging limits > > > is not in general possible. Consider: > > > lim{x -> oo} lim{y -> oo} (2x + 3y)/xy > > > > True, but is it relevant? > > Yes, because the same holds for cardinality. Could you please explain what same holds for cardinality? Regards, WM
From: mueckenh on 13 Dec 2006 14:03
Franziska Neugebauer schrieb: > 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". In *set theory* a set does not grow. In set theory a function is a set. In set theory a function cannot grow. In my view a function can grow. A function can be abbreviated as number variable or as variable number. > > >> >> > 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. Of course, I do so in order to support your understanding. > Are you still not convinced of your mathematical > revisionism? You misread. It is called mathe-*realism* not mathe-revisionism. Regards, WM |