Cantor Confusion
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From: mueckenh on 18 Oct 2006 04:03 David Marcus schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > David Marcus schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > Virgil schrieb: > > > > > > > > > My sympathies to his poor students. > > > > > > > > I will tell them your ideas about the vase and then ask them about > > > > their opinion. But don't forget: They are not yet spoiled by what you > > > > call logic. > > > > > > > > > > > > - plainly cannot > > > > > > > comprehend the difference that swapping quantifiers makes. He cannot > > > > > > > comprehend that there might be a difference between the significance of > > > > > > > "every" in "Every girl in the village has a lover" and "John makes love > > > > > > > to every girl in the village". > > > > > > > > > > > > Is the Imaginator too simple minded to understand, or is it just an > > > > > > insult? The quantifier interchange is impossible in general, but it is > > > > > > possile for special *linear* sets in case of *finite* elements. > > > > > > > > > > For example? > > > > > > > > > > Does "Mueckenh" claim that, say, > > > > > "For every natural n there is a natural m such that m > n" > > > > > and > > > > > "There is a natural m, such that for every natural n, m > n" > > > > > are logically equivalent? > > > > > > > > > > All the elements are finite and linearly ordered. > > > > > > > > The second statement is obviously wrong, because there cannot be a > > > > natural larger than any natural. > > > > The quantifier exchange however is possible for sets of finite numbers > > > > n the following form: > > > > "For every natural n there is a natural m such that m >= n" > > > > and > > > > "There is a natural m, such that for every natural n, m >= n" > > > > This natural m is not fixed. It is the largest member of the set > > > > actually considered. > > > > > > Please let people know when you are not using standard terminology and > > > when you do this, please define your terms. What does it mean to say a > > > natural number "is not fixed"? > > > > One cannot know it, cannot call it by its name, but it is provably > > present. You should be familiar with this from of existence. It is like > > the well-order of the reals: present but very. > > I'm sorry, but "cannot call it by name" is not a "form of existence" > that I've seen in any math book I've ever read. Please define what you > mean using standard mathematics. Didn't you read in your books that the well order of the reals, for example, cannot be defined better than the size of this number? The state of existence is: Existence proven but construction or definition impossible. Regards, WM
From: mueckenh on 18 Oct 2006 04:04 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Alan Morgan schrieb: > > > > > > > >> As I have inductively gone through the entire list of balls introduced > > > >> into the vase and found that each of them has been removed before noon, > > > >> why should stating that trivial fact be considered a joke? > > > > > > > >But you cannot go inductively through the cardinal numbers of the sets > > > >of balls in the vase? They are 9, 18, 27, ..., and, above all, we can > > > >show inductively, that this function can never decrease. > > > > > > You think that's bad? I have an even simpler situation! Add one ball > > > at 1 minute to noon, another ball at half a minute to noon, another > > > at 1/4 minute to noon, and so on. The number of balls in the vase before > > > noon is always finite, but somehow, miraculously, at noon the number of > > > balls in the vase becomes infinite. When, oh when, does that transition > > > from finite to infinite happen? > > > > > > I submit that this is just as wierd a result as the original problem. > > > > Weird is that adding 9 balls instead of 1 per transaction leads to zero > > balls. > > Weird is that taking off 1 ball per transaction leads to all balls > > taken off and no ball remaining, if the enumeration is 1,2,3,... but to > > infinitely many balls remaining, if the enumeration is 10, 20, 30, .. . > > This in particular is weird because there is a simple bijection between > > 1,2,3,... and 10, 20, 30, ... > > > > Right. No matter which balls you pick you are going to remove an > infinite > number of balls. So the number of balls you remove does not matter. > Which balls you remove does matter. Fine. The question is, however, is this set "all balls" or how many balls remain in the vase at noon? Regards, WM
From: mueckenh on 18 Oct 2006 04:09 David Marcus schrieb: > > > First you refer to a "relation between paths and edges". > > > > Correct. > > > > > Then you say > > > the "edge is related to a set of path". > > > > Is there any contradiction? > > Using standard mathematical terminology, there most certainly is a > contradiction. You have apparently given two different descriptions as > to what the elements of your "relation" are. In one, the elements are > paths, but in the other they are sets of paths. A path is not a set of > paths. Please state the definition of your "relation" clearly. If you > use a word in other than its standard mathematical meaning, then please > give a mathematical definition. Read this again: > > I warned you that this point is new: The edges are split in shares of > > 1/2, 1/4, and so on. But when fractions were introduced in mathematics, > > most people may have had the same problems as you today. I am sure you > > can understand it from the written text above. (Many others have > > already understood it.) If a fraction of an edge is related to a path, which is really new, as far as I know, then the whole edge is related to a set of paths. Otherwise it would be meaningless to use fractions. This new technique was exactly described in my original text. Regards, WM
From: mueckenh on 18 Oct 2006 04:11 David R Tribble schrieb: > mueckenh wrote: > >> You see, your proof is rubbish. B will have a largest element. And the > >> set of all numbers ever used in the universe in eternity also will have > >> a largest element. But it has not yet. > > > > How do you know? How do you know that the universe is of a finite age? > > > William Hughes schrieb: > >> Therefore it is unknown. However, it is not arbitrary. > > > > mueckenh wrote: > > The largest element possible with 100 bits can be very different, > > according to my arbitrary choice of representation. > > Of course. I can choose a representation whereby each bit represents > a power of 10^100, for example. Thus with 100 bits the largest number > I can represent is (10^100)^100-1. > > However, there are only 2^100 unique combinations of values using > 100 bits. So regardless of the largest _value_ I can represent, I can > represent far fewer _total_ unique values. That is correct. That is precisely the reason why the natural numbers appear to be infinitely many. There are at most 10^100 different numbers but there are far larger values. The set is finite, but has not a largest element. Regards, WM
From: mueckenh on 18 Oct 2006 04:13
William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > > > > > No. The fact that everything that is true about the infinite > > > > > must be justified in the finite, does not mean that everything > > > > > that can be justified in the finite must be true about the > > > > > infinite. > > > > > > > > > > You prove that something is true in the finite case. You > > > > > do not justify your transfer to the infinite case. > > > > > > > > Who has ever justified such a proof? In fact that is impossible because > > > > there is no infinity. Therefore all such "proofs" are false. But if we > > > > assume the existence of the infinite, then the sum of the geometric > > > > series is the most reliable entity at all. (Niels Abel: With the > > > > exception of the geometric series no series has ever been calculated > > > > precisely.) > > > > > > If you wish to assme that infinity does not exist, knock yourself > > > out. However, if you are trying to show that the assumption > > > that infinity does exists leads to a contradiction you need > > > to justify the proofs that you make using that assumption. > > > > > > > > > > > > > > The axiom of infinity > > > > > > applies to the paths. They are nothing but representations of real > > > > > > numbers. These exist according to set theory, therefore the paths > > > > > > exist too. > > > > > > > > > > > > > > > > Yes, but the question is not "do the paths exist?". > > > > > > > > There are two questions: Do the infinite paths exist and does the > > > > geometric series wit q = 1/2 have a limit? I don't need any further > > > > infinities. > > > > > > > > > > No, there is a third question: "What is the connection between the > > > infinite paths and the limit of the series?" You have only shown a > > > connection between finite paths and partial sums. > > > > Wrong. The connection between finite paths and partial sums of edges > > leads to > > (1-(1/2)^n+1)/(1 - 1/2) edges per path. > > > > And this is the only connection you have ever shown. You then > take a limit and get 2. But you have never shown that this > limit is connected to anything. Each path consists of aleph_0 edges. > The fact that you have a connection > in the fnite case is not enough to show that the same > connection holds in the infinite case. So you disagree with lim (1/2)^(n+1) = 0 for n --> oo. (At least in this special case.) Regards, WM |