Cantor Confusion
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From: mueckenh on 16 Oct 2006 16:12 Virgil schrieb: > > No. I will not have a largest number unless I have chosen *how* to use > > all the bits. In unary representation the largest number will clearly > > be less than 10^100. But I am not indebted to focus on any fixed > > representation. Therefore, there is no largest number. > > Then there is no set of numbers. 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, ... > > You have produced a finite set without an upper bound. Remember the 100 > > bits. > > That is not a set unless it is either finite with an upper bound or not > finite. But it is a set of those which really can and do exist. > > > However, you wish to do more. You want to show > > > that claiming "N does not have an upper bound and > > > N exists as a complete set" leads to a contradiction.] > > > > > That is true too. > > But that proves the your "set" of all numbers expressible with 100 bits > is also a contradic tion. Why? > > > > For all n e N we have {2,4,6,...,2n} contains larger natural numbers > > than |{2,4,6,...,2n}| = n. > > There is no larger natural number than aleph_0 = |{2,4,6,...}|. > > That would only be a contradiction if one claimed that aleph_0 were a > natural number. It is a contradiction because aleph_0 is claimed to be larger than any natural number. > > Then it cannot be an ordered set, as an ordered cannot be both finite > and unbounded. Take the largest natural number, for instance. It is a finite set with only one element but without an upper bound. > But there is nothing in the statement of the problem, nor in any part of > any standard logic which requires the number of balls at noon to be a > limit of numbers of balls at times strictly before noon. But the number of balls having been removed from the vase at noon is the limit of numbers of balls having been removed from the vase at times strictly before noon? > What relevance have properties of a function continuous for all x < 0 > have in analyzing properties of a function with infinitely many > discontinuities for x < 0? Use f(t) = 1/9t for t e N, the t-th transaction. Regards, WM
From: Han.deBruijn on 16 Oct 2006 16:13 David Marcus schreef: > Han de Bruijn wrote: > > > > How can we know, heh? Can things in the real world be true AND false > > (: definition of inconsistency) at the same time? > > That is not the definition of "inconsistency" in Mathematics. On the > other hand, I don't know of any statements in Mathematics that are both > true and false. If you have one, please state it. What then is the precise definition of "inconsistency" in Mathematics? Han de Bruijn
From: mueckenh on 16 Oct 2006 16:18 David Marcus schrieb: > I gave my translation of the ball and vase problem. I don't see any > contradictions that follow from it. If you say it implies two > contradictory conclusions, please state them and your proof. What was your result? How many balls are in the vase at noon according to your translation? Regards, WM
From: mueckenh on 16 Oct 2006 16:24 David Marcus schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > > David Marcus schrieb: > > > > > > > > Consider the binary tree which has (no finite paths but only) infinite > > > > > > paths representing the real numbers between 0 and 1. The edges (like a, > > > > > > b, and c below) connect the nodes, i.e., the binary digits. The set of > > > > > > edges is countable, because we can enumerate them > > > > > > > > > > > > 0. > > > > > > /a \ > > > > > > 0 1 > > > > > > /b \c / \ > > > > > > 0 1 0 1 > > > > > > ..................... > > > > > > > > > > > > Now we set up a relation between paths and edges. Relate edge a to all > > > > > > paths which begin with 0.0. Relate edge b to all paths which begin with > > > > > > 0.00 and relate edge c to all paths which begin with 0.01. Half of edge > > > > > > a is inherited by all paths which begin with 0.00, the other half of > > > > > > edge a is inherited by all paths which begin with 0.01. Continuing in > > > > > > this manner in infinity, we see that every single infinite path is > > > > > > related to 1 + 1/2 + 1/ 4 + ... = 2 edges, which are not related to any > > > > > > other path. > > > > > > > > > > Are you using "relation" in its mathematical sense? > > > > > > > > Of course. But instead of whole elements, I consider fractions. That is > > > > new but neither undefined nor wrong. > > > > > > > > > > Please define your terms "half an edge" and "inherited". > > > > > > > > I can't believe that you are unable to understand what "half" or > > > > "inherited" means. > > > > I rather believe you don't want to understand it. Therefore an > > > > explanation will not help much. > > > > > > In standard terminology, a "relation between paths and edges" means a > > > set of ordered pairs where the first element of a pair is a path and the > > > second is an edge. Is this what you meant? > > > > Yes, but this notion is developed to include fractions of edges. > > > > > > I am at a loss as to how "half an edge" can be "inherited" by a path. > > > > An edge is related to a set of path. If the paths, belonging to this > > set, split in two different subsets, then the edge related to the > > complete set is divided and half of that edge is related to each of the > > two subsets. If it were important, which parts of the edges were > > related, then we could denote this by "edge a splits into a_1 and a_2". > > But because it is completely irrelevant which part of an edge is > > related to which subset, we need not denote the fractions of the edges. > > > > > As > > > far as I know, a path is a sequence of edges. Is this what you mean by > > > "path"? > > > > Yes. A path is a sequence of nodes (bits, 0 or 1). The nodes are > > connected by edges. > > > > > > Is edge a the line connecting 0 in the first row to 0 in the second row? > > > > Correct. This edge is related to all paths starting with 0.0. So half > > of edge a is related to all paths starting with 0.00, and half of it is > > related to all paths starting with 0.01 > > > > > Or, is it the line connecting 0 in the first row to 1 in the second row? > > > Or, is it something else? > > > > I have tried above to make it a bit clearer. > > 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? > Then you say that "half of that > edge is related to each of the two subsets". I'm sorry, but I can't > follow this. 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.) Regards, WM
From: William Hughes on 16 Oct 2006 16:33
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > COLLECTED ANSWERS > > > > > > > > > > > Once you have chosen your set of 100 elements it will have > > > > > > a largest size. Questions as to the method of choice have > > > > > > no bearing on this. > > > > > > > > > > > Correct. Once you have chosen how to use all the bits of the universe, > > > > > then you will have a largest number. > > > > > > > > And crucial to this discussion, once you have chosen > > > > that you are going to use > > > > all the bits in the universe you know that you will have > > > > a largest number. > > > > > > No. I will not have a largest number > > > > My statement was not > > "you will have a largest number" but "you know that you > > will have a largerst number". The fact that you know > > that you will choose your largest number sometime in > > the future, and that you have not yet chosen the way > > you will choose this number does not change > > the fact that you know you will have a largest number. > > If you mean that there is a largest number ever defined, then you are > right. But we cannot know which this number will be. Independent of > this there is, as far as I see, no principle limitation to expressing a > number as large as desired. > > > > > > That is correct. It depends on the ingenuity how to define a large base > > > with few bits. There are no limits to ingenuity, therefore there is no > > > largest number. > > > > Piffle. Ingenuity cannot change the fact that there is only a finite > > time to describe the method of representation. > > That is correct. But why must the invention of a very powerful method > of representation consume much time? The invention of a very powerful method of representation may not consume much time, however, there are only a limited (though large) number of representations that can be described in the lifetime of the universe. > > > > > We could > > > > do the usual formailization in terms of Turing > > > > machines, but this would do little except free us from > > > > such paradoxes as "let K be the largest integer that can be > > > > described plus 1".(note the usual interpretation is that K has > > > > not been described, you, however, claim that K cannot exist.)) > > > > > > > > And while the largest integer that will be described is not > > > > yet known (even in theory) > > > > > > I cannot see that. > > > > > > > You cannot know what the largest integer mentioned in the > > New York times in 2007 will be (even in theory). > > Oh, I misplaced my comment. It should appear later > > > > > > the largest integer that > > > > can be described is known (at least in theory). > > namely here: I cannot see that. > In theory every possible way of describing a number is known (this includes descrbing the representation). > > Ingenuity may or may not be limited, but there is a limited amount > > of time to communicate the results of ingenuity. > > Yes. It can be communicated, however, in highly compressed form. Also > the compression depends on ingenuity. There are only a limited number of messages that can be communicated during the lifetime of the universe. This incudes descriptions and use of compression methods. > > > > > > > > > > Let N be the set of finite integers. Either N has an upper bound > > > > or it does not. If you claim that N has an upper bound we will > > > > have to go our merry ways. Otherwise let K be subset of N that > > > > does not have an upper bound. There, I have produced an > > > > infinite set without using the three "...". > > > > > > You have produced a finite set without an upper bound. Remember the 100 > > > bits. > > > > What you described is not a set. It is a large collection of sets. > > None > > of these sets is arbitrary (arbirtray describes a way of choosing > > not what is chosen) Each of these sets has an upper bound. > > I disagree. But if you are correct, then the set of natural numbers has > an upper bound. > > > > > However, you wish to do more. You want to show > > > > that claiming "N does not have an upper bound and > > > > N exists as a complete set" leads to a contradiction.] > > > > > > > That is true too. And it is easy to see: If we define Lim [n-->oo] > > > {1,2,3,...,n} = N, then we can see it easily: > > > > > > For all n e N we have {2,4,6,...,2n} contains larger natural numbers > > > than |{2,4,6,...,2n}| = n. > > > > So we have something that is true for finite sets. > > It is true for finite numbers and sets of finite numbers. Should it not > be true for all sequences of finite even numbers, then there must be > some of even finite numbers, X, in {2,4,6,...,2n, X} which care to push > the cardinality without increasing the sizes. That is obviously > impossible. Nope. Adding a single element, or a finite number of elements cannot take you from a finite set to an infinite set. You have to add an infinite number of elements. So your set X above must be infinite. Adding an infinite set to a finite set certainly changes the cardinality. > > > > > There is no larger natural number than aleph_0 = |{2,4,6,...}|. > > > Contradiction, because there are only natural numbers in {2,4,6,...}. > > > Wiithout any justification whatsoever you state something > > about infinite sets. > > I state something about finite numbers. You state something about the set {2,4,6,...} > > > > Something that is true about finite sets does not have to > > be true about infinite sets. > > That is your standard excuse. It seems that nothing can be true for > infinite sets. Piffle. If I were to say "Something that is true about sets of integers does not have to be true about sets of real numbers" would you say "It seems that nothing can be true for sets of real numbers"? > > > > > This is a (slightly) obfuscated version of "{1,2,3,...,n} is bounded > > > > for all n in N. > > > > Therefore N is bounded". > > > > > > This conclusion is nearly true, in principle, though it should be > > > stated more carefully in order to avoid misunderstanding: N is > > > potentially infinite, i.e., always finite but without an upper bound. > > > > > > > If N |