Cantor Confusion
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From: Virgil on 2 Jan 2007 17:15 In article <1167739146.564586.101570(a)s34g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > You said: > > > It is enough to show that for every finite tree, the given path is not > > > in it, in order to show that it is not in the union. > > > > Since that exactly conforms to the definition of unions, yes. > > An object is IN a union if and only if it is in at least one of the sets > > of that union, and thus is not in the union unless it is in at least one > > of the sets in the union.. > > The object N = {1,2,3,...} is not in the union of all initial segments > {1,2,3,...,n}. Not as a member, certainly, but it is a subset of that union. > > > > > > > I say: > > > Is the union of all finite trees different from the complete infinite > > > tree? > > > > Yes! > > Which nodes or edges are in one but not the other? I assume you think > the union of all finite trees contains less edges or nodes. So which > edge or node is in the complete tree but not in the union of all finite > trees? A finite tree is a set of finite paths, so every union of finite trees is a union of such sets of finite paths, and thus is limited to members which are finite paths. > > > > > > It is not what is shared by two sets of paths which separate them but > > what is not shared. To separate one path from all others requires an > > infinite set of its included edges and/or nodes. For any such infinite > > set, any other path will excude infinitely many of those. > > > > This can be seen by noting that any two infinite paths can have no more > > than a finite number of nodes and/or edges in common. So each contains > > infinitely many not in the other. > > Let me know which edges of the complete tree distinguish it form the > union of all finite trees (and all the rational numbers from the > irrational numbers). For each finite path in the union, there are infinitely many edges in any infinite path not in that finite path, so that the infinite path is separated from each finite path from some edge onwards. There is no more need to simultaneously separate one infinite path from all other paths than there is to simultaneously separate one real number from all other real numbers. Such separations need only be done for one pair at a time.
From: Virgil on 2 Jan 2007 17:17 In article <1167759460.349040.128220(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > Where is a contradiction? > > > > Read the quote below. Your answers to my questions 2 and 3 constitute a > > contradiction. Please correct me if I'm wrong: # > > You are wrong. That is not a correction, burt an unproven claim. > > > > Your answer to my question 2 states > > > > x is identical to y > > <-> there is no difference between x and y (P) > > Correct. x is an idea. It is identical to the entity which you call an > irrational number, i.e., an idea y. > > > > Your answer to question 3 states > > > > x is different from y (due to some reason) > > <-> there is a difference between x and y (Q) > > You simply fail to distinguish sharply enough between y and y'. > x is different from y', i.e., from what you think an irrational number > is., i.e., from an idea which can be approximated to any given > precision by rational numbers. That is by your definition, not anyone else's, so is only valid in your non-system, not anyone else's.
From: cbrown on 2 Jan 2007 17:25 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > <snip of many questions that remain unanswered> > > > "3/2 < 2" and "|N| < |R|" have /completely different meanings/. The > > first is a linear order applied to a field. The second is an ordering > > based on surjections / injections. > > > Both are based upon linear orderings. Yes; they are /different/ linear orderings; which are /based on/ different things. At any rate, you did not answer my question: I stated: > > > In the latter case, we are not asking "is there a surjective function > > > from 1 + 1/2 + ... onto 1?", because that would be simply nonsense. You replied: > > Only if one does not know the meaning of the arithmetic of rational > > numbers and limits. I asked: > A truly fantastic claim! Then please give a rule for determining a > surjection from sqrt(2) to 1 that is neither nonsense nor dependent on > first /assuming/ sqrt(2) >= 1. You replied by ignoring the question. I assume from your other comment "the arithmetic of rational and irrational numbers... are based on counting of ordinal numbers" that by "sqrt(2)" you mean, e.g., the equivalence class of all Cauchy sequences of rationals which, when squared, converge to the rational number 2, and that by "1" you mean, e.g., the equivalence class of all Cauchy sequences of rationals which converge to the rational number 1. By the square of a sequence of rationals (r_0, r_1, r_2, ..., r_n, ....), I mean the sequence (r_0*r_0, r_1*r_1, r_2*r_2, ..., r_n*r_n, ....). By "a rational", I mean an equivalence class of ordered pairs of integers (p,q), q not equal to 0, where (p,q) is equivalent to (r,s) iff p*s = r*q. By "p/q" I mean the unique equivalence class of (p,q). By "p", I mean the unique equivalence class of (p,1). And for two rationals x and y, we define multiplication as x*y = { (u,v) : exists (p,q) in x and (r,s) in y such that u*q*s = v*p*r}. And by "integer", I mean an ordered pair (z, s), where z is a natural, and s is in Z_2 to indicate the sign; and where (0,0) is equivalent to (0,1); and where multiplication is defined as (x, s_x)*(y, s_y) = (x*y, s_x + s_y). And by "natural", I mean finite ordinal; where multiplication of naturals is as it is usually defined for ordinals. So assuming that (or something like it) is what you are trying to say by "based on counting of ordinal numbers", I ask again: What is an example of a surjection from sqrt(2) onto 1? Why is there no surjection from 1 onto sqrt(2)? To spare us a bunch of wrangling over misunderstandings of Cauchy sequences, it would be much simpler and more meaningful if you would fumble your way through answering these questions restricted to the rationals: You say that 2 > 3/2. What is an example of a surjection from the rational number 2 onto the rational number 3/2? Why is there no surjection from the rational number 3/2 onto the rational number 2? What is the basis for your claim that there is a surjection from the rational 6/4 onto the rational 3/2, but no surjection from the rational 4/6 onto the rational 3/2? You of course understand what is meant by a surjection from a set A onto a set B, yes? Or do you now agree that asking "is there a surjective function from 1 + 1/2 + ... onto 1?" is nonsense? Cheers - Chas
From: David R Tribble on 2 Jan 2007 19:21 David R Tribble schrieb: >> Well, now I'm confused. Could you provide an example of a natural >> number that does not exist? > mueckenh wrote: > Take the first 10^100 digits of pi (if you can - but you cannot). It is > impossible to bring this number to existence in the whole universe. Can you "bring into existence" the number 1?
From: Dik T. Winter on 2 Jan 2007 20:09
In article <1167738133.535489.164800(a)n51g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1167493176.710056.286330(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > At what point is 1/3 represented? > > > > > > In the binary tree it is not represented, but it could be represented > > > by using a ternary tree. > > > > Now you are arguing differently from before. When I stated that 1/3 > > was not in your tree, you argued that it was in your tree. Do you now > > think differently? > > No. I have been knowing for several years already that the decimal or > binary representation of 1/3 does not exist. You state so, contradicting the axiom of infinity. > But if you insist that > irrational numbers exist, then you must also accept the existence of > 1/3 as a binary series. It exists. > > > There is no representation > > > of 1/3, unless it were in the infinite. But if you insist on an > > > existing decimal representation of 1/3 in the infinite, then also the > > > union of all finite binary trees contains such an existing binary > > > representation of 1/3. > > > > I still do not understand what subtleties you are making. In the > > *complete* binary tree 1/3 does exist. But that is *not* the union > > of your rational trees. > > So the complete tree must have some more edges and nodes. Where are > they? No, it has the same number of edges and nodes, but it has more paths. > By the way, if you are right, then N = {1,2,3,...} is not in the union > of all initial segments {1,2,3,...,n}? Perhaps, N does not at all > exist? Wrong again. The union of an infinite collection of sets is defined as the set of elements that is element of at least one of the sets from the collection. Each natural number is element of at least one set {1, 2, 3, ..., n}, so it is in the union. There is no infinite path in each of the finite binary trees, so there is also no infinite path in the union of those trees. It is as simple as that. > > > My question is only the following: Do you think that the union of all > > > rational binary trees is the complete infinite binary tree or not? > > > > It is not. But I already did answer that when I answered your question > > what the distinction was between the union of rational trees and the > > complete tree. So why do you ask again? I state they are distinct, > > and indicate in what way they are distinct, and you ask whether I > > think they are distinct? > > Sorry, you answered the question "What distinguishes this union of > rational trees from the complete tree?" > > That some rational numbers are missing. If the union contains all > rational numbers, there must be one of the rational trees that contains > (for instance) 1/3. There is none. Yes. The union does not contain all rational numbers. > and you asked: > > Where is 1/3? At what point does it get into the union of rational > trees? Rightly so, because 1/3 *is* in the complete binary tree, but it is not in the union, so they are different. > But you did not say which edges or nodes distinguish the complete tree > from the union of all rational trees. > Do you believe that there are such distinguishing edges but that one > cannot name them? > Or do you think that there are same edges in both trees but that some > paths can form in one tree which cannot from in the other tree? The last one. In the union there is *no* infinite path. > Wouldn't both answers point to some matheology? No, they are based on the definition of union and some elementary logic. (1) Each node is in one of the finite trees, so it is also in the union. (2) Each edge is in one of the finite trees, so it is also in the union. (3) Infinite paths are in none of the finite trees, so they are also not in the union. With which of the above three statements do you disagree? If (1), name a node not in one of the finite trees. If (2), name an edge not in one of the finite trees. If (3), name an infinite path that is in a finite tree. Apparently you agree with all three, but at the same time states that the completed infinite tree (which has infinite paths) is the same as the union of the finite trees. Do you not see a contradiction here? > > > > That is the reason why the cardinal number of the paths in both trees > > > > *is* different. > > > > > > > With a different number and configuration of nodes (or edges) or with > > > exactly the same? > > > > You lost me here. We were talking about the union of finite trees and > > the complete tree. > > And my question is: Do these two trees, namely the complete tree and > the union of all rational trees differ such that the one has edges or > nodes which are missing in the other? No. But the distinction is apparently difficult (although about first year at University for mathematics). The set of terminating binary expansions is countable, the set of non-terminating binary expansions is no countable. (You may replace terminating binary expansions with binary expansions terminating with either a continuous stream of 0'z or of 1's.) > > You are wrong. sqrt(2) has a pretty good representation: "sqrt(2)". > > That is a name. Name ist Schall und Rauch. In the same way '2' is a name, '10' is a name. '10' is clearly a name for a quantity that depends on the context. In the same way 'sqrt(2)' is the name for a quantity that depends on the context. For instance, the wedge that is used for the quantity 7 in Hyderabad Arabic is used for the quantity 8 in Devanagari and for the quantity 6 in Javanese. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |