Cantor Confusion
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From: Virgil on 3 Jan 2007 19:33 In article <1167859855.107241.239690(a)k21g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > Where is the ending of the union of all finite paths? Isn't it smilar > > > to the "ending" of all finite natural numbers? > > > > It is quite similar in that one has in both cases an infinite set of all > > whose members are finite. > > That is precisely the infinite string of finite digit indexes of > irrational numbers and similarly the infinite string of finite nodes of > infinite paths. But one has no infinite paths in the tree? Speak only for yourself. In an infinite binary tree, there are uncountably many infinite paths. It is only in unions of sets not containing infinite paths that one need look in vain for infinite paths. > > > > It is not standard that the object N = {1,2,3,...} is not in the union > > > of all initial segments {1,2,3,...,n}? > > > > It is not a member of that union because it is not a member of itself. > > > > WM again conflates being a member of a set (the uniion) with being a > > subset of that set. N is a subset of the union of all initial segments ( > > actually equals the union) but is not a member of that union. > > So the union U{n | n e N} = U{1,2,3,...,n} is not N? Wat is missing? In my book, U{n|n e N} is not equal to U{1,2,3,...,n}. And, in addition to any common sense, it appears that an "h" is missing from WM's question. > (I do not ask for a member. I do not ask whether an infinite path is a > member. I ask whether it is in the union of finite trees.) What is the "it" that you are asking about? U{n | n e N = N. but U{1,2,3,...,n} = n+1 in NBG.
From: Virgil on 3 Jan 2007 19:37 In article <1167860172.166221.257060(a)k21g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > You say that 2 > 3/2. What is an example of a surjection from the > > rational number 2 onto the rational number 3/2? > > There is an injection from 3 into 4 which is not a surjecton (namely > 3/2 and 4/2). It is equally relevant that the standard longitude and latitude of earth allows only two poles. > > > Why is there no > > surjection from the rational number 3/2 onto the rational number 2? > > There is no surjection from 3(/2) into 4(/2). It is equally relevant that the standard longitude and latitude of earth allows only two poles. > > > > > 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? > > There is a bijection from 6 onto 6 (namely 6/4 and 3*2/2*2 = 6/4). It is equally relevant that the standard longitude and latitude of earth allows only two poles. > > > > You of course understand what is meant by a surjection from a set A > > onto a set B, yes? > > Of course. Your responses would seem to indicate otherwise.
From: Virgil on 3 Jan 2007 19:51 In article <1167860527.440727.143820(a)i80g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > 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. > > So there is an infinite number of nodes in the union of finite numbers > of nodes. Only if WM can prove that one can generate the necessary infinite number of trees from a merely finite number of nodes. Which I doubt. > > > (2) Each edge is in one of the finite trees, so it is also in the union. > > So there is an infinite number of edges in the union of finite numbers > of nodes. Only if WM can prove that one can generate the necessary infinite number of trees from a merely finite number of edges. Which I doubt. > > > (3) Infinite paths are in none of the finite trees, so they are also not > in the union. > > So the set of natural numbers is not in the union of all initial > segments of natural numbers? Certainly not as a member. But it is the very set which has AS MEMBERS all of the members of all of the initial sets. > > > > 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. > > That is wrong. If an infinite number of nodes is in any path of the > union, then the path is infinite too. But it is impossible for any finite path in any finite tree to contain more than finitely many nodes, or edges. As no path in any set of the union has infinitely many nodes, and the union can contain only those paths, from which finite tree is WM getting his infinite path? > > If EVERY edge of a path is shared by another path, then both paths > cannot be distinguished. There is no "both paths" until one has chosen both paths to be compared, at which point for those two one edge in each not in the other is easy to find, the two edges branching from last node they have in common.
From: Virgil on 3 Jan 2007 19:56 In article <1167860661.713987.282460(a)k21g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > If EVERY edge of a path is shared by another path, then both paths > cannot be distinguished. That is only true in finite trees. In an infinite tree, given any path and any finite n, there is anoth path sharing the first n edges but separating after n edges. > Your quantifier magic is without any value to > veil this fact in an actually existing infinity. It is the only kind there is in ZFC or NBG. And until WM, or someone, comes up with an axiom system saying otherwise, we are stuck with it. > The reason is that I do not believe in differen trees which are > identical Except when YOU want them to be.
From: cbrown on 4 Jan 2007 03:02
mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > You say that 2 > 3/2. What is an example of a surjection from the > > rational number 2 onto the rational number 3/2? > > There is an injection from 3 into 4 which is not a surjecton (namely > 3/2 and 4/2). > Truly amazing. I can only assume from your response that you have forgotten (or perhaps just don't care) what a surjection is. A surjection is a function which maps the /members/ of one /set/ onto the /members/ of another /set/; so that for every /member/ of the latter /set/, there is a /member/ of the former /set/ which maps to that /member/. What are the /members/ of the set 2 that you mean to act as the domain of the surjection? Or is the set 2 the empty set? I don't mean some finite collection of /subsets/ of the set 2, but the /members/ of the set 2. You agree that there is in general a difference between "x is a subset of Y" and "x is a member of Y", yes? You seem to claim that 1(/2) is a /member/ of the set 2 - at least when you claim a surjection onto the set 3/2. But how then can 1(/2) cease to be a member of the set 2 when you claim a surjection f from the set 7/3 onto the set 2? If 1(/2) /doesn't/ cease to be a member of the set 2, what element of 7/3 maps to the element 1(/2) of the set 2 in a surjection? If there is no such member of 7/3 which maps to the member 1(/2) of the set 2, then you do /not/ have a surjection f. Simply because then: 1(/2) is a member of the set 2, for which there is no member x of the set 7/3, such that f(x) = 1(/2). > > Why is there no > > surjection from the rational number 3/2 onto the rational number 2? > > There is no surjection from 3(/2) into 4(/2). But there is a surjection of 6(/4) onto 4(/2). 6 is greater than 4, is it not? A function must map /every/ element in its /domain/ to /some/ element in its /range/. Either 1(/4) is a member of the set 3/2, or it isn't. It amkes non sense to arbitrarily ignore elements of a set whenever they are inconvenient, and then claim you are defining a function on the entire set. > > > > > 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? > > There is a bijection from 6 onto 6 (namely 6/4 and 3*2/2*2 = 6/4). Therefore, you agree that 1(/4) is a member of 3/2. So as noted above, 6(/4) > 4(/2), because there is a surjection from 6 onto 4. > > > > You of course understand what is meant by a surjection from a set A > > onto a set B, yes? > > Of course. It doesn't appear so to me. I think perhaps at some point you did, but perhaps not. Cheers - Chas |