Cantor Confusion
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From: mueckenh on 3 Jan 2007 16:27 Franziska Neugebauer schrieb: > > 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) Here is your error: x is different from y' <-> there is a difference between x and y' (Q) i.e., there is a difference between x and what you believe is y. The error lies in your assertion that y = y'. > > To restore the consistency*) of your universe of discourse you must > refrain from P or Q. A discussion of Q' is no remedy. > *) only wrt to the current contradiction. If that is the same logic as that which states the existence of different path systems in trees which have identical node systems then I am not surprised. Regards, WM
From: mueckenh on 3 Jan 2007 16:30 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? > > 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? (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.) Regards, WM
From: mueckenh on 3 Jan 2007 16:36 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). > 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). > > 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). > > You of course understand what is meant by a surjection from a set A > onto a set B, yes? Of course. > > Or do you now agree that asking "is there a surjective function from 1 > + 1/2 + ... onto 1?" is nonsense? No. Why should it? There is an injection which is not a bijection from 1 = 2/2 into 3/2 as well as from 4/4 into 7/4 and so on. Regards, WM
From: mueckenh on 3 Jan 2007 16:42 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. > (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. > (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? > > 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. > > Do you not see a contradiction here? Of course. It lies in the assumption of actual infinity. It lies in the assumption of a complete tree. ===================================== > Well, have a look at ZFC+V=L. Apparently a well-ordering of the reals is > automatic with V=L. Apparently it does not. I am sure otherwise you would already have posted it. > That is easy: {x} is one of those ordered sets. Here it is the first and > only element. You cannot give a well-ordering this way because you cannot reproduce the set of singletons of all real numbers. --- But you cannot do it in another way too. So maintain to assert that V=L is the magical formula which well-orders the real numbers. As I have seen, you are ready to believe in even more incredible items. > Wrong? I don't know. But I think a quantifier problem again. I state (P > is set of paths, E is set of edges, and p1 and p2 are different): If EVERY edge of a path is shared by another path, then both paths cannot be distinguished. Your quantifier magic is without any value to veil this fact in an actually existing infinity, i.e., in an infinity > (1) forall{p1 and p2 in P} thereis{e in E} e in p1 but not in p2 > or every two different paths differ in at least one edge. > (2) forall{e in E} thereis{p1 and p2 in P} such that e in p1 and in p2 > or for every edge there are two different paths that share that edge. > For some reason you conclude from this: The reason is that I do not believe in different trees which are identical > (3) forall{p1 in P} thereis{p2 in P} forall{e in p1} e in p2. > or there are different paths that share all edges. The paths obviously are not different, if they share all edges. This example shall only show you that your paragraph (2) is wrong in an actual infinity. It is, or at least could be, correct in a potential infinity. Alas, these both notions, actual and potential, are usually intermingled in set theory. > Now (3) is clearly in contradiction with (1), so I wonder how you come > to this conclusion. I come to this from (2). The reason is that I do not believe in different trees which are identical, a belief which is required by the adorants of quantifier magic. Regards, WM
From: mueckenh on 3 Jan 2007 16:44
Dik T. Winter schrieb: > > "Them" is two paths. If one path has no edge of itself, then there is > > at least one other path which shares all edges with the former. > > Wrong? I don't know. But I think a quantifier problem again. I state (P > is set of paths, E is set of edges, and p1 and p2 are different): If EVERY edge of a path is shared by another path, then both paths cannot be distinguished. Your quantifier magic is without any value to veil this fact in an actually existing infinity, i.e., in an infinity whichonly reaches up to the limit where another, larger, infinity begins. (Compare the well-ordering of all real numbers, where first all natural numbers occur and then, after "the first omega number" the other real numbers. Under this cicumstances your following paragraph (2) is wrong. > (1) forall{p1 and p2 in P} thereis{e in E} e in p1 but not in p2 > or every two different paths differ in at least one edge. > (2) forall{e in E} thereis{p1 and p2 in P} such that e in p1 and in p2 > or for every edge there are two different paths that share that edge. > For some reason you conclude from this: The reason is that I do not believe in differen trees which are identical > (3) forall{p1 in P} thereis{p2 in P} forall{e in p1} e in p2. > or there are different paths that share all edges. The paths obviously are not different, if they share all edges. This example shall only show you that your paragraph (2) is wrong in an actual infinity. It is, or at least could be, correct in a potential infinity. Alas, these both notions, actual and potential, are usually intermingled in set theory. > Now (3) is clearly in contradiction with (1), so I wonder how you come > to this conclusion. > -- I come to this from (2). The reason is that I do not believe in different trees which are identical, a belief which is required by the adorants of quantifier magic. Regards, WM |