Cantor Confusion
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From: Virgil on 21 Nov 2006 18:51 In article <1164142822.670975.248140(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > There are two types of initial segments > > > > > > > > Initial segments with a largest element > > > > Initial segments without a largest element. > > > > > > > > There is an initial segment of the diagonal that does > > > > not have a largest element. > > > > (Recall: the diagonal has a largest element if and only > > > > if there is a last line. There is no last line.) > > > > Every line has a largest element. Thus there is an initial > > > > segment of the diagonal that is not a line. > > > > > > The diagonal has only finite indexes n. Therefore it consists of line > > > ends only. Every line end is the end of a line, no? > > > > > > A classic case of > > > > > > Person 1: > > > > P_1 is true. <reasons why P_1 is true> > > P_1 implies P_2 > > Therefore P_2 is true > > > > Person 2: > > > > No > > P_1 is true > > However, P_1 does not imply P_2 <reasons why P_1 does > > not imply P_2> > > P_2 is false > > > > > > Person 1: > > > > Clearly you do not understand. > > Here is another reason why P_1 must be true. > > > > > > Yes. P_1: The diagonal consists of the ends of lines. > > > > No. The fact that > > P_1: the diagonal consists of the ends of lines > > does not imply that > > P_2: the diagonal is not longer than every line. > > > > > > > > > > > > Therefore you cannot say: > > > > > > > > The diagonal cannot be longer than every line > > > > because it consists of line-ends only. > > > > > > Of course that must be true. The diagonal has only finite indexes n. > > > Every line end is the end of a finite line. > > > As it cannot be infinite, the complete diagonal does not exist. > > > > > > > > > > If element d_nn of the diagonal is a member of line n, then all > > > elements d_mm wit m =< n of the diagonal are members of the same line > > > n. In other words: There is never more than one single line required to > > > establish a bijection with all the elements of the diagonal d_mm with m > > > =< n. > > Yes. > > And the diagonal consists of only such elements. Can WM deny that the diagonal contains for every line at least one element not in that line. > > > > However, since there is always an element (n+1), no single > > line can establish a bijection with *all* the elements of the diagonal > > (both those <=n and those > n). > > Together with the proof above this shows that "not *all* the elements > of the diagonal" exist. Not at all! WM is going though all sorts of contortions to maintain his falsehood, but nevertheless it is a falsehood, at least in ZF and NBG. And WM has no stated system of his own. > > > > > > > As the diagonal has only finite indexes n, there is never more > > > than one line required to establish a bijection with all elements of > > > the diagonal. > > > > No. No line with finite index n can establish a bijection with > > all elements of the diagonal. Every line has a finite index. > > Therefore no line can establish a bijection with all elements of the > > diagonal > > If there is a bijection with all finite line ends, then there is a > bijection with one line. Why do you prefer one of them? It is WM who is insisting on a bijection from the diagonal to one of the lines whereas we say to none of them. There is a bijection between the set of positions in the diagonal and the set of ALL lines, but not with the set of positions of any single line.
From: MoeBlee on 21 Nov 2006 18:55 mueckenh(a)rz.fh-augsburg.de wrote: > The countability of the set W of all the words of a finite alphabet is > proved by other means than the usual countability proof. The latter > consists in constructing a bijection with N, i.e., in constructing a > list. But it is impossible to construct a list of all words. It is > impossible to construct a bijection between W and N. WRONG. It is a theorem that from the set of finite sequences of members of a finite alphabet there is an injection into N. > Why should it be > possible to construct a bijection between N and N? Because ANY x is 1-1 with x, by the identity function on x. > > > Note however that in order to attach a definite cardinal number to > > > every set, well-ordering must be possible for every set. That is why > > > Cantor insisted on well-ordering of all sets. > > > > Piffle. Well-ordering has nothing to do with the existence or not of > > at least one bijection. > > > You are wrong. If a set cannot be well-ordered, it cannot be assigned a > definite cardinal number. In order to assign a cardinal, at least one > bijection to an ordinal is required. This requires at least one > well-ordering. WRONG. By using Scott's method (which requires the axiom of reqularity), we don't need the well ordering theorem to prove that every set has a cardinality. MoeBlee
From: William Hughes on 21 Nov 2006 18:55 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > There are two types of initial segments > > > > > > > > Initial segments with a largest element > > > > Initial segments without a largest element. > > > > > > > > There is an initial segment of the diagonal that does > > > > not have a largest element. > > > > (Recall: the diagonal has a largest element if and only > > > > if there is a last line. There is no last line.) > > > > Every line has a largest element. Thus there is an initial > > > > segment of the diagonal that is not a line. > > > > > > The diagonal has only finite indexes n. Therefore it consists of line > > > ends only. Every line end is the end of a line, no? > > > > > > A classic case of > > > > > > Person 1: > > > > P_1 is true. <reasons why P_1 is true> > > P_1 implies P_2 > > Therefore P_2 is true > > > > Person 2: > > > > No > > P_1 is true > > However, P_1 does not imply P_2 <reasons why P_1 does > > not imply P_2> > > P_2 is false > > > > > > Person 1: > > > > Clearly you do not understand. > > Here is another reason why P_1 must be true. > > > > > > Yes. P_1: The diagonal consists of the ends of lines. > > > > No. The fact that > > P_1: the diagonal consists of the ends of lines > > does not imply that > > P_2: the diagonal is not longer than every line. > > > > > > > > > > > > Therefore you cannot say: > > > > > > > > The diagonal cannot be longer than every line > > > > because it consists of line-ends only. > > > > > > Of course that must be true. The diagonal has only finite indexes n. > > > Every line end is the end of a finite line. > > > As it cannot be infinite, the complete diagonal does not exist. > > > > > > > > > > If element d_nn of the diagonal is a member of line n, then all > > > elements d_mm wit m =< n of the diagonal are members of the same line > > > n. In other words: There is never more than one single line required to > > > establish a bijection with all the elements of the diagonal d_mm with m > > > =< n. > > Yes. > > And the diagonal consists of only such elements. > > > > However, since there is always an element (n+1), no single > > line can establish a bijection with *all* the elements of the diagonal > > (both those <=n and those > n). > > Together with the proof above this shows that "not *all* the elements > of the diagonal" exist. > > > > > > > As the diagonal has only finite indexes n, there is never more > > > than one line required to establish a bijection with all elements of > > > the diagonal. > > > > No. No line with finite index n can establish a bijection with > > all elements of the diagonal. Every line has a finite index. > > Therefore no line can establish a bijection with all elements of the > > diagonal > > If there is a bijection with all finite line ends, then there is a > bijection with one line. No. The diagonal contains only finite line ends. Every element d_nn, of the diagonal, is the end of the line which ends at the natural number n. The bijection with all finite line ends is: The line which ends in n maps to the element d_nn. There is no bijection with one line. Such a bijection would imply that there is a largest d_nn. - William Hughes
From: Virgil on 21 Nov 2006 19:07 In article <1164143109.673986.202100(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > What about all contructible numbers or all words? They cannot be mapped > > > on N though they are a countable set. > > > > > > Absolute piffle. The question is whether the set N can > > be mapped to the set N, not whether some other > > set can be mapped to the set N. > > It is the same impossibility, but not so obvious. What is impossible about: for all n in N, n |--> n ? > > > > [Although it is interesting to note that both the constructable numbers > > and the set of all words can be mapped to the set N (please check > > the definiton of countable set)] > > The countability of the set W of all the words of a finite alphabet is > proved by other means than the usual countability proof. The latter > consists in constructing a bijection with N, i.e., in constructing a > list. But it is impossible to construct a list of all words. It is > impossible to construct a bijection between W and N. Why should it be > possible to construct a bijection between N and N? Let W be the set of all possible words , i.e., set of all possible finite letter strings in some finite alphabet of n characters. In order to establish the existence of a bijection between W and N, it is sufficient to note that expressing the members of N in base n does exactly that. > > > > > > > Note however that in order to attach a definite cardinal number to > > > every set, well-ordering must be possible for every set. That is why > > > Cantor insisted on well-ordering of all sets. > > > > Piffle. Well-ordering has nothing to do with the existence or not of > > at least one bijection. > > > You are wrong. If a set cannot be well-ordered, it cannot be assigned a > definite cardinal number. The existence of a cardinal which bijects to a set S is not a necessary condition for the identity function on S to exist. Every set, whether well orderable or not, has an identity function, at least in such relatively sane systems as ZF and NBG. If WM wishes to propose a system in which sets can lack identity functions, he will have to explicitely present the axiom system for such a system before anyone will accept its existence, much less its consistency. > In order to assign a cardinal, at least one > bijection to an ordinal is required. If one does not assume the axiom of choice, as in ZF, then there ae sets without the sort of cardinality that WM requires, but such sets still can have identity functions on them, and in ZF, must have identity functions on them. > This requires at least one > well-ordering. WM's mind needs to be well-ordered, or at least better ordered, if he is to avoid the many falsehoods he promulgates. > > Regards, wM
From: MoeBlee on 21 Nov 2006 19:13
mueckenh(a)rz.fh-augsburg.de wrote: > I > assume that part of my mathematics belongs to absolute truth which does > not rely on arbitrary axioms. And you do not formulate your absolute truths as axioms, so your absolute truths do not have objective form by which other people (not just you) can verify that a given statement is of one of your absolute truths. Your absolute truths are spoken by you only as verbiage that is at worst vague and at best unformalized. You provide no objective means by which other people (not just you) can verify that your arguments do indeed follow by a specified logic from axioms which assert your absolute truths. In this situation, all discussions can only defer to you personally, as only you personally can say what is or is not absolute truth as you have determined it and as you argue as to its implications by your unspecified logic. On the other hand, even if you disagree that certain mathematical axioms are true, at least they are given in utterly objective form as formulas in a formal language and such that at least in principle one can mechanically check whether a given sequence of formulas is or is not a proof in a given formal system (and for the most basic theorems, even in practice it wouldn't be difficult to fully formalize for mechanical checkability). In other words, at least mathematics can give you - up front, clearly, and precisely - exactly what its axioms are and exactly by which rules theorems may be proven, while your polemics offer no such intellectual courtesy or benefits. MoeBlee |