Cantor Confusion
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From: mueckenh on 22 Nov 2006 06:52 Virgil schrieb: > In article <1164126395.211430.7520(a)h54g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Tell me which point is not accepted: > > 1) Every line which contains an index of the diagonal, contains all > > preceding indexes of the diagonal too. > > 2) Every index of the diagonal is in a line. > > 3) In order to show that there is no line containing all indexes of the > > diagonal, there must be found at least one index, which is in the > > diagonal but not in any line. > > This one is flat out false, unless one assumes, a priori, a last line > and a last member of the diagonal. We assume not a last line, but we assume that eery line has finitely many indexes. And this is true. > > It is certainly false in ZF or NBG, where such an assumption is also > false. For finite indexes it is correct. > > It is quite enough to show that for every index in the diagonal, > except 1, there is some line not containing that index. Then no line > can contain every index. Name two finite indexes which cannot be in one line. Regards, WM
From: mueckenh on 22 Nov 2006 07:05 MoeBlee schrieb: > 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. Nevertheless it is impossible to construct such an injection (as it s impossible to construct a well order of the reals). > > > 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. It is an unfounded assumption to believe that all x of the set appear in this bijection, only by writing x = x. Ridiculous! > > > > > 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. But it has to do with the existence of a bijection to an ordinal number. > > > > > 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. No. But we need it to determine this cardinality. Again you think erroneously that an existence proof is of any value. Regards, WM PS: What about the binary tree in ZFC? Any results? Any idea why it can't be done?
From: mueckenh on 22 Nov 2006 07:07 William Hughes schrieb: > > > > 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. No. It implies only that all d_nn are finite. If you disagree, please give an example for two elements of the diagonal which cannot be found in one single line. Regards, WM
From: mueckenh on 22 Nov 2006 07:13 Virgil schrieb: > 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 ? It is impossible to believe that you think all n e N must be automatically included only by writing n. > > > > > 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. Is that an axiom? No. It is but an unjustified assumption. > > 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. No. Regards, WM
From: mueckenh on 22 Nov 2006 07:20
Virgil schrieb: > Consider > "For all m in N there is an n in N such that D(m) in L_n" > and > "There is an n in N such that for all m in N D(m) in L_n." > > The former is trivial, the latter is false. > > This can more easily be seen in comparing > "For all m in N there is n in N such that n > m" which is true > with > "There is an n in N such that for all m in N, n > m" which is false. This is false. And by means of the EIT we can conclude that this falsehood implies the falsehood of "There is an omega such that for all n in N : omega > n". If you do not see it, then try to find two elements of the diagonal which cannot belong to one single line. If you seek long enough, perhaps your eyes will be opened. Regards, WM |