An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: mueckenh on 11 Jul 2006 13:32 Virgil schrieb: > > But it is impossible to count over all natural numbers. Therefore the > > prescription replace a_n by b_n does not cover all natural numbers, it > > is void. > > It is only impossible to do in a way which Cantor did not attempt to do > it, by iteration of individual cases. It is quite possible, and even > trivial, to do it the way Cantor did it, by a general rule which covers > all cases simultaneously. There is no simultaneity in counting. n is not defined without counting from 1 to n. The whole set can never be exhausted by counting, therefore Cantor has no advantage over me. > > According to "mueckenh"'s way of thinking, it would be impossible to > prove that all natural numbers are greater than -1, since we would have > to do it one natural last time and would never finish. It would be > equally impossible in "mueckenh"'s way of thinking o prove that the sum > of two consecutive naturals is odd. > Both proofs, and many similar ones, are trivial if one is not > constrained to do a separate proof for each individual case, but are > allowed to do general cases. Of course these proofs can be done. But they do not prove that all natural numbers form a set with an actually infinite cardinal number. Of course you can define "replace 4 by 5", but in order to go through the list and produce the diagonal number from the single digits, you have to count. And that will never get ready. > > > I merely established a rule by which > > for an arbitrary well-ordering of all rationals one (for instance you) > > may construct an order by size without destroying the initial map of > > indices, i.e. the initial well-order. > > I see no evidence at all that your scheme will at any point even produce > the natural order in any finite initial segment of the sequence. I see that the diagonal proof will produce digits which differ from the diagonal digits of the list numbers. But I see no evidence that the diagonal proof will at any point produce even half of the digits required for an irrational number. > > Not so. You claim the final result has a certain property. I can > disprove your claim by showing that it does not have that property. I can disprove Cantor's claim by showing that the diagonal is not different from any list number because the list is by definition complete. > > You claim that the final result is in natural order and simultaneously > well ordered. between the first and second in the well ordering their > average value is a rational that in natural ordering must be between > them, but is not. This, among many other contradictions, disproves your > claim. In the well order of the rationals, all rationals have finitely many predecessors while in the order by magnitude each one has infinitely many predecessors and as many successors. Hence, a well-ordering of the rationals is impossible. This, among many other contradictions, disproves Cantor's claim. > > (1,2) > > (2,3) > > (1,2) > > For a 3 element list, this gives the wrong result in 5 out of 6 cases > and only succeeds in 1 out of 6 cases. > > For a list of length n, a similar algorithm will give give a correct > result in only 1 out of factorial n cases. > > The longer the list the worse the algorithm. Does that make it > infinitely bad for an infinite list? You have not understood the scheme. Look into my replies to Franziska. My scheme gives always the correct well-ordering - for all sets which exist. Regards, WM
From: mueckenh on 11 Jul 2006 13:34 Virgil schrieb: > > Wrong. In order to determine the n-th line, one must count from 1 to n. > > Cantor does not claim to be able to provide any such list. > And it is the provider of the list who must do that. Cantor only says > that when someone provides him with that list, that he can find a > number not listed. That is a contradiction, if the provider has listed all numbers, isn't it? Regards, WM
From: mueckenh on 11 Jul 2006 13:37 Virgil schrieb: > > Except that after each (1,2), there will only be a finite initial > subsequence in numerical order and an infinite terminal sequence not yet > ordered. > > Using 1 origin indexing, the nth occurrence of (1,2) occurs at the > (n^2 + n)/2 th position in the list of "transpostions". > > At which (n^2 + n)/2 th operation is the entire list ordered? At which number can I find the last element of Cantor's diagonal? Regards, WM
From: mueckenh on 11 Jul 2006 13:38 Virgil schrieb: > In article <1152544884.364304.157620(a)75g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Franziska Neugebauer schrieb: > > > >> > Either K is in the list (which would contradict analysis in the > > > >> > same way as my original example), then there is an n with 10^(-n) = > > > >> > 0. Or K is not in the list. > > > >> > > > >> K is not in the list. > > > >> > > > >> > Then there must be a position which cannot be enumerated by natural > > > >> > numbers > > > >> > > > >> non sequitur. All positions are indexed by definition of decimal > > > >> representation. > > > > > > > > All positions are indexed <==> K is in the list. It is a logical > > > > equivalence for linear sets like my list. > > > > > > What kind of "logic" is employed here? > > > > That one which deserves its name, and which you unfortunately seem to > > be not familiar with. The logic of a linear sequence like this sequence > > of list numbers > > > > 0.1 > > 0.11 > > 0.111 > > ... > > "Mueckenh" insists that that sequence has a double meaning, both as > unary representations of naturals and as binary, or other based, > rationals, and requires that 0.111... satisfy both interpretations > simultaneously. > Ergo, it is garbage. You would call it bright if you were able to understand it. Regards, WM
From: mueckenh on 11 Jul 2006 13:40
Virgil schrieb: > > All positions which can be enumerated are in the list, by definition. > > Hence, the decimal representation of 1/9 does not exist. > > Then how is it that so many people use it? Because they do only believe they used it. It is perhaps similar to Planck's constant, which remained undiscovered until 106 years ago. Regards, WM |