An uncountable countable set
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From: Virgil on 10 Jul 2006 13:15 In article <1152523297.422935.253480(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In Cantor's 'diagonal" proof, no numbers are transposed. > > It is merely shown that a rule of construction can be stated such that > > for an arbitrary list of reals and an arbitrary member of that list the > > constructed number is not equal to the selected number. > > This is only shown for a finite set of n numbers. The rule applies equally well simultaneously for all n in N. > It can only be proved > after having counted the digits from 1 to n without leaving out a > single one. Nonsense. The Cantor rule generates a digit to go in the nth decimal place of the number being created without any reference to any other decimal place. > > > From which one may deduce that a complete listing of reals is not > > possible. > > If one decides to belong to the set of those blind people which call > the uncountable "countable", which call the not existing "existing", > and which call themselves logicians without having a clue what logic is > (compare the binary tree). We are blind enough as not to see what is not there but not so blind as not to see what is there, both of which faults "mueckenh" repreatedly exhibits.
From: Franziska Neugebauer on 10 Jul 2006 13:24 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> >> >> Right! There are more than any finite number of finite >> >> >> naturals, indeed, an endless supply of them which collectively >> >> >> form an infinite set. >> >> > >> >> > There is no "supply". The elements of a set do exist, >> >> > instantaneously and immediately. Sets are static. >> >> >> >> Sudden insight? >> > >> > Clever application of different standpoints. >> >> So the elements of omega do exist "statically"? > > Neither nor. But if one claims their existence, I reserve the right to > let them exist at my convenience. This is acceptable when doing proofs by contradiction or when you argue within a special context. If you claim "The elements of a set do exist" and "Elements of omega do neither exist nor exist statically" in the *same* context and simultaneously then you violate the law of noncontradiction. F. N. -- xyz
From: Franziska Neugebauer on 10 Jul 2006 13:38 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 > ... > > If "all positions are indexed by list numbers" is not equivalent to "K > is in the list", Obviously not. > then you seem to imply that more than one list number is required to > index the digits of K. Every "list number" (= sequence member) is a sequence of digits indexed by every natural number. > I say: What can be indexed by different list numbers can also be > indexed by one alone. Meaning? Proof? > If you do not believe that, I do not know what you mean. > then you should be able to show an example where more > than one list number is required. Of course it must be a finite > example, because there are only finite list numbers. And note: all > list numbers are unary representations of natural numbers. But not all unary sequences are representations of natural numbers. 0,111... is a sequence which does not represent a natural. > And note: It is unimportant whether there is a definition which says > this or that. F. N. -- xyz
From: Virgil on 10 Jul 2006 13:53 In article <1152523593.592670.210570(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > No one has shown that any sequence of exchanges can convert a number > > > with finitely many digits into a number with infinitely many digits. > > > > Relevance??? > > Cantor's diagonal, of course. If you replace a_n by b_n, this cannot be > done and cannot be proved without having counted carefully from 1 to n. In mathematics, one can prove a statement for the whole of a set of objects simultaneously without having to iterate the objects individually. Cantor's rule does this for all naturals simultaneously without the need for iteration. > 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. 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. > > > > After each of his countably many exchanges Cantor has still a countably > > > infinite subset of the list numbers left which are not yet exchanged. > > > > Cantor performs no exchanges at all, he merely establishes a rule by > > which for an arbitrary list of reals one may construct a number not in > > that list. > > > That is the same I do, you see? But that is not at all what you do, as those that can see do see. > 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. > In fact, you can calculate the > complete order by size with the same pocket calculator as one can > calculate the complete diagonal. The existence of the "diagonal" depends on the existence of the list of reals, and cannot be "calculated" without it. An ordering of the rationals that is simultaneously a well ordering and natural ordering is impossible, since in the natural ordering no rational has an immediate successor but in a well ordering, every number would have to have one. Since "mueckenh" claims to have achieved an extraordinary result, and actual impossibility, he must provide extraordinarily strong evidence to support his claim. but his attempts at proof are only extraordinary in their weakness. They establish nothing of what he claims. > > > > Cantor only states a rule of construction. Those who dispute that the > > rule succeeds must show that there exists some member of the list for > > which it fails. > > So do I. It is really more comfortable to publish only the rule and let > others do the calculation if they do not believe in the rule. For the "rule" as published, it does not necessarily even correctly order by size the first two or three elements in the well-ordering. > > On the other hand, Mueckenheim must show that his rearrangement can be > > completed, even though it can't be, in order to double order the > > rationals. > > Your arguing does not prove that there is any number in that list for > which > my rule does not apply, and the finding of such a number is > required to invalidate me. 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. 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. > > > > If there could be any exchange in the list which did > > > not face the initial problem all over again, there would have to be a > > > first one, but that is impossible. > > > > What "initial" problem is that? Cantor's rule can apply to any member of > > the list of reals independently of any other member, so applies to all > > simultaneously. So there is no "initial" problem at all. > > > In order to apply the rule to a given list, you must count its lines. That is the problem of the person providing the list in the first place. Cantor does not require himself to provide any lists at all, he merely says what he can do with such a list after it is provided. > > > Mueckenheim's transpositions must be applied sequentially, so cannot be > > applied simultaneously. > > They must not be applied at all. IThe order-by-size is > established with the simple rule: Order all rationals by size which can > be reached within the well-order. Or somewhat formalized: > > (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?
From: Virgil on 10 Jul 2006 14:33
In article <1152523821.186836.300340(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > You have posited an infinite sequence of infinite sequences of > > > > transpositions to perform your alleged miracle. As you can never even > > > > finish the first subsequence in finite time, you must reorder your > > > > transpositions into a single sequence to even consider its effect. > > > > > > I did so. > > > > When? Since reordering disrupts the effect of sequences of overlapping > > transpositions (non-commutativity), you have not done so to anyone's > > satisfaction but your own. > > (1,2) > (2,3) > (1,2) > (3,4) > (2,3) > (1,2) > What does this do to a list that starts out (a,b,c,d,e,f,... )? It merely reverses the positions of the first 4 elements, giving (d,b,c,a,e,f,...) If that pattern continues, after each (1,2) transposition one will merely have the original list with a finite initial segment reversed. This is not a sorting algorithm at all. > > > > > No I have only one infinite sequence of transpositions. They > > > are as well defined and as fast to be executed as Cantors sequence of > > > replacements of digits. But even if they could all be completed, they do not produce an ordering, > > > E2 = (a2,1, a2,2, . . .,a2,nu, . . .), > > > . . . . . . . . . . . . . . . . . . . . . . . . . . . . . > > > E??? = (a???,1, a???,2, . . .,a???,???, . . .), > > > . . . . . . . . . . . . . . . . . . . . . . . . . . . . . > > > > That is not a list so much as an N by N matrix over the decimal digits > > into which any list can be fitted. > > It is an abstract list. An N by N matrix over the decimal digits is quite abstract enough. >You should learn to think in abstract terms. Teaching your grandmother to suck eggs? |