An uncountable countable set
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From: Virgil on 26 Jun 2006 14:29 In article <1151319667.312599.252200(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > >Daryl McCullough schrieb: > > > > > >> mueckenh(a)rz.fh-augsburg.de says... > > >> > > >> >Such a function exists from |N --> |R for all real numbers which can be > > >> >individualized, i.e. which are really real numbers. > > >> > > >> If such a function exists, then demonstrate it. Give us a function > > >> f from N to R such that every real number that is "individualized" > > >> is in the image of f. > > > > > >Cantor's general proof breaks down > > > > I'm not asking about Cantor's proof. I'm asking you to demonstrate > > a function f from N to R such that every "individualized" real number > > is in the image of f. Can you demonstrate such a function? If not, > > why not? > > The diagonal number is not an individualized number, unless it is of > such a simple kind I demonstrated. A random Cantor list can never be > read to the bottom, the diagonal number can never be individualized. Since for non-constructivists, there exist real numbers which are not capable of being "individualized" in the "mueckenh"'s sense, any such number will not be in any of in "mueckenh"'s lists. This shows that unless one allows in "mueckenh" to set the rules, Cantor is right.
From: Virgil on 26 Jun 2006 14:39 In article <1151319926.736596.18840(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > Which of the real numbers so determined are not really real numbers? > > > > > > Irrationals, for instance, are not numbers at all, because Cauchy's > > > epsilon cannot be made arbitrarily small. > > > The members of the sequence, 1, 1/2, 1/4, 1/8, ... are all positive > > but contain members smaller than any given positive , thus rationals can > > be made arbitrarily small. > > And natural can be made arbitrarily large. But not all of them do > exist. Take [pi*10^10^100], change the last digit to 6. You will not be > able to find out whether one of both natural numbers is larger than the > other. They do not exist. Nevertheless 10^10^10000 is an existing > natural number. If you can describe the number as you have done, you have made it exist, even if you cannot necessarily compare it for size with another number. > > > > > But we will keep this > > > disussion at the reordering of the well-ordered set of rationals. If > > > you have understood that there are not all rationals, you will > > > understand easily, that there is no irrational. > > > > In my imagination, all of the rationals exist and all of the irrationals > > as well. I am sorry to hear that "mueckenh"'s imagination is so > > self-limiting. > > It recognizes the reality. Mine recognizes the reality of the reals, as well as the rationals and the naturals. We do not wish to keep those who choose to follow constructivist philosophy from doing so, so long as they do not try to keep us from following our own, as "mueckenh" is trying to do here.
From: Virgil on 26 Jun 2006 14:49 In article <1151330489.853370.287220(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > p=2E 214: "Die Frage, durch welche Umformungen einer wohlgeordneten > > > Menge > > > ihre Anzahl ge=E4ndert wird, durch welche nicht, l=E4=DFt sich einfach > > > so > > > beantworten, da=DF diejenigen und nur diejenigen Umformungen die Anzahl > > > unge=E4ndert lassen, welche sich zur=FCckf=FChren lassen auf eine > > > endliche > > > oder unendliche Menge von Transpositionen, d. h. von Vertauschungen je > > > zweier Elemente." > > > Cantor admitted infinitely many transpositions. I think he meant > > > countably many. But more aren't needed. > > > > No he did not. Pray read again. He was talking about the kind of > > transformations of well-ordered sets for which the *number* of elements > > would change. Or is well-orderedness part of the number of elements? > > Number (Anzahl) is only defined for well-ordered sets. Number (Anzahl) > here is *not* meant as cardinality. > > The previous page reads: (Collected works, p. 213): "Durch Umformung > einer wohlgeordneten Menge wird ... nicht ihre M?chtigkeit > ge?ndert, wohl aber kann dadurch ihre Anzahl eine andere werden. > > Compare the letter from Cantor to Mittag-Leffler (17.12.1882): "Den > gr??ten Vortheil gewinne ich jetzt durch die Einf?hrung ... der > "Anzahl einer wohlgeordneten Menge".... Die Anzahl einer > wohlgeordneten Menge ist also ein Begriff der in Beziehung steht zu > ihrer Anordnung; bei endlichen Mengen findet er sich offenbar als > unabh?ngig von der Anordnung; dagegen jede unendliche Menge > verschiedene Anzahlen im Allgemeinen hat, wenn man sie auf verschiedene > Weise als "wohlgeordnete" Menge denkt. > > Hence "Anzahl" is connected with the special kind of well-ordering. In > particular different wel-orderings lead to different numbers > (Anzahlen). > > > I think not. And indeed any number of interchanges can be performed on > > a set without changing the number of elements. > > > > I think what he is arguing here (about a field that needed new exploration) > > that at least for well-ordered sets any number of transpositions (and in > > fact any finite or infinite permutation) would not change the number of > > elements. Apparently it was not clear to him if that was also the case > > for sets that could not be well-ordered. That is the best I can make out > > of your quote. > > He may have thought as follows: One transposition does not destroy the > well-ordering. n transpositions do not destroy the well-ordering. > Infinitely many will not destroy the well-ordering, because there is no > first one, which did so (as Virgil argued correctly). > > But, independent of what were Cantor's actual thoughts, if infinity > would exist, we could order the rationals by magnitude and maintain the > well-order. Actually, that would require that no infinity exist, or at least that the set of rational be finite, so that the rationals in their natural order be well-ordered, as is every finite ordered set. That an infinite set of rationals in their usual order are not, and cannot be, well ordered is trivial: In order for a set to be well ordered, every non-empty subset must have a smallest member The non-empty set of positive rationals has no first element, since for every positive rational, 1/2 of it is strictly between it and zero.
From: Virgil on 26 Jun 2006 15:06 In article <1151330868.199805.89070(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > > >> How are you saying anything different from that? But the > > >> latter statement immediately implies the statement > > >> "forall f from N to P(N), f is not surjective". > > > > > >forall f from N to any set, K(f) is not in the image of f . > > > > Yes, that's true. And that implies that "forall f from N > > to P(N), f is not a surjection". > > It is not a problem of surjectivity. This set K may be in P(|N), but it > very *defintion* is nonsense. Your definitions often are, but I see no nonsense other than yours above. And below! > > Define a bijective mapping from {1, a} on P({1}) = {{}, {1}}. a is not > a natural number. There are two bijections possible. The set K cannot > be mapped by a number although K is in the image of both the possible > mappings. > f: 1 --> {1} and a --> { } with K_f = { }, > g: 1 --> { } and a --> {1} with K_g = {1}. > Here we have certainly no problem with lacking elements. Nevertheless > Hessenberg's condition cannot be satisfied. The set {K_f, m, f} is an > impossible set. For the function f, 1 in f(1), and a not in f(a) means K_f = {a} but then K_f is not in {{}, {1}}, the codomain of f. For the function g, 1 not in g(1) and a not in g(a) means K_g = {1,a} but then K_g is not in {{}, {1}}, the codomain of g. So neither of your f or g "examples" above is anything but nonsense.
From: Virgil on 26 Jun 2006 15:29
In article <1151331606.221643.81550(a)u72g2000cwu.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1151079833.849194.73370(a)c74g2000cwc.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > A *surjective* mapping does exist, but it is not f. There exists > > > > > a > > > > > surjective mapping g -> M(f). > > > > > > > > Indeed. But your short list does not show anything at all, I wonder > > > > what > > > > I have to do with it. I think you intend to show something about the > > > > diagonal number of a list of reals. But we were talking about a > > > > mapping > > > > from N to P(N). So please keep to the subject. > > > > > > Don't you recognize that K(f) takes exactly the same function as the > > > diagonal number D in Cantor's second argument? f enumerates the list > > > numbers. If f can arbitrarily be replaced by g, then this proof is > > > invalid and shows only what it does show in fact, namely the > > > countability of all list numbers including all diagonal numbers which > > > can be constructed. > > > > I have diffculty making sense from this. But let me try, and the answer > > is (in my opinion) no. > > > > In the case of K(f) and M(f) I start with a bijection between N and S, > > where > > S is the set of finite subsets of N. You state: but K(f) is not in it. I > > answer: no obviously not, K(f) is not in S, and I constructed f as a > > bijection between N and S. > > That is the same as enumerating by f the numbers of Cantor's list, > which could , e.g., contain all terminating rationals. It is not "Cantor's list" but ANY list. You are challenged to find a list which lists all reals, then Cantor presented a rule for showing any list you provide is incomplete. > > > Not you state: so M(f) is not countable, and > > I say, but it is, and I construct a bijection (g) between N and M(f). > > So you utter, but K(g) is not in it, and I answer, no obviously, g was a > > bijection between N and M(f) and K(g) is not an element of M(f). So there IS a set, K(g), not in the image of what you have claimed must be a bijection. So you alleged bijection crashed. Next > > you state, but g is not a bijection between N and M(g), and my answer is: > > yes, that is obvious, because g is not even a mapping between N and M(g) > > because K(f) (which is in the image of g) is not an element of M(g). > > Let's be generous and let you have said that it was not a bijection between > > N and M(f, g) (where M(f, g) is appropriately defined). And I state, of > > course not, that was not the requirement, the requirement was a bijection > > between N and M(f). So each time I come with a bijection you give a > > different set to which it is not a bijection. That is obvious. Actually you have it backwards: Every time you come up with an f: N --> M(f), which is not a bijection, we find a g: N --> M(f) which IS a bijection. For every f: N --> M(f) there is a bijection g:N <--> M(f) > > And each time I implement the diagonal number K(f) into the list f > (insert it before the first ordinary list number, for instance), you > will construct another diagonal number K(g) and say, look here, this > diagonal is not contained in your g. As that K(g) was not in M(f) , that is irrelevant. > > > > Now consider the case of a bijection between N and P(N). Now consider the case where 2 = 1, .... Why consider impossible cases? You state: f is > > a bijection, and I state: no, it is not a bijection because H(f) (H > > standing for Hessenberg set) is not in the image, while it is in P(N). > > It may be *as a set* in P(N), but it is not in P(N) by that very > definition of Hessenberg's. Wrong, provided f is any possible function from N to P(N). If one wants to consider all possible properties of impossible functions, one should not do it in sci.math. > > > OK you state, lets reformulate to g such that H(f) is in the mapping, and > > I say, no, now H(g) is not in the image. You can continue, but everytime > > I can state a set that is not in the image and should be there. Note that > > in this case the target does *not* change. > > Consider one more time the bijective mapping from {1, a} on P({1}) = > {{}, {1}}. a is a symbol but not a number. There are two bijections > possible. The set M of all numbers which are non-generators cannot be > mapped by a number although M is in the image of both the possible > mappings. > f: 1 --> {1} and a --> { } with M_f = { }, > g: 1 --> { } and a --> {1} with M_g = {1}. {x in {1,a}: x not in f(x)} = {a}, which is not in P({1}) {x in {1,a}: x not in g(x)} = {1,a}, which is not in P({1}) So your examples are both cooked. > Here we have certainly no problem with lacking elements. Nevertheless > Hessenberg's condition cannot > be satisfied. The set {M_f, m, f} is an impossible set. > > Regards, WM |