An uncountable countable set
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From: Dik T. Winter on 26 Jun 2006 20:33 In article <1151331606.221643.81550(a)u72g2000cwu.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1151079833.849194.73370(a)c74g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > 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. Possibly. But again you are switching arguments. I was not talking about diagonal numbers and whatever. I was talking about mappings from N to sets of subsets of N. Can you please stay with the argument at hand? > > 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). 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. > > 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. Stay with the argument, please... > > Now consider the case of a bijection between N and P(N). 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. I have really *no* idea what you are stating here. By Hessenberg's definition H(f) *is* in P(N). How do you come at the idea that it is not? > > 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}. > 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. It can not be satisfied in *your* interpretation of Hessenberg's condition. Hessenberg's condition applies *only* if the source set is a subset of the natural numbers, which is not the case here. Consider Hessenberg's argument, given in steps: (1) Let us assume a surjection f from N to P(N). (2) Construct the set K(f), {x | x in N and x !in f(x)} (3) This set is well-defined and element of P(N). (4) As the set is an element of P(N) it should be (because the assumed surjectivity) in the image of f. (5) Because it should be in the image of f, there should be an element of N that is the source of K(f). (6) There is no such source, so the assumption is invalid. Now translate this to A = {1, a} and B = {{}, {1}}: (1) Let us assume a surjection f from A to B. (2) Construct the set K(f), {x | x in A and x !in f(x)} K(f) = {}, K(g) = {1}. (3) This set is well-defined and element of B. (4) As the set is an element of B it should be (because the assumed surjectivity) in the image of f. (5) Because it should be in the image of f, there should be an element of A that is the source of K(f). (6) There is such source, so f and g are (possibly) surjections. All around you are assuming that the Hessenberg requirement is that the source element should be a natural number. That is false. The requirement is that the source element should be an element of the source set. And in the case of N -> P(N) that source set *is* the set of natural numbers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: briggs on 27 Jun 2006 11:36 In article <vmhjr2-27492D.13460426062006(a)news.usenetmonster.com>, Virgil <vmhjr2(a)comcast.net> writes: > In article <1151332937.425608.129380(a)i40g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >> Dik T. Winter schrieb: >> >> > In article <1151135467.925502.14460(a)b68g2000cwa.googlegroups.com> >> > mueckenh(a)rz.fh-augsburg.de writes: >> > > Dik T. Winter schrieb: >> > > > > This mapping is impossible. (See my >> > > > > article >> > > > > "On Cantor's Theorem", arXiv, math.GM/0505648, 30 May 2005.) >> > > > >> > > > Consider two cases: >> > > > 1. >> > > > a -> {} >> > > > 1 -> {1} >> > > > K =3D {} >> > > > so K is in the image > > What does 3D {} mean? It's an artifact of the MIME quoted-printable encoding. The original message containing the "=" was encoded in quoted-printable. And the person who replied and quoted the passage including the "=" did not use software that understood quoted-printable. The offending line should read: >> > > > K = {} In quoted-printable the sequence "=xx" is used to denote the character whose ASCII code is xx in hex. "=3D" denotes the equal sign, "=". In many cases, the only evidence that a bit of text has been encoded in quoted-printable is the fact that all the equal signs have been converted to "=3D" Once a bit of text has been quoted in standard Usenet fashion the MIME headers that would have indicated that the bit of text was supposed to be interpreted as quoted-printable are no longer present. So any number of subsequent posters can quote and requote the passage and the "=3D" will never be decoded.
From: mueckenh on 27 Jun 2006 16:41 Virgil schrieb: > In article <1151333635.771069.62830(a)y41g2000cwy.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Daryl McCullough schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de says... > > > > > > >Daryl McCullough schrieb: > > > > > > >> 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. > > > > > > Forget about Cantor for now. 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? > > > > I cannot demonstrate a function f enumerating all reals, but I can > > demonstrate a form containing all real numbers, the binary tree: It is > > briefly described in my paper > > http://www.fh-augsburg.de/~mueckenh/Infinity/P2%20R4%20final.pdf > > > Actually an infinite binary tree ( in which each path stating from the > root node passes through an endless seqeuence of child nodes) can most > easily be bijected with P(N), the set of all subsets of N. > We consider only infinite paths. The set of all nodes is countable. The set of paths separated up to level n is less than the set of nodes up to level n. Every separation of paths requires two further nodes. Hence the set of separated paths cannot be larger than the set of nodes, i.e., the set of paths is countable. Regards, WM
From: mueckenh on 27 Jun 2006 16:43 Dik T. Winter schrieb: > So, I now assume that "Anzahl" is the "ordinal number" and "Mächtigkeit" > the cardinal number. That is correct. > But I still can be wrong, because the "ordinal > number" *can* change with a countably infinite number of interchanges. > Consider N with ordinal number w. Interchange 0 with 1, 0 with 2, ... > at the end you will have a well-ordered set with ordinal number w+1. > So to me the quote remains quite unclear. That is correct too, but the quote is irrelevant, because it is clear that a countable number of transpositions cannot change a well-ordered set into a not well-ordered set: There would have to be a first one to destroy the well-ordering. Perhaps Cantor meant that. > > Hence "Anzahl" is connected with the special kind of well-ordering. In > > particular different wel-orderings lead to different numbers > > (Anzahlen). > > Yes, this leads to the conclusion that it is the ordinal number of a set. > Every infinite set that can be well-ordered can be well-ordered in > different ways, leading to different ordinal numbers. But also, every > infinite set that can be well-ordered can also be ordered in such a way > that it is no longer well-ordered. But not by a countable number of transpositions. > > > 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). > > I do not know how to interprete it. A finite number of changes does not > change the ordinal number and well-ordering. A countably infinite number > of changes, with order type w, does not change well-ordering (I think) but > changes the ordinal number (but it depends on the meaning of a countably > infinite number of interchanges). But it is unclear to me how to interprete > an infinite number of interchanges. That is unimportant, because for my example there is only a countable number of transpositions required. > > 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. > > Under two conditions: No. These conditions are irrelevant. > 1. Cantor was right. > 2. Your interpretation of what Cantor wrote was right. > I think (2) is wrong, I do not know whether (1) is correct or wrong, because > I do not know how Cantor would interprete an infinite number of transpositions. Interprete it as a mapping. For n = 1 to oo: (q_2n-1, q_2n) --> (q_2n-1, q_2n) if q_2n-1 < q_2n, else: (q_2n-1, q_2n) --> (q_2n, q_2n-1) For n = 1 to oo: (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, else: (q_2n, q_2n+1) --> (q_2n+1, q_2n) And again: For n = 1 to oo: (q_2n-1, q_2n) --> (q_2n-1, q_2n) if q_2n-1 < q_2n, else: (q_2n-1, q_2n) --> (q_2n, q_2n-1) For n = 1 to oo: (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, else: (q_2n, q_2n+1) --> (q_2n+1, q_2n) .... repeat aleph_0 times. Of course we can also define all mappings in a closed loop, such that they are executed in zero time. > But your picking the rationals is really too much. If we accept the intuitive > "definition" of the limit and infinite "permutation" given above, we could > define the following: > P(k) = lim{n->oo} (k,k+1)(k,k+2)...(k,k+n) > as a permutation, and with the same method we could define: > lim{k->oo} P(0)P(1)...P(k) > which would reverse the ordering of the naturals and so is not a well-order. > On the other hand, the number of transpositions is still countably finite > (but the order-type is not w, but w * w). I am not sure whether you can define k+n for n --> oo. However, my mapping, given above, has always a well defined first element, and each subset has such a well-defined first element. Regards, WM
From: mueckenh on 27 Jun 2006 16:45
Virgil schrieb: > In article <1151332059.355968.171920(a)p79g2000cwp.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > > But whether or not this is accepted: If infinity exists, then also > > infinitely many transpositions do exist, and none of them will destroy > > the well-ordering. > > It remains for "mueckenh" to provide any transposition, or sequence of > transpositions, which will convert a non-empty well ordered set into one > not having a first element. My mapping is chosen, on purpose, so that it does not convert well-order into non-well-order. That supplies the contradiction. Regards, WM |