An uncountable countable set
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From: Virgil on 26 Jun 2006 15:32 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. His last effort failed.
From: Virgil on 26 Jun 2006 15:46 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? if K = { x: x is not in the value of x } then a is not in {}, the value of a, but 1 is in {1}, the value of 1, so K = {a}, but the image is {{},{1}}, so K is not a member of that image.. > > > > 2. > > > > a -> {1} > > > > 1 -> {} > > > > K =3D {1} > > > > so K is in the image. a is not in {1} 1 is not in {} K = {x: x not in value of x} = {a,1} Image = {{1},{}} So K is NOT in the image. > > > > > > K is not in the image of *a natural number*. a is not a natural number! > > > > Why should it be? The question was about a bijection between {a, 1} and > > {{}, {1}}. The pre-image does not consist of natural numbers, so why > > should > > K be the image of a natural number? > > Because we require it in order to show you that it is impossible, > independent of cardinalities. What you require, we see as nonsense. Such analyses for any set and its power set do not depend in any way on the type of members of the original set.
From: Virgil on 26 Jun 2006 16:03 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. To get a nice bijection of subtree with the reals, first biject the reals to the interval [0,1) = {x in R: 0 <= x < 1}. Then omit from the tree all paths with only finitely many left branches. This corresponds to omitting one representation of reach real in [0,1) having two binary representations. For each remaining path, P, in that subtree construct the sum of all terms 1/2^n for which the nth branch of P is a right branch.
From: Dik T. Winter on 26 Jun 2006 20:01 In article <1151330489.853370.287220(a)b68g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > p=3D2E 214: "Die Frage, durch welche Umformungen einer wohlgeordneten = > Menge > > > ihre Anzahl ge=3DE4ndert wird, durch welche nicht, l=3DE4=3DDFt sich e= > infach so > > > beantworten, da=3DDF diejenigen und nur diejenigen Umformungen die Anz= > ahl > > > unge=3DE4ndert lassen, welche sich zur=3DFCckf=3DFChren lassen auf ein= > e 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=E4chtigkeit_ > ge=E4ndert, wohl aber kann dadurch ihre _Anzahl_ eine andere werden. So, I now assume that "Anzahl" is the "ordinal number" and "M?chtigkeit" the cardinal number. 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. > Compare the letter from Cantor to Mittag-Leffler (17.12.1882): "Den > gr=F6=DFten Vortheil gewinne ich jetzt durch die Einf=FChrung ... 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=E4ngig 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). 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. > 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. We can look at it as a limiting process where we define something like limits on a sequence of sets, but even there I have no good idea how to define it. Or perhaps we can look at it as a limiting process in permutations, but I am not really able to get a good understanding about infinite permutations. Intuitively, lim{n->oo} (0,1)(0,2)...(0,n) should be (1,2,...,0), but that is not a definition. > 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: 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. 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). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 26 Jun 2006 20:08
In article <1151330868.199805.89070(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. What is nonsense with the definition? Given a particular f, K(f) is well-defined and a member of P(N). So, where is the nonsense? > 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. In this case Hessenberg's condition *is* satisfied because K is in the image of the function (although the pre-image is not a natural number, but *that* is not a problem, because the source set is not a set of natural numbers). > The set {K_f, m, f} is an > impossible set. It is impossible in the case N -> P(N) because the source set is the set of natural numbers, so if f is surjective, K(f) *must* be in the image of f, and so there *should* be a pre-image of K(f) that is a natural number. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |