An uncountable countable set
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From: Dik T. Winter on 22 Jun 2006 23:03 In article <1151009835.167756.67970(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Daryl McCullough schrieb: > > That case is impossible. To prove a case impossible, you only > > need to show that it leads to a logical contradiction. > > 1) Consider a bjective mapping from the set {a, 1} --> P({1}) = {{}, > {1}}, where a is *not a natural number*. This mapping is arbitrary but > has to satisfy only one condition, namely that the set K = {k e |N & k > /e f(k)} is in the image. That set (as written) is impossible. f is define on 'a' and '1', so what is f(2) and so on? It makes no sense as stated. I think you mean here K = {k in {1} & k !in f(k)}. Some annotation for the Hessenberg set, considering a mapping f between sets Q and R. K = {k in S & k !in f(k)} requirements for this to be an actual requirement (and not nonsense) are: 1. S subset of Q (the source, otherwise f(k) need not exist) 2. R set of subsets of Q (otherwise the condition k !in f(k) makes not sense) So rather than K = {k in N & k !in f(k)} I assume K = {k in {1} & k !in f(k)}. > This mapping is impossible. (See my article > "On Cantor's Theorem", arXiv, math.GM/0505648, 30 May 2005.) Yes, that article is plain nonsense. Consider two cases: 1. a -> {} 1 -> {1} K = {} so K is in the image 2. a -> {1} 1 -> {} K = {1} so K is in the image. > 2) Consider a mapping |N --> P(|N) which need not be surjective but has > to satisfy only one condition, namely that the set K = {k e |N & k /e > f(k)} is in the image. This mapping is impossible. Right. > In both cases there is this impredicable request {f, k, K} which is > impossible to satisfy. But in the proof by Hessenberg, you insist, it > would prove non-surjectivity? Wrong. In the first case the condition can be satisfied, in the second case it can not be satisfied. > It will be enough to use the positive rationals only. > > > > 0/1, > > -1/1, +1/1, > > -2/1, -1/2, +1/2, +2/1 > > -3/1, -1/3, +1/3, +3/1 > > -4/1, -3/2, -2/3, -1/4, +2/3, +3/2, +4/1 > > -5/1, -1/5, +1/5, +5/1 > > etc. > > > > (The first row has |p| + q = 1, the next row has |p| + q = 2, etc.) > > > > But notice that this is not ordered by magnitude, since 1/2 < 1/1, but > > 1/1 comes before 1/2 in the ordering. > > > > The ordering by magnitude in not a well-ordering. > > > Here is how we obtain it: > Cantor said: A well-ordered set remains well-ordered, if finitely many > or infinitely many transpositions are executed. Let's see what happens. Give a quote please. Where did he state that? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Daryl McCullough on 22 Jun 2006 23:34 mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: >> That case is impossible. To prove a case impossible, you only >> need to show that it leads to a logical contradiction. >> > >1) Consider a bjective mapping from the set {a, 1} --> P({1}) = {{}, >{1}}, where a is *not a natural number*. This mapping is arbitrary but >has to satisfy only one condition, namely that the set K = {k e |N & k >/e f(k)} is in the image. This mapping is impossible. Right. If f is any function from any set A to any set B, then we can define K(f) = { x in A | x is not an element of f(x) }. Then K(f) is not in the image of f. Furthermore, if K(f) is an element of B, then f is not a surjection from A to B. In the particular case in which B = P(A), we therefore have, for any function f from A to P(A): 1. K(f) is not in the image of f. 2. K(f) is an element of P(A). 3. Therefore, f is not a surjection. Therefore, there is no surjection from A to P(A). >2) Consider a mapping |N --> P(|N) which need not be surjective but has >to satisfy only one condition, namely that the set K = {k e |N & k /e >f(k)} is in the image. This mapping is impossible. Right. There is no f such that K(f) is in the image of f. There is no f that is a surjection to any set containing K(f). >In both cases there is this impredicable request {f, k, K} which is >impossible to satisfy. But in the proof by Hessenberg, you insist, it >would prove non-surjectivity? Yes. If f is a function from A to B, and K(f) is an element of B, then f is not a surjection from A to B. From this it follows that for any function f from A to P(A), f is not a surjection. >In a bijective mapping |N --> P(|N) We just proved that for any set A, it is impossible for there to be a surjection from A to P(A). >there is every element of |N and >every subset of |N. But no set is defined as K above, because this >very *definition* leads to an impossible result Yes, the assumption that there is a bijection from N to P(N) leads to a contradiction. >> This well-ordering gives the following first few terms of the sequence: > >It will be enough to use the positive rationals only. >> >> 0/1, >> -1/1, +1/1, >> -2/1, -1/2, +1/2, +2/1 >> -3/1, -1/3, +1/3, +3/1 >> -4/1, -3/2, -2/3, -1/4, +2/3, +3/2, +4/1 >> -5/1, -1/5, +1/5, +5/1 >> etc. >> >> (The first row has |p| + q = 1, the next row has |p| + q = 2, etc.) >> >> But notice that this is not ordered by magnitude, since 1/2 < 1/1, but >> 1/1 comes before 1/2 in the ordering. >> >> The ordering by magnitude in not a well-ordering. >> >Here is how we obtain it: >Cantor said: A well-ordered set remains well-ordered, if finitely many >or infinitely many transpositions are executed. I don't care whether Cantor said it or not (and I don't believe he did), it's not true. If you have a well-ordering, then a *finite* number of transpositions leaves it a well-ordering, but an *infinite* number of transpositions may not. To see this, start with the naturals in their usual ordering: 0 < 1 < 2 < 3 < 4 < 5 ... This is a well-ordering, in the sense that every natural has only finitely many naturals that are smaller. However, suppose we do the following infinite sequence of transpositions: 1. Put 1 before 0. 2. Put 2 before 1. 3. Put 3 before 2. 4. Put 4 before 3. ... Then the result is a reverse-ordered set of naturals: ... < 5 < 4 < 3 < 2 < 1 < 0 This is no longer a well-ordering, because every number has infinitely many smaller numbers, according to the new ordering. >Let's see what happens. > >Let {q_1, q_2, q_3, ...} be the well-ordered set of all positive >rationals. >1) Consider for n e |N the rationals q_n and q_n+1. If q_n > q_n+1, >exchange q_n and q_n+1, otherwise let the succession unaltered. This >requires at most aleph_0 transpositions. Yes, after doing infinitely many transpositions, you no longer have a well-ordering. >Of course, this all is defined in zero time, like the definition of >Cantor's diagonal. >After aleph_0 * aleph_0 = aleph_0 transpositions, the set {q_k_1, >q_k_2, q_k_3, ...} of all positive rationals is well-ordered. No, it's no longer well ordered. >This can happen, if the well ordered set of all positive rational does >exist. No, I just gave you a well-ordering of all the positive rationals. Of course the well-ordering exists. >But we know, it cannot happen. Because well-orderings are not preserved by an infinite sequence of transpositions. >Hence, the well ordered set of all positive rational does not exist. I just gave you one. Of course it exists. When you prove something that you know to be false, then you have to suspect something is wrong with your proof. In your case, you know that there is a well-ordering of all the rationals, because I just showed you one. So the question is: what is wrong with your proof. The mistake is that well-orderings are not preserved by an infinite sequence of transpositions. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 22 Jun 2006 23:37 mueckenh(a)rz.fh-augsburg.de says... >Cantor said: A well-ordered set remains well-ordered, if finitely many >or infinitely many transpositions are executed. I don't believe that Cantor ever said that. But whether he did or not, it's a false statement. -- Daryl McCullough Ithaca, NY
From: mueckenh on 23 Jun 2006 12:14 Daryl McCullough schrieb: > No, that doesn't follow. What follows is that M(f) is *not* a subset > of S. Since S contains all finite subsets of N, it follows that M(f) > contains some element that is *not* a finite subset of N. > Consider the mapping f : |N --> M, where M contains a countable set of *infinite* subsets of |N and K. What follows now? Regards, WM
From: mueckenh on 23 Jun 2006 12:16
Daryl McCullough schrieb: > If an assumption leads to a contradiction, then that assumption must > be false. The assumption includes that |N and P(|N) do exist as complete sets. >The assumption that there is a surjection from N to P(N) leads > to a contradiction. Therefore, it is false that there is a surjection > from N to P(N). Therefore, P(N) is uncountable. Therefore, |N and P(|N) do not exist as complete sets. Regards, WM |