An uncountable countable set
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From: Jens Kruse Andersen on 22 Jun 2006 15:14 mueckenh wrote: > Jens Kruse Andersen schrieb: > > > By definition, an infinite set M is uncountable if and only if: > > There is no bijective mapping f from N to M. > > > > Note that M is given first, and *then* it is said that no bijective f > > exists for that M, with no other condition on f than it has to be > > bijective. > > Oh, I see. Let's apply this new insight: > > 2) 0.12324389 > 3) 0.23123123 > 4) 0.85348714 > 5) 0.11133333 > 6) 0.31415161 > .. > > 1) 0.24446... > > So this is the proof, that the proof that a surjective mapping f: |N > --> [0,1] does not exist, does not exist, isn't it? As others have said: Nonsense. A small list of numbers with no definitions, assumptioms, explanations or anything else is, well, a small list of numbers. Your final claim is hard to parse but if you think your small list of numbers with no comments proves that Cantor's diagonal proof is wrong, then: Nonsense. After seeing your replies here and elsewhere, I have decided to not spend more time on this discussion. Others have already said several times what I would say but you appear to ignore it. I suggest rereading the replies carefully. > Please note: My name is Mueckenheim or short WM. If you want to be referred to by some name then use it as poster name instead of your e-mail address. Many people reply with a program which automatically inserts your poster name. -- Jens Kruse Andersen
From: mueckenh on 22 Jun 2006 16:52 Jens Kruse Andersen schrieb: > mueckenh wrote: > > Jens Kruse Andersen schrieb: > > > > > By definition, an infinite set M is uncountable if and only if: > > > There is no bijective mapping f from N to M. > > > > > > Note that M is given first, and *then* it is said that no bijective f > > > exists for that M, with no other condition on f than it has to be > > > bijective. > > > > Oh, I see. Let's apply this new insight: > > > > 2) 0.12324389 > > 3) 0.23123123 > > 4) 0.85348714 > > 5) 0.11133333 > > 6) 0.31415161 > > .. > > > > 1) 0.24446... > > > > So this is the proof, that the proof that a surjective mapping f: |N > > --> [0,1] does not exist, does not exist, isn't it? > > As others have said: Nonsense. A small list of numbers with no definitions, > assumptioms, explanations or anything else is, well, a small list of numbers. I think everyone her does know this list of numbers. It is not a small list, but a countably infinte one. (One point was missing, it should read "...".) > Your final claim is hard to parse but if you think your small list of numbers > with no comments proves that Cantor's diagonal proof is wrong, then: Nonsense. Comment: The first mapping read f = 1), 2), 3), .... The second mapping reads 2), 3), 4), ..., 1). Isn't that enough? > > After seeing your replies here and elsewhere, I have decided to not spend more > time on this discussion. > Others have already said several times what I would say but you appear to > ignore it. I suggest rereading the replies carefully. I have never stated the following problem. So you might be interested: Cantor said: A well-ordered set remains well-ordered, if finitely many or infinitely many transpositions are executed. 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. 2) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2, exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This requires at most aleph_0 transpositions. 3) 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. 4) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2, exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This requires at most aleph_0 transpositions. 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 and simultaneously ordered by magnitude, starting with the smallest positive rational and ending with the largest. This can happen, if the well ordered set of all positive rational does exist. But we know, it cannot happen. Hence, the well ordered set of all positive rational does not exist. Regards, WM
From: mueckenh on 22 Jun 2006 16:57 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. (See my article "On Cantor's Theorem", arXiv, math.GM/0505648, 30 May 2005.) 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. 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? In a bijective mapping |N --> P(|N) 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 as can be seen by the examples above. > 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. 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. 2) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2, exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This requires at most aleph_0 transpositions. 3) 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. 4) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2, exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This requires at most aleph_0 transpositions. 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 and simultaneously ordered by magnitude, starting with the smallest positive rational and ending with the largest. This can happen, if the well ordered set of all positive rational does exist. But we know, it cannot happen. Hence, the well ordered set of all positive rational does not exist. Regards, WM
From: mueckenh on 22 Jun 2006 16:59 Daryl McCullough schrieb: > In article <1150975362.277573.39860(a)b68g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de says... > > > > > >Virgil schrieb: > > > > > >> So the resolution is that either f, K_f and M_f are self-contradictory > >> and do not exist at all > > > >if f is defined to be a surjection. That is right. And why do you > >believe that a self contradictory approach should prove anything (for > >instance about the surjectivity of certain mappings)? > > If an assumption leads to a contradiction, then that assumption must > be false. 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. > 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. {f, k, K} is impossible to satisfy. But in the proof by Hessenberg, you insist, it would prove non-surjectivity? In a bijective mapping |N --> P(|N) 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 as can be seen by the examples above. Regards, WM
From: mueckenh on 22 Jun 2006 17:12
Daryl McCullough schrieb: > mueckenh(a)rz.fh-augsburg.de says... > > >Maybe. It is, however, the same approach as yours. If it is impossible > >to find a direct surjective mapping, then first define the set and then > >try to find a surjection. I can proudly declare in your words: > > > > A *surjective* mapping does exist, but it is not f. There exists a > >surjective mapping g -> M(f). > > And those words are perfectly correct: f is not a surjective mapping > from N to M(f), but g *is* a surjective mapping from N to M(f). So > M(f) is countable. > > In contrast, there is no surjection from N to P(N). So P(N) is *not* > countable. This is proven by three invalid proofs. Do you know Canto's first proof of 1874? Probably not, so we need not discuss it. You certainly know his second one. It is wrong, because all it shows is the construction of a number of a set of countably many constructible numbers. It is ridiculous to believe this to be a proof of uncountability. There are simply not more than countably many constructible and individualizable real numbers, because there are not more than countably many different symbols or names in any language. And I pointed out already that your arguing invalidates this proof completely: > 2) 0.12324389 > 3) 0.23123123 > 4) 0.85348714 > ... > 1) 0.244... (I hope, you understand, what I mean.) The third proof is that by Hessenberg, which utiizes an impredicable definition. This set K does *exist*, if every subset of |N does exist, but it cannot be *defined* as it is done. But I discussed that with you in a previous posting. Uncountably many sets, however, cannot be distinguished, because of laking names. Regards, WM |