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From: David R Tribble on 18 Oct 2005 20:15 David R Tribble wrote: >> Come on, now. It's up to you to prove your own claim, especially >> when it contradicts established mathematics. I know you cannot show >> a bijection between N and P(N). >> >> [Example snipped] > Albrecht S. Storz wrote: > Second, your above argueing is not clear to me, since both sets are > well-ordered. But it's nice, so I give this: > > N > {1},{2},{3}, ... > N/{1},N/{2},N/{3}, ... > {1,2},{1,3},{2,3},{1,4},... > N/{1,2}, ... > ... > > Now count in diagonal sequence. You may think of Cantor's first > diagonal proof. > > Which subset of N is not included? I don't know what you mean by "count in diagonal sequence". Let's try a simpler approach... Let S0 = {}, the empty set. Let S1 = {{0}, {1}, {2}, {3}, ...}, so it is the set of all singleton subsets of N. Let S2 = {{0,1}, {0,2}, {1,2}, {2,3}, ...}, so it is the set of all two-member subsets of N. And so on, defining S3, S4, S5, etc., so that Sn(N) is the set of all n-member subsets of N. Now let U = S0 u S1 u S2 u S3 u ... It seems to me that N is as large as S1, since for every member n in N, there is a member {n} in S1, right? So shouldn't U be even larger than N? And isn't U just P(N)? [I realize this might not be completely correct as per standard set theory; but I'd like to see Albrecht's response.]
From: William Hughes on 18 Oct 2005 20:24 albstorz(a)gmx.de wrote: > David R Tribble wrote: > > Albrecht S. Storz wrote: > > >> Cantor proofs his wrong conclusion with the same mix of potential > > >> infinity and actual infinity. But there is no bijection between this > > >> two concepts. The antidiagonal is an unicorn. > > >> There is no stringend concept about infinity. And there is no aleph_1, > > >> aleph_2, ... or any other infinity. > > > > > > > David R Tribble wrote: > > >> For that to be true, there must be a bijection between an infinite > > >> set (any infinite set) and its powerset. Bitte, show us a bijection > > >> between N and P(N). > > > > > > > Albrecht S. Storz wrote: > > > At first, you should show, that bijection means something to > > > notwellordered infinite sets. > > > > > > Bijection is a clear concept on finite sets, it also works on > > > wellordered infinite sets of the same infinite concept. > > > Aber: Show me a bijection between two infinite sets with the same > > > cardinality, where one of the sets is still not wellorderable. > > > Than I will show you a bijection between N and P(N) or N and R or P(N) > > > and P(P(N)) or what you want. > > > > I see. I'm supposed to show you a proof before you can show me your > > proof. Okay, I give up, you win, so your proof must be correct. > > > > Come on, now. It's up to you to prove your own claim, especially > > when it contradicts established mathematics. I know you cannot show > > a bijection between N and P(N). > > > > > > P.S. > > > > Let > > D(n,i) = floor(n/2^i) mod 2, for all i=0,1,2,3,... > > This is the i-th binary digit of natural n > > L(n) = i where D(n,i) = 1 and D(n,j) = 0 for all j > i > > This is the number of binary digits of natural n, or ceil(log2(n)) > > Let > > M(n) = sum{i=0 to L(n)} 1/2^(i+1) where D(n,i) = 1 > > This reverses the binary digits of n. > > > > Then M(n) is a mapping N -> N, from all n in N to M(n) in N, > > but the set of all M(n) is not a well-ordered set. > > Happy? > > > Sad! > First of all you don't argue on my claim of the thread. > Second, your above argueing is not clear to me, since both sets are > well-ordered. But it's nice, so I give this: > > N > {1},{2},{3}, ... > N/{1},N/{2},N/{3}, ... > {1,2},{1,3},{2,3},{1,4},... > N/{1,2}, ... > ... > > Now count in diagonal sequence. You may think of Cantor's first > diagonal proof. > > Which subset of N is not included? Sigh. Try {2,4,6,8,...} Or any other infinite subset that is not the complement of a finite set. It is common for people to note that the finite subsets of N are countable and incorrectly claim that the subsets of N are countable. Adding in the complements of the finite subsets does not change things very much. Those who do not study anti-cantor cranks are doomed to repeat their idiocies. - William Hughes P.S. No TO, going to TO-infinite rows is not going to help us. The problem is that for any TO-finite row, the subsets listed will have an bounded finite number of elements, or be the complement of a subset with a finite number of elements. Yes, you can claim (without a shred of motivation) that when you get to TO-infinite rows the subsets will suddenly have an unbounded number of elements, but all this tells us is that a bijection from the TO-naturals to P(N) exists. What we need is a bijection from the finite TO-naturals to P(N).
From: albstorz on 18 Oct 2005 21:11 David R Tribble wrote: > > Because I can prove it (and it's a very old proof). A powerset of > a nonempty set contains more elements that the set. Can you prove > otherwise? This argument is stupid. Is there any magic in the powerfunction? A hidden megabooster for transcendental overflow? What is the very special aspect of the powerfunction to be so magic? Why should all operations with transfinite numbers lead to results with the same "level" of infinity, but only powerfunction beams up to the next level? Is not true: a^2 = a*a? 2a = a+a? Is the powerfunction something other than a very shortcut for multiple additions? Which amount you are able to reach with powerfunction which is unreachable by succesor operation? I'm very sensible about this because this argument is found in very much books although it's total meaningless. (Weak minds might be impressed by the big numbers which are easily produced by powerfunction.) What in finity holds may not (or do not) hold in infinity. Regards AS
From: albstorz on 18 Oct 2005 21:16 Virgil wrote: > In article <1129622094.019699.233300(a)g44g2000cwa.googlegroups.com>, > albstorz(a)gmx.de wrote: > > > David R Tribble wrote: > > > At first, you should show, that bijection means something to > > notwellordered infinite sets. > > > > Bijection is a clear concept on finite sets, it also works on > > wellordered infinite sets of the same infinite concept. > > Aber: Show me a bijection between two infinite sets with the same > > cardinality, where one of the sets is still not wellorderable. > > If either of two sets is well-orderable and there is a bijection between > the sets, then that bijection induces well-order-ability on the other. > > So that you cannot have a bijection between two sets when one is > well-orderable and the other is not. That's right. So biject two non-well-orderable sets. Regards AS
From: William Hughes on 18 Oct 2005 21:33
albst...(a)gmx.de wrote: > David R Tribble wrote: > > > > > Because I can prove it (and it's a very old proof). A powerset of > > a nonempty set contains more elements that the set. Can you prove > > otherwise? > > This argument is stupid. Is there any magic in the powerfunction? A > hidden megabooster for transcendental overflow? What is the very > special aspect of the powerfunction to be so magic? > Why should all operations with transfinite numbers lead to results with > the same "level" of infinity, but only powerfunction beams up to the > next level? > Is not true: a^2 = a*a? 2a = a+a? Yes, but the powerfunction does not look like a^2 but 2^a. > Is the powerfunction something other than a very shortcut for multiple > additions? Yes. You cannot represent 2^x as multiple additions. > Which amount you are able to reach with powerfunction which is > unreachable by succesor operation? Infinity for one. You cannot get from a finite quantity to an infinite quantity by using the successor operation (unless like TO you are willing to wave a circular magic wand and apply the successor operation an infinite number of times). > > I'm very sensible about this because this argument is found in very > much books although it's total meaningless. > I suspect that you mean "sensitive" not "sensible". > (Weak minds might be impressed by the big numbers which are easily > produced by powerfunction.) Strong minds are impressed with the fact that there is no bijection between X and P(X). > > > What in finity holds may not (or do not) hold in infinity. Words to live by. Start by noting that a finite set has a "number of elements" that can be described by a natural number while an infinite set (e.g. the set of natural numbers) does not have a "number of elements" that can be described by a natural number. However, some things are true for both finite and infinite sets. e.g. the fact that there is no bijection between X and P(X). -William Hughes |