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From: David R Tribble on 18 Oct 2005 19:52 Albrecht S. Storz wrote: >> At first, you should show, that bijection means something to >> notwellordered infinite sets. > David R Tribble said: >> 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? > Tony Orlow wrote: > Is that supposed to make me happy? It was supposed to make Albrecht happy, since I was responding to him. Slight correction: M(n) = sum{i=0 to L(n)} 2^L(n)2^(L(n)-i+1) where D(n,i) = 1 This reverses the binary digits of n.
From: David Kastrup on 18 Oct 2005 19:57 "David R Tribble" <david(a)tribble.com> writes: > Because I can prove it (and it's a very old proof). A powerset of a > nonempty set contains more elements that the set. Drop the "nonempty", it is not necessary. P({}) = {{}} contains more elements than {}, and the old proof works even for that case. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: albstorz on 18 Oct 2005 19:59 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? Regards AS
From: David R Tribble on 18 Oct 2005 20:03 David R Tribble writes: >> Because I can prove it (and it's a very old proof). A powerset of a >> nonempty set contains more elements that the set. > David Kastrup wrote: > Drop the "nonempty", it is not necessary. P({}) = {{}} contains more > elements than {}, and the old proof works even for that case. Yep, I forgot about that one!
From: stephen on 18 Oct 2005 20:12
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}, ... > ... Is this supposed to be a list? My reading of this is that your list is: N, {1}, {2}, {3}, .... N/{1}, Right here we have a problem. What is the element before N/{1} in your "list"? There is no end to the list {1}, {2}, {3}, ... so you cannot put N/{1} after the end of that list. Remember each element in a list must be indexed by a natural number. > Now count in diagonal sequence. You may think of Cantor's first > diagonal proof. What diagonal? > Which subset of N is not included? N/{1} for one. What is the index of N/{1} in your "list" above? The index of N is 0. The index of {1} is 1. The index of {2} is 2. What is the index of N/{1}? Also your "enumeration" above only includes finite sets, or sets whose complement with respect to N is finite. That excludes an awful lot of sets. In fact it excludes "most" of them. For example, where is the set of primes going to show up? Stephen |