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From: David R Tribble on 19 Oct 2005 13:15 Tony Orlow wrote: >> I already showed you the bijection between binary *N and P(*N). >> What didn't you like about it? It is valid. > David R Tribble said: >> No, you showed a mapping between *N and R, which is equivalent >> to a mapping between *N and P(N). That's easy. > Tony Orlow wrote: >> What do you want me to try, anyway, and infinite mapping, >> element-by-element? A bijection's a bijection, right? > David R Tribble said: >> Yes, that would be nice. Please show us your bijection. Tony Orlow wrote: > f(0) = ...000 = {} > f(1) = ...001 = {0} > f(2) = ...010 = {1} > f(3) = ...011 = {0,1} > f(4) = ...100 = {2} > f(5) = ...101 = {0,2} > f(6) = ...110 = {1,2} > f(7) = ...111 = {0,1,2} > > etc. Any questions? Where are the infinite subsets, such as the set of even numbers?
From: Randy Poe on 19 Oct 2005 13:24 Tony Orlow wrote: > David R Tribble said: > > Tony Orlow wrote: > > Yes, that would be nice. Please show us your bijection. > f(0) = ...000 = {} > f(1) = ...001 = {0} > f(2) = ...010 = {1} > f(3) = ...011 = {0,1} > f(4) = ...100 = {2} > f(5) = ...101 = {0,2} > f(6) = ...110 = {1,2} > f(7) = ...111 = {0,1,2} Since this infinite strings are supposed to represent subsets of *N, I presume that they have |*N| bits. There is one bit for each element of *N. So what you are claiming is that this is a bijection between the bit strings of length *N, and the elements of *N. I claim that there is no x such that f(x) = *N. Here's a TO-ish argument: For initial subsets {0,1,...,x} for any x, this is mapped by f(y) where y = 2^x - 1. Clearly y > x for all x, and the list f(1), f(2), ... f(y) is longer than the list 1, 2, ..., x. Since this is true for all finite x, it is also true for x = *N. So there are not enough elements in *N to complete the list f(1), f(2), ..., f(2^*N - 1). Another argument by TO-induction: The elements mapped by f(1),f(2), .., f(x) only include elements among {0,1,...,log2(x)}. Thus the list f(1), f(2), ..., f(*N) only includes elements among {0,1,..., log2(*N)}. It isn't long enough to include {0,1,..., *N}. - Randy
From: William Hughes on 19 Oct 2005 13:44 Tony Orlow wrote: > William Hughes said: > > > > 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). > > > > > Look, it's certainly not my position that the pwoer set is the same sie as the > set. It's clearly not. I just see a bijection between them, which only bolsters > my argument that bijection alone is not sufficient to equate the sizes of two > sets. > > When it comes to the evens (let's start with 0), the value 0:010.......1010101 > represents such a subset, and is essentially binary N/3. Well the binary notation is certainly convenient. Let's lay things out in a sort of a square Take the set X How many columns do we need. One for every element of X How many rows do we need. One for every element of X Pity about the complement of the diagonal. Lets try X = N*. How many columns do we need. One for every element of N* How many rows do we need. One for every element of N* Pity about the complement of the diagonal. - William Hughes
From: stephen on 19 Oct 2005 13:42 Tony Orlow <aeo6(a)cornell.edu> wrote: > David R Tribble said: >> Tony Orlow wrote: >> >> >> > What do you want me to try, anyway, and infinite mapping, >> > element-by-element? A bijection's a bijection, right? >> >> Yes, that would be nice. Please show us your bijection. > f(0) = ...000 = {} > f(1) = ...001 = {0} > f(2) = ...010 = {1} > f(3) = ...011 = {0,1} > f(4) = ...100 = {2} > f(5) = ...101 = {0,2} > f(6) = ...110 = {1,2} > f(7) = ...111 = {0,1,2} If we define w= { x : x not in f(x) } then we get w = {0, 1, 2, 3, ...... } So for what y does f(y) = { 0, 1, 2, 3, ..... }? And is y in f(y)? Let me guess. You will claim that F(N) = { 0, 1, 2, 3, ..... }. So is N in { 0, 1, 2, 3, ..... }? If it is, then N is not in w, and F(N) does not equal w. If it is not, then N is in w, and F(N) does not equal w. <snip> > But I have constructed a bijection between the two using an intermediate binary > representation. What is the specific rule I have broken concerning the > construction of bijections. If I haven't broken any such rule, then is it true > that a bijection between two sets means that have the same size, or even > cardinality? No you have not. Stephen
From: David Kastrup on 19 Oct 2005 13:47
Tony Orlow <aeo6(a)cornell.edu> writes: > David Kastrup said: >> albstorz(a)gmx.de writes: >> >> > Virgil wrote: >> >> In article <1129684276.251366.121150(a)g47g2000cwa.googlegroups.com>, >> >> albstorz(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? >> >> >> >> "Proofs" are not stupid until they can be refuted. The proof that for an >> >> arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. >> > >> > >> > Even if you think that the powersets of finite and infinite sets >> > have both a greater cardinality than their starting sets, you would >> > not really think it depends on the same cause in both cases. >> > >> > You must proof it independently for finite and for infinite sets. In >> > this sense the argument is stupid. >> >> Uh, no. The proof depends merely on the fact that some value has to >> be either a member of a set, or not. And if some value is in one set, >> bur not another, then those two sets are different. > Oh! So, N is a different size from N/{1}? Wrong. N is a different set from N\{1}, but that does not mean that the sets have different cardinality. To illustrate this for the really dense people with a simpler example: 2 is in {2} and not in {1}, and that means that the set {2} is different from the set {1}, even though both sets have the same cardinality, namely 1. Whenever one thinks that your stupidity has reached bottom, you still manage to come up with something even more harebrained. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |