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From: Randy Poe on 21 Oct 2005 13:42 Tony Orlow wrote: > stephen(a)nomail.com said: > > It is not a bijection. N is a subset of N, and is therefore > > in the powerset of N, and if no natural maps to N, then > > there is no bijection. > There is always a natural for every subset, This contradicts your earlier statement. You have said there is NO natural for the subset consisting of the entire set. Remember? Shall we go through the argument again? There may be two different discussions here, one on *N <-> P(*N) and one on N <-> P(N). Everything I say here can apply to either one using your map definition. Let w = {y in N: y not in f(y)} You agree that w = N, that w is in P(N), and that there is no natural y such that f(y) = w. Therefore you agree that there is AT LEAST ONE subset, namely w, for which there is no natural y that maps to it. That is a direct contradiction to "There is always a natural for every subset." There's at least one for which that is not true. - Randy
From: William Hughes on 21 Oct 2005 13:44 Tony Orlow wrote: > William Hughes said: > > > > Tony Orlow wrote: > > > William Hughes said: > > > > > > > > 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 > > > > > > > > > > > Ha ha ha. Nice try, William. Let's get real. You want a sqaure? It's not > > > square. > > > > > > We have a set with N elements. > > > > Right we will have N rows (each row represents the > > value of the "enumeration" at a given element and there > > are N elements). > > > > >We will enumerate the elements of the powerset, > > > represented as binary strings with 1 bit per N elements, or N bits, per line of > > > the table. > > > > Right there are N columns > > > > > Now, how many such subsets are there? We have a power set with 2^N > > > elements, > > > > And you are somehow going to fit these 2^N elements into N > > rows. > No, you can't. You can only traverse N rows with the diaginal, since you only > have N columns. Your antidiagonal does not exist in the first N rows. But, > there are 2^N-N rows left. No. There are N rows. Where are you getting 2^N rows? Each row represents the value of the supposed bijection evaluated at an element of the set. There are N elements in the set, therefore there are N rows. - William Hughes
From: Tony Orlow on 21 Oct 2005 13:45 stephen(a)nomail.com said: > Tony Orlow <aeo6(a)cornell.edu> wrote: > > stephen(a)nomail.com said: > >> Tony Orlow <aeo6(a)cornell.edu> wrote: > >> > 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. > >> > >> So which number does that first 1 represent? > > > log2(N-1). Perhaps we should say we have 2^N subsets and strings of N digits. > > That's what I should have said. Then that number is 2^N/3, and the first 1 in > > it represents N-2, since the bits are numbered from 0 through N-1. > > So this is not the set of all even numbers then, because > according to you there are even numbers larger than log2(N-1). This is the imagined completed set. The number of digits is variable according to what infinity you are working in. You have to choose a unit to measure. Do you have a ruler with no marks on it? Excuse me while I notch my infinite number line. These numbers are relative. > > >> Presumably that is > >> the "last" infinite even number. > > > The last one before N. There is no ultimate last even, as we all know, silly > > canilly. > > So this is not the set of all even numbers and you were > wrong when you claimed it was. Why did you claim that > that string represented all the even numbers when you in > fact knew that it did not? You know my approach at this point. The variable infinity is in the "...". You may consider it to have log2(N), N, 2^N, or 3N^2+2N-4 digits. It really doesn't matter. Pick an infinity, any infinity.... > > >> And which 0 represents > >> 0:010........10101? It is an odd number afterall, so one of those > >> 0's must represent it. > > No, that number is 2^N/3, and would be represented by the 2^N/3th bit, which is > > not in the first N bits, right? This is the heart of your power set proof. > > That makes no sense, but that is besides the point. You claimed > that 0:010......10101 represents the set of all even numbers. > You now are claiming that it only represents some of the even numbers. No, I am claiming it is a representation of the set of even numbers. You are asking about specific numbers of them, which I have said is a nonsensical approach to unending sets. There is no specific number of them. For any infinity you choose as a number of bits N, the evens are a subset of that set of binary numbers with N bits, and has a set size of 2^N/3. N can be finite or infinite, really. I mean, the ellipses could be 10 digits long, and it would still hold. Or it can be N digits, or R digits, or aleph_3000 digits. It don't gives a hoot. > > So where is the string that represents all the even numbers? .....101010101. Is that better? oo/3. > > Your arguments would be a lot more consistent and sensible > if you just dropped the whole notion of "infinity". It > is clear that you do not actually believe in infinite sets. > For you all sets must have a first and last element, and > all your arguments require all sets to be finite. No, all measurements require two endpoints. The endpoint is variable, and can be finite or infinite. > > Stephen > -- Smiles, Tony
From: Tony Orlow on 21 Oct 2005 13:54 stephen(a)nomail.com said: > Tony Orlow <aeo6(a)cornell.edu> wrote: > > stephen(a)nomail.com said: > >> Tony Orlow <aeo6(a)cornell.edu> wrote: > >> > stevendaryl3016(a)yahoo.com said: > >> >> albstorz(a)gmx.de says... > >> >> > >> >> >I don't know what you are talking about. The proof for finite sets > >> >> >needs just a complete induction. This will not hold for infinity I > >> >> >think. > >> >> > >> >> There is no induction involved in the finite case, and the > >> >> infinite case is *exactly* the same proof as the finite case. > >> >> > >> >> Here it is once again: > >> >> > >> >> Let A be any set whatsoever, finite or infinite, it doesn't matter. > >> >> Let f be any function from A to P(A). > >> >> Let w = { x in A | x is not an element of f(x) }. > >> >> Let x = any set in A. > >> >> Let u = f(x). We prove that u is not equal to w. > >> >> > >> >> By definition of w, we have x in w <-> x is not an element of f(x). > >> >> So x in w <-> x is not an element of u. That means that there are > >> >> two cases: Case 1: x in w, and x is not in u. In that case, u cannot > >> >> equal w. Case 2: x is not in w, and x is in u. In that case, u cannot > >> >> equal w. > >> >> > >> >> So what we have proved is that forall x, w is not equal to f(x). So > >> >> w is not in the image of f. So f is not a bijection between A and P(A). > >> >> > >> >> There's no induction. There's no assumption that A is finite. > >> > But there is an assumption that y is in S. > >> > >> Of course there is an assumption that y is in S. The > >> goal is to find a bijection f from S to P(S). That > >> means that for every element x in P(S), there must be > >> an element y in S such that f(y)=x. w is an element of P(S). > >> In order for f to be a bijection, there must exist an > >> element y in S such that f(y)=w. > >> > >> If y is not in S, then it is irrelevant to the question > >> of whether or not f is a bijection from S to P(S). > >> > >> Do you consider the following a bijection from S={a,b,c} to its > >> power set? > >> > >> f(a) = {} > >> f(b) = {a} > >> f(c) = {b} > >> f(d) = {c} > >> f(e) = {a,b} > >> f(g) = {a,c} > >> f(h) = {b,c} > >> f(i) = {a,b,c} > >> > >> If w= { x : x in S and x not in f(x) } > >> then w= {a,b,c}. > >> > >> Is there a y in S such that f(y) = {a,b,c}? No. > >> > >> Is there a y such that f(y) = {a,b,c}? Yes, f(i)={a,b,c}, but > >> i is not in S, and is not part of a bijection from S to P(S). > >> > >> The above function is a bijection from {a,b,c,d,e,g,h,i} > >> to P({a,b,c}). It is not a bijection from {a,b,c} to P({a,b,c}). > >> > >> Stephen > >> > > But, if this set went on forever, then i WOULD be in the set, and for any > > subset, you could identify an x such that it mapped to that set. > > No it would not. Sure it would. > > > that no bijection is possible in the finite case. In the proof, a last element > > of the set is assumed when we assume there is some string representing the > > entire set. > > There is no mention of strings in the proof. Can't you > read a simple proof? There is no mention of strings or > last elements. Just read the proof. Stop making > up stuff. I am referring also to my bijection, to which no one has raised any serious objection except on pronciple. > > The proof really only makes use of four things. > A set S, its power set P(S), a function f:S->P(S), > and the set w = { x : x not in f(x) }. And the question as to what element maps to the completed set. This is a "largest natural" argument. > > Given that P(S) is determined by S, and that w is > determined by f, the proof really only depends on > two things. An ended unending set and a largest natural number. > > There are no strings. There is no largest element. > > > However, to whatever extent we have considered the set, say to N > > elements, we can always consider it to N+1, or 2^N elements. If this is the set > > of all naturals, for instance, then N in the set implies 2^N in the set. There > > is no end to the bijection. The proof assumes one. > > No. The proof does not assume an end to the bijection. > If you think it does, please point out where it makes > such an assumption. It asks for the largest natural when it asks for the natural that maps to the entire completed set. It may be masked somewhat, but that's precisely what's going on. The first element maps to the null set, the last to the entire set. Is there a last element in the set? No? Then I guess it cannot map to anything. > > Remember, the rest of us are talking about the entire set, > and the entire power set, not just an "extent" of it. Yes, and you are asking which element maps to the entire set, which is like asking which is the last element. Think about it. > > > So tell me, what rule of bijection did I break, and at what point does the > > mapping through the infinite binary strings break down. For lack of answers to > > thses questions, my bijection stands. > > The set of even naturals does not show up in your bijection. What are you talking about? I offered a bijection between *N and P(*N), remember? With the binary strings? (sigh) Of course you forget. Everyone wishes it would just go away. All the hand waving only makes the bijection mad. Calm down or you might get stung. > You even admitted it. You can only account for the evens > that are less than some N, but that is not all the evens. How did you get on the subject of the evens? Take a vitamin. > > Stephen > -- Smiles, Tony
From: Tony Orlow on 21 Oct 2005 13:59
Virgil said: > In article <MPG.1dc1b1d591e93a9a98a50e(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > David R Tribble said: > > > 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? > > > > > > > > Well, infinitely far down the list of course! > > And where in that list is {x in *N: x not in f(x)}? > At the bottom. -- Smiles, Tony |