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From: stephen on 19 Oct 2005 15:53 stephen(a)nomail.com wrote: > 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. I am revising my guess. You will claim that F(N-1) = { 0, 1, 2, 3, ....}. Afterall f(N-1) = ..1111111 = { 0, 1, 2, 3, ....} Presumably N-1 is in f(N-1), so N-1 is not in w, and f(N-1) does not equal w. It follows that N-2, N-3, etc are not in w also. So it looks like w might be the set of all finite natural numbers. I am not sure as I do not know if there are infinite numbers that are not expressible as N-k where k is some finite number. I would think there would have to be, but you never talk about them. Anyway, for what y does f(y) = w? And is y in w or not? Stephen
From: Virgil on 19 Oct 2005 15:58 In article <MPG.1dc045529466f7c998a4e9(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > 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}? Does TO now claim that different ssets cannot have the same size? Nothing was said about the sizes of sets not having exactly the same members, only that they are not the same sets. Pay attention, TO. > Finally we reach agreement. Welcome > to reality, David. David is still safe outside the twilight zone of TOmatics. > > > The proof just constructs a set which differs by the membership of at > > least one particular value with every target set in the assumedly > > complete mapping of set to powerset. > And we all know, that one element makes a world of difference! Good job, > David! It is a good enough job to show how wrong TO is, despite TO's pretending otherwise.
From: Tony Orlow on 19 Oct 2005 16:33 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. If you are assuming you have the complete set of naturals, that you have identified the last, and can therefore identify the element that maps to the entire set, then you indeed run into a contradiction. However, both the infinite set of naturals and the infinite power set go on forever, so you never run out of naturals to map to subsets, nor subsets to map to naturals. Despite the fact that, within any range up to S, you cannot map every subset to an element within that range, the lack of a largest element makes it so there DOES exist an element that maps to all n through S, but it is more than S. This is a prime example of where the value range matters in the bijection. > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony
From: Tony Orlow on 19 Oct 2005 16:38 William Hughes said: > > albstorz(a)gmx.de wrote: > > David Kastrup wrote: > > > 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. > > > > > > That's all. Finiteness or infiniteness does not even play into it. > > > The proof just constructs a set which differs by the membership of at > > > least one particular value with every target set in the assumedly > > > complete mapping of set to powerset. > > > > > > -- > > > David Kastrup, Kriemhildstr. 15, 44793 Bochum > > > > > > > > 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. > > Perhaps you are thinking about proving that n<2^n for all natural > numbers n? This can be done by induction, and indeed this proof > doen not extend to infinite numbers. However, this is not the > proof usually used to show that there is no bijection between a > set and its powerset. The standard proof does not use induction, > nor does it require finiteness. I don't understand why an inductive proof of an equality relationship does not extend to infinite numbers. Only relationships which disappear at n=oo are limited to the finite case, as far as I can see. I think your statement is the result of the definition based on natural numbers, and the belief that all naturals are finite, but I just don't see why induction in general should be limited to finite iterations. It seems to me that the inductive proof of that equality works without flaw. > > -William Hughes > > > > > > Regards > > AS > > -- Smiles, Tony
From: William Hughes on 19 Oct 2005 16:53
Tony Orlow wrote: > William Hughes said: > > > > albstorz(a)gmx.de wrote: > > > David Kastrup wrote: > > > > 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. > > > > > > > > That's all. Finiteness or infiniteness does not even play into it. > > > > The proof just constructs a set which differs by the membership of at > > > > least one particular value with every target set in the assumedly > > > > complete mapping of set to powerset. > > > > > > > > -- > > > > David Kastrup, Kriemhildstr. 15, 44793 Bochum > > > > > > > > > > > > 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. > > > > Perhaps you are thinking about proving that n<2^n for all natural > > numbers n? This can be done by induction, and indeed this proof > > doen not extend to infinite numbers. However, this is not the > > proof usually used to show that there is no bijection between a > > set and its powerset. The standard proof does not use induction, > > nor does it require finiteness. > I don't understand why an inductive proof of an equality relationship does not > extend to infinite numbers. Because we are discussing standard mathematics here and there are no infinite natural numbers (only infinite ordinals and cardinals which are not the same thing). Yes it is true that the semi-mythical TO naturals have infinite members and that induction sometimes holds, but that is irrelevent here. >Only relationships which disappear at n=oo are > limited to the finite case, as far as I can see. I think your statement is the > result of the definition based on natural numbers, and the belief that all > naturals are finite, but I just don't see why induction in general should be > limited to finite iterations. It seems to me that the inductive proof of that > equality works without flaw. Well, as you have not even defined 2^N for infinite TO naturals, you have some work to do. - William Hughes |