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From: Daryl McCullough on 18 Oct 2005 21:16 albstorz(a)gmx.de says... >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'm not sure exactly what your list is supposed to mean. I think you're trying to say that row 2 = all sets with 1 element row 4 = all sets with 2 elements row 6 = all sets with 3 elements etc. row 1 = the set containing all elements row 3 = the sets missing 1 element row 5 = the sets missing 2 elements etc. If that's what you mean, then you have left out the set of all even numbers. You've left out the set of all odd numbers. You've left out the set of all prime numbers. You've left out the set of all powers of two. You've left out a whole bunch of sets. -- Daryl McCullough Ithaca, NY
From: Virgil on 18 Oct 2005 21:39 In article <MPG.1dbf2c34a59f67bc98a4d9(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > stephen(a)nomail.com said: > > > The proof that there does not exist a bijection between a set and > > its power set is quite short. > Then it shouldn't take too much looking to see where it goes > wrong.... > > > > Let f be a function from S to P(S). > Our proposed mapping bijection.... > > > > Define the set w as follows: > > > > w= { x : x in S and x is not in f(x) } > So, w is the set of all elements which are not members of the subsets > which they map to through f(x).... > > > > Clearly w is a subset of S, and so w is an element of P(S). > Clearly...but properly? > > > > We now show that w is not in the image of f. That is, there does > > not exist a y such that f(y)=w. > So, there can be no y such that it maps to the subset of all elements > which do not map to subsets containing themselves? We'll see..... > > > > Suppose such a y exists. If such a y exists, it must either be an > > element of w, or not. > One or the other. I'll accept the excluded middle.... > > > > If y is an element of w, then y is in f(y), which means it is not > > an element of w. > If y is in w, this means y is a member of the set of elements which > map to subsets which do not contain themselves. This means that y is > in S but not in f (y). So, indeed, it IS a member of w. There is no > reason why both x and y cannot map to subsets which do not contain > themselves. > > If y is a member of w, then all you can say is that y does not map to > any subset which contains itself. Y does not map to w. Neither does > x. > > > > If y is not an element of w, then y is not in f(y), which means it > > is an element of w. > If y is not a member of w, then y maps to a subset which contains y > as a member, and y IS in f(y). Is this possible? We'll see what you > think..... > > > > > These are both contradictions. So y cannot be an element of w, and > > it cannot not be an element of w. So y cannot exist. > Neither of those possibilities causes a contradiction. You are > getting confused with your double negatives. If w is the set of all > elements which do not map to subsets containing themselves, then > being a member of w means simply that y does not map to a subset > containing y, which is perfectly possible. Not being a member of w > means that an element maps to a subset which DOES contain itself. > Where is the contradiction? What x has f(x) = w? If there is no such x, then f is not a surjection. > > > > So there is at least one element in P(S) which is not in the image > > of f, so f is not an onto function, and it is not a bijection. > Sorry, not so. Sorry, but it is only "not so" in TOmatics, it is so everywhere else. > > > > What is wrong with this proof in your opinion? > You confused yourself with double negatives. If I misinterpreted any > of what you said, please clarify. On the face of it, you ain't got no > proof. > > Let's put this in terms of the bijection between *N and P(*N). Given > the common binary string representation of the binary naturals and > the subset specifications, and starting from 0, let's see what's in > w. TO assumes what is not in evidence here, that there is a bijection between *N and P(*N) through his non-existent double-ended endless sequences. Having assumed it, TO has no difficulty in proving it follows from its assumption. Psuedo-proof deleted.
From: Virgil on 18 Oct 2005 21:42 In article <MPG.1dbf2de38cf6ec898a4da(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: > > > No, it was specifically a bijection between two sets of infinite binary > > > strings representing, on the one hand, the whole numbers in *N starting > > > from 0, both finite and infinite, in normal binary format, and on the > > > other > > > hand, the specification of each subset of whole numbers in *N, where each > > > bit which, in the binary number, represents 2^n denotes membership of n > > > in > > > the subset. This is a bijection between the whole numbers in *N and > > > P(*N), > > > using an intermediate bijection with a common set of infinite binary > > > strings. > > > > But that's an incomplete mapping, because there are not enough infinite > > binary strings in *N to enumerate all of the subsets of *N. Try it, > > if you don't believe me. > > > > > Not enough infinite binary strings? Precisely.
From: albstorz on 18 Oct 2005 21:43 David R Tribble wrote: > 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". It's a usual visualisation of Cantors first diagonal proof. You can go through diagonally e.g. N, {1}, {2}, N/{1}, {1,2}, N/{2}, {3}, {4}, N/{3}, {1,3}, ... It's a good way to understand, that the most subsets of the powerset are this, which are infinit and and also their counter parts. And at the end, are unconstructable. But this is far away from the intend of this thread. Exists the powerset? Exists the sum of all naturals? The definition of infinity is wrong. I had shown an easy understandable argument. Regards AS
From: Virgil on 18 Oct 2005 21:47
In article <MPG.1dbf2f2d4e8547598a4db(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > David R Tribble said: > > 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). > The only reason you cannot show a bijection between those two is that > you claim to have an infinite number of finite naturals, so you would > have infinite bit strings in P(N) and only finite bit strings in N. > But, of course, this is an artificial and incorrect interpretation of > the situation, as I have repeatedly tried to convey. But that peculiar situation only holds in the wild woolly world of TOmatics where everything is both provable and disprovable, and definitely does not hold in the standard world where the negaation of any provable statement is necessarily unprovable. |