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From: imaginatorium on 21 Oct 2005 14:44 Tony Orlow wrote: > 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: <skip proof - in part mathematics> Tony: > I am referring also to my bijection, to which no one has raised any serious > objection except on pronciple. Indeed. I think it's fair to say that if (counterfactually) mathematics were postmodern stamp-collecting, or art appreciation, there would be bound to be at least one person in the class who said "I really like your bijection - it's rather cute". And over the objections of those who pointed out that it isn't a bijection, we could all say "Well done, Tony - you've shown some creativity, some determination not to be rulebound, to break the mould, cast the die, mix the metaphor. What is a metaphor, anyway? I've got one, and I'd love to use it." OK, we'll meet next week, and look at Phil's worms in a new light... Brian Chandler http://imaginatorium.org
From: Virgil on 21 Oct 2005 14:46 In article <MPG.1dc2b65218313f9398a521(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Randy Poe said: > > > > Tony Orlow wrote: > > > stephen(a)nomail.com said: > > > > What is element "2^S-1"? S is a set. Element "2^S-1" means > > > > nothing to me. > > > If set S has S elements, from number 0 through S-1, then the element > > > which maps > > > to the entire set is a string of 1's which is S long, which corresponds > > > to a > > > value of 2^S-1, which would be element number 2^S (not 2^S-1, sorry - > > > damned > > > error of 1!). > > > > So you would say that no element x is in f(x), correct? > Not given any particular x, no. > > > > Therefore for this mapping, the set > > > > w = {x in S: x not in f(x)} > > > > contains every element of S, right? That is, w = S. > Yes. > > > > Now let's consider every element y of S. Obviously y is > > in w, since w = S. But then y is not in f(y), since that's > > what it means for y to be in w. Then f(y) can't be S. > That's true, and this is an interesting argument. You certainly cannot > identify > any natural which maps to the entire set of naturals. That would require > identifying some last element and and establishing a particular number of > bits > N for the subset representation, so as to say w maps to (2^N)-1. Of course, > since w only includes N elements, it doesn't include y, and y is not in w, > within the original range of N. This is why one needs infinite sets for their > bijections. Since neither set ends, there is always a corresponding element > for > each element, even though the numbers in the subsets are always smaller than > the numbers representing the subsets. That doesn't matter. There is always > another one larger, so you never run out of corresponding elements. At least, > this is the way your bijections normally work. > > What rule am I breaking with this bijection? The rule that for any function, f, from any set, S, to its power set, P(S), there is no member, s, of S such that f(s) = {x in S:x not in f(x)} > > > > > So this means that no matter what y you pick, f(y) can't > > be S. So S is not mapped by your "bijection". > Not within the range of S, no, but over the infinite range, yes. So TO is saying that there are elements OUTSIDE of the domain of the function that map onto all those subsets that are not images of elements of the domain. But that does not establish anything like a bijection from the domain set to anything. > > > > What's wrong with your "proof"? Simple. You have f(z) = S, > > where z is NOT a member of your original set S. You don't > > have f(y) = S for any y IN your set S. > That is true. Then TO does not have his claimed bijection fron S to anything. <Extensive TO doubletalk snipped>
From: imaginatorium on 21 Oct 2005 14:52 imaginator...(a)despammed.com wrote: <snip to proof> > Consider an arbitrary set A. [ ] (Thus any entity in the universe > [proper class] of all entities either is or is not a member of A [ ].) > > Consider the power set of A. [ ] (The set of all subsets of A, > including A itself; thus for any particular one of these subsets B, any > element of A either is or is not a member of B [ ].) > > Now consider any mapping f from A into P(A) [ ]. (At least one such > mapping exists [ ]: the canonical map that takes any element x e A [e > means 'element of' OK?] to the subset {x} e P(A). So no worries about > existence.) > > We want to show that whatever mapping f is chosen, there will be at > least one element q of P(A) (q is a subset of A, right?) which the > mapping f does not map anything onto. [ ] > > Consider an arbitrary element x e A [ ], and suppose f(x) = y e P(A). [ > ] > Well, y is a subset of A, so x either is or is not an element of y. [ ] > Thus we can divide the elements of A into two classes: [ ] > CLASS I: those which map to a subset of which they are a member [ ] > CLASS II: those which map to a subset of which they are not a member [ > ] > > Call w the subset of A which includes exactly those elements in CLASS > II. [ ] > > [-----end prologue-----] > > Now I assert that the particular mapping f we are considering at the > moment does not map any element of A to the subset w in P(A). [ ] > > Every element p inside w is mapped to a subset that does not include p, > so it cannot be mapped to w. [ ] > Every element p' outside w is mapped to a subset that _does_ include > p', which thus includes an element outside w, and thus p' cannot be > mapped to w. [ ] > > Therefore the mapping f is not a surjection. (We called that "onto" > when I was doing this stuff.) > > Thus there is no bijection. Forgot to add in: not only does this proof not mention "finite" or "infinite", it doesn't mention "empty" or "non-empty" either. So it works for the empty set too (how many 'bits' is that, I wonder)... The empty set E has no elements and one subset. Therefore the only mapping from E -> P(E) is the null mapping (doesn't map anything to anything). This does not therefore map anything to the only element of P(E) = {{}}, which is the empty set. Brian Chandler http://imaginatorium.org
From: David R Tribble on 21 Oct 2005 14:58 Albrecht Storz wrote: >> So, if there is an infinite set there is an infinite number. > David R Tribble wrote: >> Do you mean that an infinite set (or natural numbers) must contain an >> infinite number as a member (which is false)? Or do you mean that >> the size of an infinite set is represented by an infinite number >> (which is partially true)? > Albrecht Storz wrote: > Not only partially. If there is no infinite number there is no infinite > set. > If sets consist of discret, distinguishable, individual elements, and > sets are definde like this, the natural numbers are just representative > for the elements and also for the sets. > {1,2,3} means a set with element #1, element #2, element #3, this is > the ordinal aspect of numbers. The set with cardinality 3 is just this > set {1,2,3}, and at the same time it represents all sets with 3 > elements. That's the open secret of the numbers. > > ordinal = cardinal = natural Only for finite sets or numbers. For infinite sets we must use infinite ordinals and infinite cardinals. But these ordinals and cardinals do not equate to any naturals because there are no infinite natural numbers. What is the number for set N = {0,1,2,3,...}? If it's w, then how can w be a member of N? If it's w+1, then how can w+1 be a member of N? If it's w-1 (whatever that means), then how can w-1 be a member of N?
From: Tony Orlow on 21 Oct 2005 15:03
David R Tribble said: > Stephen said: > >> Do you consider the following a bijection from S={a,b,c} to its > >> power set? > >> [...] > >> > >> 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}). > > > > Tony Orlow wrote: > > 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. It is true > > 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. > > Again with the "last element" red herring. Just let it go, Tony. > > Consider the set of naturals > N = {0, 1, 2, 3, ...} > which obviously has no last element. > > I can map an infinite binary natural to it, so that f(x) = N: > x = 2^0 + 2^1 + 2^2 + 2^3 + ... Yes, that's the same one can do with *N. ....11111 maps to the entire set. > > So I've managed to identify an x such that it mapped to the entire > (infinite) set N. Granted, x must be an infinite natural, but it > still works. And all without resorting to any "last" element in > either the set or in the series. Actually, since I consider the set of finite naturals to be finite, that number woudl have a finite number of bits and a finite value, but anyway... > > > Tony Orlow wrote: > > 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. > > If there is no end to the bijection, then the bijection is not > complete, and it therefore not a bijection. If we can't account for > all of the mappings of numbers to subsets, then we can't say the > mapping is a bijection. Oh. What is the end of the mapping between naturals and evens? What is the even corresponding to the last natural, or for that matter, to any of the last half of the naturals? > > None of the standard proofs assume a last element. Just let it go. It assumes a completed set with a completed string and a natural that equals that string, corresponding to the completed set. For any completion you consider, than number is beyond it. So, don't assume a completion to the set. > > -- Smiles, Tony |