infinity ...
in [Math]
|
Prev: math
Next: The proof of mass vector.
From: Virgil on 20 Oct 2005 18:52 In article <MPG.1dc1b5532d34307b98a50f(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Randy Poe said: > > > > 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} There is no *n in this 'listing', or any other, for which f(*n) = { x in *N: x not in f(x)}. > and > I accept the inductive proof that shows it is 2^n for set size n, finite or > infinite. My point is that a bijection can be formed, without apparently > breaking any rules of bijection. If this bijection is disallowed, I would > like > to know by which rule this occurs. What makes this bijection invalid? The fact that we can specify a member of P(*N) which is not the image of any member of *N.
From: Virgil on 20 Oct 2005 18:55 In article <MPG.1dc1b69de12df94698a510(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > stephen(a)nomail.com said: > > 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 > > > Notice above that no natural maps to a subset which contains it, so w > is all of N. Imagining any completed w leads to a contradiction, > since the natural that would map to it is always bigger than every > natural in that set. That's okay though, because for every natural, > there's a larger one. If x exists, 2^x exists. The sets are infinite, > so the bijection continues, even though there is a difference between > the values which are in the subsets and the values which denote the > subsets. None of this handwaving and doubletalk designates any member of *N which maps to {x in *N: x not in f(x)}. And without it, TO's "bijection " is not even a surjection.
From: Virgil on 20 Oct 2005 18:59 In article <MPG.1dc1b877f4ab567098a511(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Is it > impossible to biject an infinite ordered set with its power set? Nope For any set S, finite or infinite, and any function f:S -> P(S), the set {x in S: x not in f(s)} is not of form f(s) for any s in S.
From: David R Tribble on 20 Oct 2005 19:00 Tony Orlow: >> 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: >> 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: >> What do you want me to try, anyway, and infinite mapping, >> element-by-element? A bijection's a bijection, right? > David R Tribble: >> Yes, that would be nice. Please show us your bijection. > Tony Orlow: >> 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? David R Tribble: >> Where are the infinite subsets, such as the set of even numbers? > Tony Orlow: > Well, infinitely far down the list of course! > . > . > . > f(N/3) = 0:010101.....010101 = {0,2,4,6,8,....} > f(2N/3)= 0:101010.....101010 = {1,3,5,7,9,....} All of the sets you listed, finite and infinite, are composed of only finite naturals. Which means that they are all subsets of N. You're using N (and N/3, 2N/3, etc.), which is a member of *N. So all you're doing is showing a bijection between *N and P(N), which is what I said you were doing in the first place. That's easy, and is equivalent to a bijection between R and P(N). But we're asking for a bijection between *N and P(*N), or between N and P(N). You haven't adequately shown those yet. > Oh, now comes the list of primes (sigh) > Uh, um.... > > f(N/pi) = ....1010001010110 = {2,3,5,7,11,13,...} > > PS, I'm kidding about the N/pi part (I think). The infinite natural x in *N that maps f(x) to the infinite set of primes is: x = 2^2 + 2^3 + 2^5 + 2^7 + 2^11 + 2^13 + ... But like all the other subsets your f mapping defines, the set of primes contains only finite numbers, which means that it is one of the subsets of N. Do keep trying, though.
From: Virgil on 20 Oct 2005 19:01
In article <MPG.1dc1b877f4ab567098a511(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Is it > impossible to biject an infinite ordered set with its power set? Nope Wrong again, TO! At least everywhere except possibly in that wonderland of TOmatics. |