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From: Virgil on 27 Oct 2005 16:01 In article <MPG.1dcac0e5423aa36398a583(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > # O O O O O O O O O ... 1 > > > # # O O O O O O O O ... 2 > > > # # # O O O O O O O ... 3 > > > # # # # O O O O O O ... 4 > > > # # # # # O O O O O ... 5 > > > # # # # # # O O O O ... 6 > > > # # # # # # # O O O ... 7 > > > # # # # # # # # O O ... 8 > > > # # # # # # # # # O ... 9 > > > . . . . . . . . . . . > > > . . . . . . . . . . . > > > . . . . . . . . . > Every natural is represented by a horizontal string of #'s. The > vertical string of zeroes directly to the right of the last # in each > natural denotes the size of the set through that natural. With the > addition of each natural, both the horizontal number of #'s and the > verticle number of 0's are incremented in tandem, so this equality is > preserved. The slope of -1 shows that there is a 1- 1 correspondence > between element value and set size. This is a good graphical > illustration of what I have been trying to say regarding the > naturals. You only have an infinite number of them when you allow > infinite values for them. Albrecht is correct in this regard. The representations are equally valid if all 0's are deleted and each number is represented by a finite string of #'s. Then nothing infinite, except the Dedekind infiniteness of the set of finite naturals, is needed. That one can use infinite strings to represent finite numbers does not mean that one has to.
From: Virgil on 27 Oct 2005 16:18 In article <MPG.1dcaca55cb16499798a584(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > > If I am bijecting *N and P(*N), which was my original bijection, then *N > contains infinite elements with infinite bit strings. *N is not limited to > finite values. To keeps claiming that he can bijects *N to P(*N), but never demostrates the validity of that claim, merely reitierates it. As if repetition will make it eventually valid. > Yes, if P(S) has N members, then there must be log2(N) elements in S. Very > good. Keep in mind, I am by no means trying to say the power set is the same > size as the set, simply that there is a bijection. TO has things backwards. Sameness of "size", at least as TO defines it, implies bijectability but not conversely. TO cannot present any instance of a provable bijection between sets of unequal "size", but there are many bijections between sets of unequal "sizes". For example x -> 2*x bijects [0,1) to [0,2), even though he latter set is twice the size of the former. > You are not forced to conclude that this prevents a bijection. Yes we are! > The entire > infinite set would be the final subset enumerated, but the set does not end, > nor does the set of its subsets, so the bijection will never, ever get to > this > point. Then it is not a bijection. In order for a mapping to be a bijection, it must "get to" every point! Or perhaps any old mapping can be called a bijection in the strange world of TOmatics. Outside of TOmaics, a function from a set S to a set T is a bijection ONLY IF for every s in S, f(s) is a unique member of T, AND for every t in T, there is a unique s in S such that f(s) = t. TO's alleged "bijections" do not meet these criteria, so that, outside TOmatics, they are not bijectinos at all. > You used "TOmatics" fairly well. What if we apply it to the evens? > Whatever n you choose, you have 2n, but 2n is a natural too, so you have to > map > it to 4n, etc. If you were to assume any such "last element" TO is the only one making that false assumption of "last element".
From: Randy Poe on 27 Oct 2005 16:33 Tony Orlow wrote: > David R Tribble said: > > We are forced to conclude that there is no natural s that maps to > > *N, and that therefore your mapping scheme is not a bijection > > between *N and P(*N). > You are not forced to conclude that this prevents a bijection. The entire > infinite set would be the final subset enumerated, but the set does not end, > nor does the set of its subsets, so the bijection will never, ever get to this > point. You used "TOmatics" fairly well. What if we apply it to the evens? > Whatever n you choose, you have 2n, but 2n is a natural too, so you have to map > it to 4n, etc. The objection above is that there exists at least one element of P(*N) which is not equal to f(s) for any s in *N. That objection says nothing about "last". The same objection can not be raised about f:N->E, f(x) = 2x. Every element in E is mapped by some element in N. Every single one. Nobody but you says you have to consider "the last" in order to justify the claim that this statement is true for every single member of the set of evens. The surjection part of my simple proof of the bijectivity of f(x)=2x had this simple structure: Let y be any element of the codomain. Then there exists x such that f(x) = y. If you wanted to establish surjectivity of your map f:*N->P(*N) you need the same structure: Let y be any element of P(*N). Then (TO can show) there exists x in *N such that f(*N) = y. But of course that last part isn't possible. Because we know at least one example of a y in P(*N) such that x does NOT exist. Your claim that surjective proofs are impossible without identifying a last element is pure nonsense. Where is this "last element" in my now-TO-approved proof of the bijectivity of f(x) = 2x? - Randy
From: Virgil on 27 Oct 2005 16:33 In article <MPG.1dcacd0dcc9011f898a585(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > David R Tribble said: > > It would help if you listed the mappings for a few infinite > > naturals, i.e., show us f(k) for some infinite k, k+1, k+2, in *N. > > They are not present in your list above. > Okay, for N=1:000...000, the set mapped is {2^N}. For 1:000...001 > it's {0,2^N}. Here's the set of all multiples of 3: .....1001001001. > The entire set? ...1111. This still does not show that there is anything mapped to the set of all elements not in their image sets. Until TO can show that some element can be so mapped, his mapping does not meet the definition of a bijection, at least not outside TOmatics. > > > > Except for the tiny little fact that each of your infinite naturals > > is an infinite bitstring of finite powers of two, so they can only > > be mapped to infinite subsets of finite naturals. Please prove > > otherwise, if you can. > I'm sorry, but you make no sense. If the string is infinitely long, > then it has bits in infinite positions which represent infinite > powers of 2 and infinite values. Outside of TOmatics, where infinite means Dedekind infinite, all the bit positions in an infinite string are in finite positions, each corresponding to a finite natural from the infinite set of finite naturals. What goes on in TOmatic trolldom is of no interest to any non-trolls. > I have been through N=S^L with you, > and the only thing blocking you understanding is the bigus proof that > all naturals are finite. Outside to TO's trollhaven, they are! > There is little more I can do for you in that regard. TO giving help in mathemaics is like TO throwing an anchor to a swimmer to help him float. > > Orlow: > > > Like which, for instance? Keep in mind that my bits never end. > > > WHich other mappings do you want? What would satisfy you? Proof that any complete mapping from any set to its power set maps some element to the set of elements not in their images.
From: David R Tribble on 27 Oct 2005 19:31
Albrecht Storz wrote: >> My argumentation is very easy: >> Every nat. number represents a set. If you look at the first 100 nat. >> numbers, the 100th nat. number "100" represents the set {1, ... , 100}. >> As this holds for every nat. number, if there are infinite nat. numbers >> there must be a infiniteth nat. number representing this set. > Randy Poe wrote: >> No, there musn't. There is no logical basis on which to draw >> this conclusion from your premise. > Albrecht Storz wrote: > Yes, it must. You can derive from the peano axioms or settle an equal > definition: nat. numbers are this, what count elements of sets. > If there is a congregation which elements you can't count you have > either no size or no set. Or the set has a size that is not equal to a natural number. Consider the set of reals in the interval [0,1], that is, the set S = {x in R : 0 <= x <= 1}. The elements of this set cannot be enumerated by the naturals (which is why it is called an "uncountably infinite" set). But all sets have a size, so this set must have a size that is not a natural number. It is meaningless (and just plain false) to say this set "has no size" or "is not a set". |