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From: Virgil on 21 Oct 2005 23:40 In article <MPG.1dc2f31edfe14f0398a52a(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> 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: > > >> > Look, it's certainly not my position that the pwoer set is the same > > >> > sie as the > > >> > set. It's clearly not. I just see a bijection between them, which only > > >> > bolsters > > >> > my argument that bijection alone is not sufficient to equate the sizes > > >> > of two > > >> > sets. > > >> > > >> > When it comes to the evens (let's start with 0), the value > > >> > 0:010.......1010101 > > >> > represents such a subset, and is essentially binary N/3. > > >> > > >> So which number does that first 1 represent? > > > > > log2(N-1). Perhaps we should say we have 2^N subsets and strings of N > > > digits. > > > That's what I should have said. Then that number is 2^N/3, and the first > > > 1 in > > > it represents N-2, since the bits are numbered from 0 through N-1. > > > > So this is not the set of all even numbers then, because > > according to you there are even numbers larger than log2(N-1). > This is the imagined completed set. Outside of TOmatica, one works only with complete sets, as incomplete things are not sets. > The number of digits is variable according to what infinity you are > working in. The 'number' of digit positions in an endless sequence of digits is constant , and is equal to the 'number' of finite naturals, everywhere except in TOmatica. > You have to choose a unit to measure. TO might but nobody else has to.Do > > So this is not the set of all even numbers and you were > > wrong when you claimed it was. Why did you claim that > > that string represented all the even numbers when you in > > fact knew that it did not? > You know my approach at this point. To retreat into the ambiguity of TOmaitics! > > > > You claimed > > that 0:010......10101 represents the set of all even numbers. > > You now are claiming that it only represents some of the even numbers. > No, I am claiming it is a representation of the set of even numbers. How does it differ from 0:010......101010, as "a representation of the set of even numbers? > > > > Your arguments would be a lot more consistent and sensible > > if you just dropped the whole notion of "infinity". It > > is clear that you do not actually believe in infinite sets. > > For you all sets must have a first and last element, and > > all your arguments require all sets to be finite. > No, all measurements require two endpoints. The endpoint is variable, and can > be finite or infinite. TO obviously cannot distinguish between counting and measuring. While measuring may require units, one is not never measuring how many, only how much. Counting, which determines only how many, and not how much, is quite different. TO is trying to measure what he should be counting.
From: Virgil on 22 Oct 2005 00:09 In article <MPG.1dc2f545d4ef244198a52b(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > I am referring also to my bijection, to which no one has raised any > serious objection except on pronciple. I and others have raised serious objections, some of us many times, to which TO has offered no serious counters. > > > > The proof really only makes use of four things. A set S, its power > > set P(S), a function f:S->P(S), and the set w = { x : x not in f(x) > > } .. > And the question as to what element maps to the completed set. As nothing but completed sets are involved. Outside TOmatica there are only completed sets, anything uncompleted is not a set in the outside world. > This is a "largest natural" argument. Why does TO keep insisting that any such thing exists, when he knows it does not? > > > > Given that P(S) is determined by S, and that w is determined by f, > > the proof really only depends on two things. > An ended unending set and a largest natural number. Those things only exist in TOmatica. Why TO ever comes out of his ambiguous land of TOmatica to face the harsh winds of reality is a mystery. > > No. The proof does not assume an end to the bijection. If you > > think it does, please point out where it makes such an assumption. > It asks for the largest natural when it asks for the natural that > maps to the entire completed set. But it does not do that. TO is the only one who insists that it maps to the entire completed set. WE only say what it does NOT map to, namely the set {x in S: x not in f(x)} which can NEVER, NEVER be the "completed set". > It may be masked somewhat, but that's precisely what's going on. The > first element maps to the null set There need not be a first, as S need not be ordered at all, much less ordered so as to have any first. S could be the naturally ordered set of ratinals for instance whichdoes not have anything like a first. , the last to the entire set. What is the "last" rational? Nobody but TO needs S even to be ordered at all, much less that it be ordered so as to have a first and a last. TO is clearly off in his cloud cuckoo land of TOmatica again. > Is there a last element in the set? No? Then I guess it cannot map to > anything. A circle has no last member but can map into the plane. What sort of stuff is TO getting off on today? > > > > Remember, the rest of us are talking about the entire set, and the > > entire power set, not just an "extent" of it. > Yes, and you are asking which element maps to the entire set False, we are asking which member of the entire set S maps to {x in S: x to in f(x)}. Answer: none! > I offered a bijection between *N and P(*N), remember? TO has said there is such a thing, but has not given anything more subtantial than that claim of existence for us to go on. TO must give a specific rule, f: *N -> P(*N), telling how each member of *N determines a subset of *N in such detail as to allow a disinterested person to determine whether there is any *n in *N for which f(*n) = {x in *N: x not in f(x)}. Until he has done that, TO is just handwaving. > > > You even admitted it. You can only account for the evens that are > > less than some N, but that is not all the evens. > How did you get on the subject of the evens? How does TO avoid it?
From: Virgil on 22 Oct 2005 00:16 In article <MPG.1dc2f69a940a440998a52c(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dc1b1d591e93a9a98a50e(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: > > > > >> What do you want me to try, anyway, and infinite mapping, > > > > >> element-by-element? A bijection's a bijection, right? > > > > > > > > > > > > > David R Tribble said: > > > > >> Yes, that would be nice. Please show us your bijection. > > > > > > > > Tony Orlow wrote: > > > > > 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? > > > > > > > > Where are the infinite subsets, such as the set of even numbers? > > > > > > > > > > > Well, infinitely far down the list of course! > > > > And where in that list is {x in *N: x not in f(x)}? > > > At the bottom. If there is *n in*N such that f(*n) = {x in *N: x not in f(x)}, is *n in {x in *N: x not in f(x)} or is *n not in {x in *N: x not in f(x)}? In either case, it can only happen in the looking glass land of TOmatica where things can be sinmultaneoulsy true and false.
From: Virgil on 22 Oct 2005 00:22 In article <MPG.1dc2f7d4729e142f98a52d(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > 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. > > > So, it is required that we determine the very last element in an infinite > bijection? Who says anything about "last"? Only TO! Order of elements is not relevant in finding an element of *N which f will map to {x in *N: x not in f(x)}. TO's continuing obsession with such irrelevances to the issue as order can only be to support his attempts to avoid facing the issue squarely and admitting his error. Why must I name the end of the set in order for the bijection to > be > valid? You cannot name anything except some conceptual end of the unending > set > as a point where the bijection breaks down. I swear this theory just seems > more > and more insane to me. How do you tolerate it? Ugh! Please line up by the > basketball courts for your head-whacking. Maybe that'll work. At least it'll > be > fun.
From: Virgil on 22 Oct 2005 00:36
In article <MPG.1dc2f981910d288e98a52e(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > David R Tribble said: > > > 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. > Incorrect. See the digits to the left of the ellipses? Those are in > infinite positions, representing infinite evens and odds. So that 0:010101.....101010 and 0:101010.....010101 must both represent sets of half evens and half odds. > No, it's P(*N). The sets include elements of infinite value. If TO is saying that each 1 bit in x:xxxxx.....xxxxxx represents the presents of a member of *N in the set represented by that string, and each 0 represents the absence, then each 1 member set can be represented by a single 1 bit, with all others zero, so that it member can also be represented with a single 1 bit as well, and all the zeros on one side or the other of that 1 digit are extraneous. Then TO is saying that every member of *N can be represented by a single 1 preceded (or followed) by an appropriate string of 0's. s this represents the set whose only element is that number, it equally represents that number! > > > > 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. > It's done. Not outside TOmatica. > If there are no infinite primes, then the subset would correspond to > a finite value. Not all subsets have to have infinite values. I have > only listed finite numbers here, but do take note of the ellipses. > Are there infinite primes? If not, then there arent an infinite > number of them. There are infinitely many finite primes outside of TOmatica, where the sun still shines and infinite still means Dedekind infinite. |