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From: albstorz on 30 Oct 2005 16:42 David R Tribble wrote: > David R Tribble said: > >> [Alrecht] are claiming that one of the members of the infinite set is > >> equal to the size of the set. This is not obvious to us because it > >> contradicts proven set theory. It is your responsibility to prove > >> that your claim is true, and that standard set theory is wrong. > >> Saying that your claim is "obvious" is not a proof. > > > > Tony Orlow wrote: > >> His claim is obvious based on the diagram he offered, which graphically > >> shows the element values along one side of a growing square, and the > >> element count along the other side. They are obviously, graphically, > >> inductively, always equal. One is not infinite while the other is finite. > >> It's impossible. > > > > Here's his second diagram: > > > >> # 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. > > If that's the case, at what point do the naturals on the right side > stop being finite and start being infinite? > > All I see is an infinite list of rows, where each row is labeled by > a finite natural k and contains k leading # marks and an infinite > number of trailing O marks. > > >From the other angle, I see an infinite list of columns, where each > column contains a finite number of O marks on top and an infinite > number of # marks on the bottom. > > > But why can't there be an infinite number of finite naturals? > Do you have some kind of proof of this? You have never offered one. > You've offered plenty of statements about "unidentifiable largest > naturals" and "set ranges" and "unit infinities", but you've never > actually pinpointed how you get from the finites to the infinites. Think about the numbers which follows in horizontal direction as sets of Os as natural numbers and the numbers in vertical direction which are sets of #s as cardinal numbers. Or think of the example with the infinite triangle http://groups.google.com/group/sci.math/msg/44426a670822d8c2?dmode=source it might be clearer to you since it is very impressive. I understand, it could not be easy to end indoctrinated thinking, and it is not done in one day. Regards AS
From: Tony Orlow on 31 Oct 2005 09:40 Virgil said: > In article <MPG.1dc9588314461aa298a572(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Virgil said: > > > > If it did suffice, TO would be able to give the argument that is > > > mapped into the set of all elements not mapped into their images. > > > The set of all elements not mapped to sets which contain them is the > > set of all elements, since none map to sets which contain them. > > A perfectly legitimate mapping from any set, S, to its power set, P(S), > is the function that maps each element to the singleton set which > contains it, f(x) = {x}. For this function, {x in S: x not in f(x)} is > the empty set, even though TO in his perpetual state of confusion claims > it must be the whole of P(S). Are you able to stick to the bijection I offered? Distraction and irrelevance are not valid modes of argument. > > Since the composition of {x in S: x not in f(x)} depends critically on > the nature of the function f, all one can say about such sets, in > general, is that for each f from any set S to its power set P(S), there > {x in S:x not in f(x)} exists as a subset of S and member of P(S). In the bijection I offered, the natural bijection between an ordered set and its power set, every element maps to a set which contains a greatest element no greater than log2 (order-wise) of that number. > > > The natural which maps > > to it is represented by an unending string of bits. > > According to TO's description of "TOnaturals", they are all infinite > strings of bits. Yes, but not all have infinite numbers of significant bits. This has all bits equal to 1, with no end. Where does it fail? > > > > > This is the number which maps to the entire set. :D > > But why does TO insist that the set of naturals that map to sets not > containing themselves contains all naturals, How does he know that no > natural maps to any set containing itself? Such a wild claim requires > proof, or al least some sort of evidence beyond the mere claim itself. I already explained before, and now above. The singleton {x} maps to 2^x. it will be 2^(x+1)-2^x elements before we get the next element, x+1, as a singleton set. It should be obvious to you that no element can ever be a member of the set to which it maps. I gave a listing to illustrate this. Demanding rigorous proof of the obvious is a waste of time. With your experience, i am sure you can whip up a proof of this in five minutes. > > If TO can prove this, there is a lot about his mapping from *N to P(*N) > that he is carefully not telling us. Like what? > > > > > > > > > > > Your list shows an infinite number of finite naturals mapping > > > > > to an equally infinite set of finite subsets. That's easy. > > > > > > > > There is no infinite set of finite naturals, as Albrecht has > > > > tried valiantly to illustrate. > > > > > > The set of finite naturals is Dedekind infinite, which is quite > > > infinite enough outside TOmatics. > > > Sufficiently imponderable? Good enough. > > > > > > > > > > > > But it's not a bijection between *N and P(*N). It's not a > > > > > bijection between *N and P(N), or even between N and P(N). It > > > > > is a bijection between N and some of the members of P(N). > > > > > Obviously that's not good enough. > > > > > > > That's not what it is. I have declared it to be between *N and > > > > P(*N). > > > > > > To declare something does not make it so, at least outside of > > > TOmatics. > > > It's my bijection, after all. > > > While it is undoubtedly yours, whether it is a bijectin is a matter of > contention. > > > I can declare it to be whatever I wish. > > Such a declaration is invalid outside TOmatics without evidence to back > it up, and so far, the evidence all runs counter to that claim. Like what? > > > Go ahead and rain on my parade, but don't move the police barriers. > > > > > > Outside of TOmatics, declarations require proofs before they need > > > be accepted. > > > > Not when you are defining a bijection. > > To declare that something is a bijection is to claim it has certain > properties. If it does not, in fact, have those properties, then no > declaration will make it into what it is not. > > Unless TO has a function from the complete, entire, finished *N to the > complete, entire, finished P(*N), he does not have a relevant function > at all. You tell me how *N finishes, and I'll give you the natural which maps to it. Fair enough? "No largest natural! (gong) Huyah huyah Ommmm....." and the Catorians begin their dance again. > > And unless such a function,f:*N -> P(*N), once defined and "complete", > can be shown to map some member of *N to {x in *N: x not in f(x)}, he > does not have what can be called a bijdection outside TOmatics, > regardless of what TO chooses to call it inside TOmatics. Except that there is no point at which the bijection fails, since both *N and P (*N) are unending. > > > > > It includes such numbers in *N as 0:01.....0101 for the evens and > > > > 0:1010...1010 for the odds. You have no reason to think this set > > > > is limited to finite values. > > > And we have no reason to think it a bijection until TO can find > > > some > > > member of *N which maps to the set of all members of *N not in the > > > set they map to. > > .........................1111111111111111111111111111111111111111!!!!! > > !!!!!!!! ! > > > > > > > > > > > Even if we were to grant you that the list somehow includes > > > > > infinite naturals in *N mapping to subsets, you would still be > > > > > missing a lot of subsets in P(*N). > > > > > > > Like which, for instance? > > > > > > The set of all members of *N not in the set they map to. > > > That's the entire set. > > TO keeps saying that, but never gives any reason why, even though > examples have been given showing that it can even be the empty set. See above, keep in mind my general bijection, and read, for god's sake. No element is in the set to which it maps. Therefore, the set of all elements not in the subsets to which they map includes every element in the entire unending set. > > Considering TO's track record with unsupported statements, and > considering that there are examples showing that TO's claim need not be > true, we must regard any such TO claim as false, at least outside > TOmatics, until proven otherwise. Considering that Virgil can't even stay on topic, I don't know why I even bother responding to his posts. > > > > > > > > > > > > > So fill in the blanks, and show us those other mappings. > > > > > Otherwise your "bijection" is incomplete and is not, in fact, a > > > > > bijection between *N and P(*N). > > > > > > > WHich other mappings do you want? What would satisfy you? > > > > > > A mapping which maps some member of *N to the set of all members of > > > *N not in the set they map to. > > ...1111111111111111111111111111111111111111111111111111111111111111111 > > 1, with oo^oo^oo^oo^oo......... 1's. Now blow out the candles, > > Virgil. > > How are we to know whether the above abomination does or does not map to > the set of all members of *N not in the set they map to ? It's got a 1 for every element in the set. I already explained (multiple times, now) that my bijection does not have any elements that are included in the sets to which they map, so the number you are asking for is the natural which maps to the entire unending set. Here is your unending string of 1's, 1 for every element in the set. Tada! > > TO's claims, being untrustworthy, require supporting evidence if they > are to be acceptable outside of TOmatics. And what happens inside is > irrelevant to sane mathematics anyway. Sanity, like evil, is often relative. > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 09:42 David R Tribble said: > David R Tribble wrote: > >> My set is: > >> S = {0, 2^0, 2^2^0, 2^2^2^0, 2^2^2^2^0, ...} > >> S = {0, 1, 2, 4, 16, 65536, ...} > > > > Tony Orlow wrote: > > This actually is interesting stuff, called tetrations. [...] > > As I understand it, these kinds of formulas are not well > > understood yet. Kind of hard to figure an inverse to this function. > > The elements of S are: > s(0) = 0 > s(i) = 2^s(i-1) > > Recursion, no less. > > As a guess, the inverse would probably be something like: > si(0) = 0 > si(i) = si(log2(i)) + 1 > > Oh yes, it's certainly possible to describe the relationship recursively. I meant as an algebraic formula. I should have been specific. -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 09:52 Virgil said: > In article <1130352741.718980.94490(a)f14g2000cwb.googlegroups.com>, > "David R Tribble" <david(a)tribble.com> wrote: > > > David R Tribble said: > > >> So I'm giving you set S, which obviously does not contain any > > >> infinite numbers. So by your rule, the set is finite, right? > > > > > > > Tony Orlow wrote: > > > If it doesn't contain any infinite members, it's not infinite. Those terms > > > differ by more than a constant finite amount, but rather a rapidly growing > > > amount greater than 1. There is no way you have an infinite number of them > > > without achieving infinite values within the set. > > > > Yes, you and Albrecht keep saying that repeatedly. Please demonstrate > > why it must be so, because it's not. > > When TO says "infinite" he intends that to exclude such Dedekind > infinite sets as the set of finite naturals. > > TO has a very Humpty Dumpty attitude towards terms like that, they mean > for him exactly what HE intends them to mean, regardless of what they > might mean to everyone else. yes, I regard the quantitative aspects of infinity to be as important as the set-related aspects, and am not one-sided about this. Boo hoo. > > But even with that meaning of infinite, TO is wrong, at least outside of > his twilight zone of TOmatics, since the set of reals between 0 and 1 is > infinite even by his peculiar standards but has no infinite elements. Again I must repeat myself to you, Virgil. Take a look at what I wrote above. In the set under discussion (was it the set of tetrations of 2?), the difference between successive members is not only at least a constant finite value, but a rapidly growing value. I have said innumerable times, and you still can't get this straight, if there is a finite (non-zero) lower bound on the difference between members, then a finite value range can only contain a finite number of elements, and an infinite value range is required for an infinite set of such elements. If there is NO finite (non-zero) lower bound on the differences between elements, and there is at least one condensation point, or point of infinite density, then there can exist an infinite number of elements in a finite range. Otherwise, it is impossible to ahev an infinite number of elements ina finite range of values. > > Lest TO object, TO's version of infinitesimals puts equal spacings > between "consecutive (in TOmaitcs) reals just as there is for naturals > outside TOmatics, so that argument won't wash. Those equal spacings are infinitesimal, not finite. Infinitesimals are defined, more or less, as the result of dividing a finite by an infinite, so there can be an infinite number of them in finite range. > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 10:02
David R Tribble said: > David Kastrup said: > >> This does not make aleph_0 a member of the set of naturals. > > > > Tony Orlow wrote: > > A decree does not make it correct either. The first option is better, to admit > > that the set size is equal to the largest member, but that there is no > > identifiable largest member and therefore no idnentifiable set size, which > > makes sense given the lack of endpoint for measure. > > Why is this better than just saying that the size of the set is larger > than any element in the set? Because that statement is demonstrably false, as Albrecht and I have been trying to explain. For every element you add, you add one to the largest element, and if you start with 1, these two numbers are equal. With this 1-1 correspondence between set size and value range, there is no way to achieve an infinite set size without infinite values in the set. The system of ordinals, with its discontinuities, is faulty. You define the set size to be the set of all elements less than that size. Basically, you are saying the set size is the largest element, plus 1, which works if you start the naturals with 0. However, you have no largest finite element, so you say, okay, we'll use the NEXT ordinal, which we will invent. I'm sorry, but the first limit ordinal is not 1 greater than the largest element of the set as required. You are better off starting the naturals with 1 in this case, noting that the set size and largest element are always equal, and that as long as one is finite, so is the other. The system of ordinals is trash in my opinion. It's a kludge. Sorry. I don't buy it. > > Every set has a size, but not every set size is a natural number. That's just wrong, on both counts. Oh well. > > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/ |