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From: Tony Orlow on 27 Oct 2005 11:36 Virgil said: > In article <MPG.1dc956a36ff838c098a571(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > David Kastrup said: > > > Tony Orlow <aeo6(a)cornell.edu> writes: > > > > > > > David Kastrup said: > > > > > > > >> Every member "counts" itself. And every member "counts" a set > > > >> which ends with itself. The set of natural numbers does not end > > > >> with any member, and so it is not "counted" by any of its > > > >> members. Only the last member of such a set "counts" the set, > > > >> and there is no last natural number to "count" it. So you have > > > >> the options of declaring that the set can't be counted, or you > > > >> have to invent an unnatural number which, while not counting the > > > >> set in the customary way, is supposed to represent the count of > > > >> the members of a set obeying the Peano axioms. Not by actually > > > >> doing the counting, but by checking whether the axioms hold, and > > > >> then declaring the count of the set to be aleph_0, by decree. > > > >> > > > >> This does not make aleph_0 a member of the set of naturals. > > > > > > > > A decree does not make it correct either. > > > > > > Well, one knows that there is no natural number depicting the size. > > > > > > > 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. > > > > > > This has, of course, the advantage that one need not deal with > > > number-like entities which behave rather peculiarly when doing > > > arithmetic with them. > > > > > > It has the disadvantage that there happens to be a hierarchy of > > > surjectability even between infinite sets. And those sets can be > > > compared in the context of surjectability, and it turns out that > > > one can group them in equavalence classes. > > > > > > And that makes it convenient to also have a _name_ for the > > > cardinality of the naturals and other infinite sets. This name can > > > be used for sorting; and a few rules concerning the arithmetic can > > > be derived, too, when deriving those rules from what happens to > > > cardinality when you form the union of disjoint sets. > > > > > > It _is_ a valid stance to leave the cardinality unnamed or > > > unspecified or unidentifiable or whatever else. But it is stopping > > > short of what _can_ be achieved in a consistent manner. > > > > > > > > Perhaps. Cardinality is better than nothing. > > And 'nothing' is what TO's set sizes amount to. Ther are a number of > ways of measuring different aspects of sets, any of which can be clled > "size". None of them, except for cardinality, allow comparison of > arbitrary sets. For example, set inclusion cannot compare sets neither > of which is a subset of the other. similarly, non-numeric sets cannot be > compared bases on "value ranges" since non-numeric objects cannot be > subtracted one from another to get "ranges". Correct, and that's where N+S^L comes in for symbolic sets, trees, and decision systems. > > > And yet, I think that > > noting that the size really is not a number with an absolute value > > but needs to have some value applied to it in order to measure it, > > leads to more fruitful ways of dealing with the set. > > Their is no such thing as 'the' size of the set, there are many sizes, > depending on which proprties one is looking at. But only the cardinality > size can be applied to every set. all other types of size are limited in > what sets they can compare. Do you have one tool in your tool drawer? Is it a hammer, a screwdriver, or a wrench? Probably just a rock. > > > I hope that's okay, even if it sounds like babbling > > hogwash. Does deep hogwash run still? Hmmm.... > > If TO is content to wallow in his hogwash, ... Come wallow, O Virgil, for the sun is hot, and the hogwash is refreshing! Cool thy meaty rump in the mud, as you bask in the afternoon breeze. :D > -- Smiles, Tony
From: Tony Orlow on 27 Oct 2005 11:48 David R Tribble said: > 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. It is > > these kinds of pictures the axiomatikers need to practice seeing. > > His second diagram is an infinite list of member sets, where each set > contains a finite number of leading #s and an infinite nubmer of > trailing 0s. Each member set is associated with a finite natural, so > the entire set is the same size as the set of finite naturals (N). > > There's not much to say about his diagram, though, because it doesn't > show any contradictions in standard set theory. It also does not > show that the size of the set is a member of the set. > > 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. Albrecht is correct in this regard. -- Smiles, Tony
From: imaginatorium on 27 Oct 2005 12:04 Tony Orlow wrote: > William Hughes said: > > > > Tony Orlow wrote: > > > William Hughes said: > > > > > > > > Tony Orlow wrote: > > > > > David R Tribble said: > > > > > > Tony Orlow wrote: > > > > > > >> So, which element of the power set does not have a natural mapped to it? > > > > > > > > > > > > > > > > > > > Virgil said: > > > > > > >> {x in S:x not in f(x)} > > > > > > > > > > > > > > > > > > > Tony Orlow wrote: > > > > > > > Oh yeah, the entire set, last element and all. What element was that again? > > > > > > > > > > > > Your mapping does not include any natural that maps to the entire set > > > > > > *N, which it must do in order to be called a bijection, since *N is one > > > > > > of the members of P(*N). Show us that one single mapping, please. > > > > > It would obviously be the natural with an infinite unending strings of 1's, > > > > > right? Ah, but you want to know, how MANY 1's? > > > > > > > > No, this is irrelevent. No natural has a binary representation that > > > > consists only of 1's. > > > I assume you mean in the infinite string, otherwise 1, 11, 111, etc would fit > > > the bill. Why do you say I cannot declare that the 1's go on indefinitely? We > > > are talking about *N, the set including infinite naturals, with infinite > > > numbers of bits. When will I run out of bits for my subset elements? Remember, > > > there is no last element, as you remind me again, below. > > > > > > You do not "run out of bits". > > The reason that no natural number can > > have a binary representation composed only of ones > > is that every natural has a successor which also > > has a binary representation. > > > > Suppose that a natural n > > had a binary representation that is composed only of ones. > > What could the representation of n+1 be? It cannot consist > > only of ones (this is the binary representation of n). It cannot > > have a 0, as a binary representation containing a 0 is the binary > > representation of a number smaller than the number with a binary > > representation consisting only of ones. Thus n+1 cannot have a binary > > representation. Contradiction. > Well, golly gee. A contradiction. And all you asked me for was a natural with a > 1 bit for every natural in the set. What did you expect? Of course there are > problems with the answer. There are problems with the question. That's why you > get a contradiction. You want the complete mapping to every element of an > endless set. I want a flying pink unicorn that pees single-malt scotch. Oh > well. Oh, but I'm sure that's easy. Of course no definite flying pink unicorn exists, but if it's flying, pink, and a unicorn, it must be what we need, even if its existence is tenuous. And everything pees something, so we declare it to pee single-malt scotch. There you are! Brian Chandler http://imaginatorium.org
From: Tony Orlow on 27 Oct 2005 12:29 David R Tribble said: > Tony Orlow wrote: > >> So, which element of the power set does not have a natural mapped to it? > > > > Virgil said: > >> {x in S:x not in f(x)} > > > > David R Tribble said: > >> Your mapping does not include any natural that maps to the entire set > >> *N, which it must do in order to be called a bijection, since *N is one > >> of the members of P(*N). Show us that one single mapping, please. > > > > > It would obviously be the natural with an infinite unending strings of 1's, > > right? Ah, but you want to know, how MANY 1's? > > Perhaps 'N' ones? > > So your infinite natural that maps to the entire set *N is: > s = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ... > s = 1 + 2 + 4 + 8 + 16 + ... > > Thus s is a sum of an infinite number of finite powers of two (2^p), > each term being twice the previous term. So s is your 2^N-1, where > N is your "unit infinity": > N = 1 + 1 + 1 + 1 + 1 + ... The dawn dawns on David Tribble.... > > (We'll pretend for the moment that s and N are different, and ignore > the fact that they are actually the same value.) .....and the clouds enshroud the sun. Do you really think sum(n=0->N:1) is the same value as sum(n=0->N:2^n), when every term of the second (except the first) is greater its corresponding term in the first? Oy..... > > So the subset that s maps to is: > Ps = { 0, 1, 2, 3, ... } > > Ps should look familiar - it's just N, the set of finite natural > numbers. So you can see that s, an infinite number, maps to the > infinite set of all finite naturals, or N. But s does not map to > *N, because Ps does not contain any infinite naturals as members. 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. > > It's fairly easy to show this to be the case using TOmatics. > We choose an s of all 1 bits, N bits in length, as the infinite > natural to map to Ps, where each natural member k in Ps corresponds > to the 2^k bit of s. If s is composed of N bits, then Ps contains > N members. But s itself cannot be a member of Ps, because the > members of Ps are all less than log2(N) (which is also an infinite > value). 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. > > Fine, you say, let each member in Ps have N bits. But then s must > now have 2^N bits, and s is still too big to be a member of Ps. > You can keep doing this and choose even longer bitstring lengths > for s and the members of Ps, but you'll find an s that is a member > of Ps, all because N < log2(N) for any value of infinite N. Exactly. We can go on forever, fnding every larger naturals to satisfy the two sides of the relation in alternation. > > 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. If you were to assume any such "last element", as you have for the power set of the unending set, you would draw a contradiction, since any natural maps to a natural larger than it. Does this mean the bijection fails? No, it means more information is required than the simple existence of a bijection. > > -- Smiles, Tony
From: Tony Orlow on 27 Oct 2005 12:48
David R Tribble said: > David R Tribble said: > >> Your scheme listed these mappings: > >> f(0) = {} > >> f(1) = {0} > >> f(2) = {1} > >> f(3) = {0,1} > >> f(4) = {2} > >> f(5) = {0,2} > >> f(6) = {1,2} > >> f(7) = {0,1,2} > >> f(8) = {3} > >> ... > >> > >> This list defines only the mappings of finite naturals in *N to > >> the finite subsets in P(*N). Your list does not show any infinite > >> naturals, nor does it show any infinite subsets, nor does it show any > >> subsets containing infinite naturals. > > > > Tony Orlow wrote: > > I was asked about that and gave examples for the sets of odds and evens, > > which map to 2N/3 and N/3. There are other examples. It should suffice to > > say the same patter continues without end. > > Which is not sufficient to prove that every element in P(*N) is mapped > to an element of *N. Please show us how _all_ the subsets are mapped, > not just some of them. > > 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. > > > Tribble: > >> Your list shows an infinite number of finite naturals mapping to an > >> equally infinite set of finite subsets. That's easy. 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. > > > > Orlow: > > That's not what it is. I have declared it to be between *N and P(*N). 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. Sorry. > > 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. 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. There is little more I can do for you in that regard. > > > Tribble: > >> 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). 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). > > > > Orlow: > > Like which, for instance? Keep in mind that my bits never end. > > WHich other mappings do you want? What would satisfy you? > > Oh, like {...111}, for example, or {0,...110}. Your mapping omits > every subset containing an infinite natural. Again, please prove > otherwise, if you can. Why don't you try proving YOUR statement? There are an infinite number of bits in each of *N, even if only a finite number of significant bits in some of them. The infinite values require and have infinitely long bit strings. Prove otherwise. > > -- Smiles, Tony |