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From: David R Tribble on 26 Oct 2005 14:54 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 + ... (We'll pretend for the moment that s and N are different, and ignore the fact that they are actually the same value.) 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. 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). 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. 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).
From: David R Tribble on 26 Oct 2005 15:11 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. 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. 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.
From: Virgil on 26 Oct 2005 15:14 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). 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). > 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. > 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. If TO can prove this, there is a lot about his mapping from *N to P(*N) that he is carefully not telling us. > > > > > > > > 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. > 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. 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. > > 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. 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. > > > > > > > > > > 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 ? 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.
From: David R Tribble on 26 Oct 2005 15:23 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
From: Virgil on 26 Oct 2005 17:58
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. 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. 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. |