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From: Tony Orlow on 31 Oct 2005 11:17 Robert J. Kolker said: > Tony Orlow wrote: > >>betters indicates strongly that you will never learn to be one. > > > > You have no clue what I'm doing here, do you Bob? > > Yes I do. You are a troll. > > > > I know what they are, but they don't make sense. > > They don't make sense to YOU. But what does? > > Bob Kolker > > Things well beyind you, apparently, Bob. You don't even believe in mind, so it's not surprising. -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 11:27 William Hughes said: > > 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. Guess we're both SOL. > > > > > > I am a bit confused as to exactly what you are claiming. > > Do you agree with the statment? > > Given any natural number n, the binary representation > of n does not consist only of ones. No, any binary string will represent some kind of natural, including the string of all 1's. I assume, by the way, that you are viewing any binary representation as including an implied infinite string, which may include trailing zeroes. > > Do you agree with the statement? > > If we specify a natural number n, the binary representation > of n does not consist only of ones. Unless you specify the natural number to consist only of unending 1's. Any particular nameable natural will have some zeroes somewhere, except for this one defined to not have any. > > Do you agree with the statement? > > For all natural numbers n, the binary the binary representation > of n does not consist only of ones. True except for the amorphous ....11111. It's not a specific number, really, but an unending string of bits representing a number without limit. > > > Do you agree with the statement? > > The natural number n which maps to the entire set > of natural numbers cannot be specified. It certainly cannot be spcified any better than ....11111. The unedning set of naturals cannot have a mapping which ends. > > Do you agree with the statement? > > The natural number n which maps to the entire set > of natural numbers does not exist. Essentially this is true, since the size of the set and the last element do not exist. It is not a requirement of an infinite bijection that the end be declared, which is precisely what you are doing in the standard proof of the unbijectibility of the power set with the set. This is the root of the contradiction. You want the entire set, which is the last in an unending set of subsets of an unending set. And, you don't expect a contradiction? You get the same contradiction from assuming the end of any endless set. > > -William Hughes > > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Randy Poe on 31 Oct 2005 11:41 Tony Orlow wrote: > Randy Poe said: > > > > 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. > > > > Um, yes you are. If a bijection requires that there exist such > > an s, and no such s exists, you are forced to conclude that > > you do not have such a bijection. > And yet, the bijection between 2 unending sets does not require you to be able > to name the end of the sets, does it? No, but it requires you to prove that there exists x such that f(x) = y for every in the image set. The word "end" does not appear there for, nor should it, especially for unending sets. But the word "every" does. If there exists y such there is no x with f(x) = y, then you provably do not have a bijection. By definition. > > > The entire > > > infinite set would be the final subset enumerated, > > > > Your usual red herring. But it's nonsense. This can clearly > > be seen since your scheme CAN be used to establish a > > bijection between *N and P(N), and no recourse to "last > > subsets enumerated" is needed. It is provable that > > EVERY subset of N is mapped, just as it is provable that > > SOME subset of *N is not mapped, ever, by anything, in this > > mapping. > Look again at the bijection I offered. Your mapping between *N and P(N) is not > valid in Bigulosity Theory. In what way? I have raised a specific objection to your map: a bijection between A and B is required to have some x in A such that f(x) = y for every y in B. You don't. Do you have a specific objection to my map? > In the bijection I offered, no element is inthe set > mapped to it, so the set of all elements not in the set they map to is the > entire unedning set, and you want a natural that maps to the end of it? No, I don't. I just want a member of *N to map to every element of P(*N), since you claim you have a map that maps some element of *N to map to every element of P(*N). 1. Do you have a bijection from *N to P(*N)? 2. Do you have a map that maps some element of *N to every element of P(*N)? 3. Are you aware you can't have #1 without #2? - Randy
From: William Hughes on 31 Oct 2005 11:42 Tony Orlow wrote: > William Hughes said: > > > > 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. Guess we're both SOL. > > > > > > > > > > I am a bit confused as to exactly what you are claiming. > > > > Do you agree with the statment? > > > > Given any natural number n, the binary representation > > of n does not consist only of ones. > No, any binary string will represent some kind of natural, including the string > of all 1's. I assume, by the way, that you are viewing any binary > representation as including an implied infinite string, which may include > trailing zeroes. > > > > Do you agree with the statement? > > > > If we specify a natural number n, the binary representation > > of n does not consist only of ones. > Unless you specify the natural number to consist only of unending 1's. Any > particular nameable natural will have some zeroes somewhere, except for this > one defined to not have any. > > > > Do you agree with the statement? > > > > For all natural numbers n, the binary the binary representation > > of n does not consist only of ones. > True except for the amorphous ....11111. It's not a specific number, really, > but an unending string of bits representing a number without limit. > > > > > > Do you agree with the statement? > > > > The natural number n which maps to the entire set > > of natural numbers cannot be specified. > It certainly cannot be spcified any better than ....11111. The unedning set of > naturals cannot have a mapping which ends. > > > > Do you agree with the statement? > > > > The natural number n which maps to the entire set > > of natural numbers does not exist. > Essentially this is true, since the size of the set and the last element do not > exist. At this point, according to the standard definition you have "Essentially", no bijection. >It is not a requirement of an infinite bijection that the end be > declared, No, but it is a requirment of a bijection (infinite or not) from X to Y, that for every element y in Y, there exists an element x in X which maps to y. >which is precisely what you are doing in the standard proof of the > unbijectibility of the power set with the set. This is the root of the > contradiction. You want the entire set, which is the last in an unending set of > subsets of an unending set. And, you don't expect a contradiction? You get the > same contradiction from assuming the end of any endless set. You might want to consider which TO natural maps to the binary representation of all bits representing finite n, equal to one, and all bits representing infinite n, equal to 0. Note that this one gets skipped. -William Hughes
From: Daryl McCullough on 31 Oct 2005 11:35
Tony Orlow says... >Look again at the bijection I offered. It's not a bijection. As you say, no element is in the set mapped to it. It follows that no element is mapped to *N. From that, it follows that the mapping is not a bijection. By definition. -- Daryl McCullough Ithaca, NY |