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From: Tony Orlow on 25 Oct 2005 15:46 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? > > But none of us can figure out where you keep pulling this "last > element" gibberish from. Just let it go, Tony. The entire set includes the last element, and the mapping includes a bit for it. What position should I put this bit in? Isn't there another bit after it? > > -- Smiles, Tony
From: Tony Orlow on 25 Oct 2005 15:47 William Hughes said: > > Tony Orlow wrote: > > Virgil said: > > <snip> > > > > > So, which element of the power set does not have a natural mapped to it? > > > > > > {x in S:x not in f(x)} > > Oh yeah, the entire set, last element and all. What element was that again? > > > No, just the entire set. The set does not have a last element. > > - William Hughes > > Then there is no last bit to the natural that maps to it, so it cannot be truly identified, can it? -- Smiles, Tony
From: Tony Orlow on 25 Oct 2005 15:53 Virgil said: > In article <MPG.1dc70e511038079d98a54e(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Virgil said: > > > In article <MPG.1dc1c0b757ad81cd98a518(a)newsstand.cit.cornell.edu>, > > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > > > Virgil said: > > > > > > > > For what set is TO assuming that each member of its power set > > > > > can be represented by a single bit in an infinite sequence of > > > > > bits? > > > > I never said that. > > > > > A member of the power set is a subset of the entire set. You are > > saying that I am saying that each of these subsets can be represented > > by a single bit in an infinite sequence? > > NO! I am saying that you say that each singleton set of P(S), set with > exactly one member, must be so represented in TO's argument. Well, duh, it is. For any element n, the element which maps to the singleton {n} is 2^n, all zeroes with a 1 in the nth position. I already said that multiple times. > > > > If you take the power set of > > the power set, then you have sets of subsets, each of which can be > > assigned a bit corresponding to its order in the enumeration of the > > power set. But, that is not what you seem to be saying here now. > > If TO were not functionally illiterate, he could read what I actually > say. If you wrote clearly, I would have been able to give the above response posts ago. > > > > > If each subset of *N is to be represented by an infinite binary > > > sequence of digits with 1 in some position representing the > > > presence of a member *N and 0 representing its absence, then one > > > element sets must be represented by strings with one 1 in them. > > > This makes sense, but doesn't jibe with what you said before, as far > > as I can tell. > > It does, but TO can't tell. Because what you first wrote was gramatically broken and made no sense even with analysis. > > > Yes, each singleton set in P(*N) will map to a natural > > whose binary representation has a single bit. This only includes N > > out of 2^N subsets. > > But you already have a bijection from all of *N to the singleton members > of P(*N), and that leaves most of P(*N) out. And you already have a bijection from the evens to the evens in N, but that leaves half of N out. Big difference. > > > > > > So that, while he may not have been aware of it, TO was saying > > > precisely that. That TO is often unaware of the meaning of what he > > > is saying has long been apparent. > > > Precisely what? You are being particularly opaque, even for Virgil. > > If TO is too illiterate to read what I have said, it is not my > responsibility to send him back to kindergarten. Nice clarification. Go back to your hamsters. > > > > To do that in a string, TO must not only well-order the reals, but > > > order them isomorphically to the set of finite naturals. > > > But the finite naturals have finite numbers of bits, so what do you > > mean? > > Each finite naturally individually requires only a finite number of > bits for its expression, but there is no finite number of bits that > suffices for all of them. Sure there is, depending on what you mean. > > Those who fail to understand this have a form of quantifier dyslexia. You wish. > > > > > > > So, which element of the power set does not have a natural mapped > > > > to it? > > > > > > {x in S:x not in f(x)} > > Oh yeah, the entire set, last element and all. What element was that > > again? > > > What "last element" is talking about? The one that completes the set and corresponds to the last bit in the mapping natural. > > The set S need not even be an ordered set, as order is not essential for > the existence of {x in S:x not in f(x)}, so how can an unordered set be > though to have a last element? In the mapping to the naturals order is inherent. Next. > > > The conents of {x in S:x not in f(x)} depends entirely on the nature of > the function f:S -> P(S). Any ordering of S is irrelevant to the > existence, or membership, of {x in S:x not in f(x)}. I gave a bijection, and am discussing your foolishness in the context of that bijection, between *N and P(*N). > > The set {x in S:x not in f(x)} is defined soon as f is defined, but > until f is defined for all of S, the issue of whether it can be a > bijection on S is irrelevant. > > So that TO's complaints are out of order! What, are you going to bang your gavel now? > -- Smiles, Tony
From: Tony Orlow on 25 Oct 2005 15:56 albstorz(a)gmx.de said: > > David R Tribble wrote: > > David R Tribble wrote: > > >> I've got a set S = {0, 2^0, 2^2^0, 2^2^2^0, ...}, which contains > > >> all the powers of 2 of the form 2^p, where p=0 or 2^q. > > >> 1) If it is not an infinite set, tell me how many members it has. > > >> 2) If it is an infinite set, tell me what the smallest (first) > > >> infinite number is a member of it. > > > > > > > Albrecht Storz wrote: > > > A short view upon this makes me think that you are writing sensless > > > symbols. S don't contain all numbers of the form 2^2^q if you think > > > about the sequence 0, 2^0, 2^2^0, 2^2^2^0,... . If this is your > > > intention you may have a infinite sequence 0, 1, 1, 1, ..., and a set > > > {0,1}, so > > > 1) 2 > > > 2) ? > > > > Obviously, you do not understand. > > > > 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, ...} > > > > > I don't know what kind of math you apply here. Tribble-O-Math? > I don't want discuss your very interesting system. > Look at my starting posting if you want to know in what I'm interested > to discuss. > Start a new thread if you search for people which want talk about > Tribble-O-Math. > > Regards > AS > > This actually is interesting stuff, called tetrations. It is closely related to the baseless enumeration of the reals, which I am finishing writing up (should be posted soon). As I understand it, these kinds of formulas are not well understood yet. Kind of hard to figure an inverse to this function. -- Smiles, Tony
From: Tony Orlow on 25 Oct 2005 16:01
David Kastrup said: > albstorz(a)gmx.de writes: > > > David R Tribble wrote: > > > >> The size (cardinality) of the set is infinite, but this set measure > >> is not a natural number, and is not a member of the set itself. > >> Infinite cardinal numbers are not natural numbers, so they are not > >> members of sets of naturals. > >> > >> The "largest natural number" is undefined because it does not > >> exist; there is no such number. On the other hand, the size of the > >> set is well-defined as the smallest infinite set size (Aleph_0). > >> But the size is not a natural number. > >> > >> It is reasonable to say that the number of elements in the set is > >> greater than any member in the set. Therefore, the size cannot be > >> a member of the set, and it cannot be a natural number. > > > > It's not reasonable in the case of the natural numbers. The size of > > a set could not be greater than any member in the case if the > > members count themself. This is so obvious. > > You did not read _any_ of what David wrote above, right? Or at least > you did not understand a word of it. 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. 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. > > -- Smiles, Tony |