Uncountability of the Rationals?
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From: Tony Orlow on 14 Jun 2010 17:29 On Jun 14, 12:15 pm, Brian Chandler <imaginator...(a)despammed.com> wrote: > Tony Orlow wrote: > > On Jun 14, 1:21 am, Brian Chandler <imaginator...(a)despammed.com> > > wrote: > > > David R Tribble wrote: > > > > Tony Orlow wrote: > > > > > Actually, in Bigulosity, there are much larger countably infinite > > > > > sets. The square roots of the naturals have a higher "density" than > > > > > the naturals. > > > > > How is that? The two sets should be exactly the same size > > > > (your size(), not just cardinality). Every natural has a square root, > > > > and every natural square root has a corresponding natural. > > > > Can't you see? You're just using "cardinality" to point out a > > > bijection between naturals and their square roots. Bigulosity Theory, > > > or BT, simply rejects such arguments, in its search for the Answer. > > > > Consider any large number Q (perhaps a quillion): it is clear that in > > > the range 1-Q there are Q naturals, but there are (roughly, at least) > > > Q^2 real values x such that x^2 is a natural. Now appealing to ICI, > > > PCT, QD, and any other TLAs to hand, declare Tav* to be our unit > > > infinity, let Q tend to Tav, and lo! the answer we wanted. > > > > * Last letter of the Hebrew alphabet according to Wikipedia. > > > > It is true that BT, of which I am but a beginning student, leaves > > > nagging doubts. Suppose, for example we thought about the bigulosity > > > of the set NQ > > > > NQ = { {p,q} | p^2 = q e N} (set of pairs of naturals with their > > > roots) > > > Done. You have an omega X omega^2 matix. You omega^3 elements. > > Is this the answer, Dear Leader? "You omega^3 elements"? This sentence > no verb? Explication possible? > > I would have thought that since their is one pair {p,q} for each > natural q, that there might be omega of these pairs. Or possibly, > since their is one pair {p, q} for each root of a natural p (of which, > DL, you say there are omega^2), then there might be omega^2 of these > pairs. But you tell us there are omega^3?? > > I see no matrix, by the way. I also wonder what the bigulosity is of > the set of pairs {p, p} where p is a natural? > > Brian Chandler > Don't you make any connection between a set of pairs and, say, spatial coordinates? When you talk about points in n dimensional space, do you not define them as unique n-tuples in the spatial set of points? Hmmmm..... Tony
From: Tony Orlow on 14 Jun 2010 17:41 On Jun 14, 2:48 pm, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > >> However omega cannot exist within the set. Consider this. > >> Are the following two statements logically equivalent? > >> AneN n<omega > >> ~EneN n>=omega > > >> Does either imply, actually, that such a size exists? No. Both assert > >> that no natural number will suffice. However, this is not to say that > >> there exists ANY number which does. Logically, you must concede, > >> neither statement proves that omega exists as a number. > > David R Tribble wrote: > >> Yes, that's correct. Neither of your two statements implies that > >> there exists any number equal to card(N). > >> Neither statement implies card(N) does not exist, either. > > Tony Orlow wrote: > > Nope. They simply say that IF N has a size, it cannot be a member of N. > > Neither statement says anything about the size of N, unless > you're taking omega to indeed be the definition of the size of N. > In which case you're attempting to disprove the definition of > omega. But then, definitions are not "proved" or "disproved" > in logic. But you already knew that. I know nothing of the sort. If a definition contradicts itself, it's worse than circular. It's impossible. > > Also, neither statement says anything about "If" N has a size. No, they both say what the size can't be. Neither says what it is. > > I'm of course assuming we're talking about ZFC and standard > arithmetic and so forth. In your BO ("Bigulosity Ordering"), I > guess not all sets have a size, so it might very well be possible > for statement in BO to say "if set N has a size, then ...". If it's a countably infinite set then it is expressed with relation to standard N+. If it is an uncountable number of points it begins to have Lebesgue measure, and can start to be quantified with respect to zillions. > > Do you have a test in BO for "size" that determines when a > set has a size and when it does not? Countably infinite sets as normally defined have only parametric size in relation to N+. > > Tony Orlow wrote: > > I never said the two statements I gave above prove the nonexistence of > > |N|, but that the statements also do not state that it exists, only > > that it cannot be any natural number, if it exists. > > Yes, omega is not a natural. We already knew that. I suppose then > that you had some bigger point to make about omega not being a > natural? That it doesn't necessarily exist according to those elementally logical statements, and that I am not obliged by the axioms or that argument to believe that any such absolute size exists for such a set. > > > Bigulosity of countably infinite sets is always in functional relation > > to omega, but omega is more of a limit than a number. > > Now you're saying that omega is not a number. How then can > you say (in a previous post) that there must be an omega-th > (indexed) member of N+? That appears to be two contradictory > points of view about how "omega" acts. You're being dense, and somewhat deliberately so. You're smarter than that David. I am saying that a set of that description, with value equal to position for each element, of size n must include an nth element. If the size is omega, then there is an omega=th, but there cannot be an omega-th becuase that cannot be part of the set. Therefore, omega as a definition is self contradictory, and connot "particularly" exist. > > Tony Orlow wrote: > > With IFR, N is hardly the smallest caountably infinite set, consider > > the evens, squares of naturals, natural powers of 2, etc. > > Yes, in BO that may be the case. It's hard to tell right now, > though, because we still don't really know what you mean > by "size" and "less than" in BO. And does the old bijection > cardinality still work in BO or not? Maybe only sometimes?- Hide quoted text - > Bijection is required, as well as the application of IFR and ICI to extend the ordering to the infinite. Tony
From: Tony Orlow on 14 Jun 2010 17:52 On Jun 14, 3:08 pm, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > >> Actually, in Bigulosity, there are much larger countably infinite > >> sets. The square roots of the naturals have a higher "density" than > >> the naturals. > > David R Tribble wrote: > >> How is that? The two sets should be exactly the same size > >> (your size(), not just cardinality). Every natural has a square root, > >> and every natural square root has a corresponding natural. > >> (I assume you're only dealing with the positive roots.) > > Tony Orlow wrote: > > Holy cannoli! There exists a bijection, f:N->S and g:S->N, so indeed > > they have the same cardinality. They do not share the same bigulosity, > > since f and g are not the identity function. The number of square > > roots over omega is omega^2. > > Sorry, but it's not clear what you mean by "the number of square > roots over omega is omega^2". Do you mean: > 1. the number of naturals in omega that are square roots is omega^2? > 2. the number of naturals in omega that have square roots is > omega^2? > 3. the set of the squares roots of all the naturals in omega has > omega^2 members? > 4. or something else? Number 3. The set of square roots of naturals has a Bigulosity of omega^2. Over the range of omega or the range of R (same thing, basically) the Bigulosity of the set of reals mapped from neN+ using sqrt(n) is equal to omega^2. > > When you said > | The square roots of the naturals have a higher "density" than > | the naturals. > I assumed you meant that the "square roots of the naturals" is > the set > S = { x | x*x = n, for all n in N } Correct. For the positive naturals, N+ > > Is this not what you mean by the "square roots of the naturals"?- Hide quoted text - > Hope that was helpful. Tony
From: Tony Orlow on 14 Jun 2010 17:55 On Jun 14, 3:09 pm, FredJeffries <fredjeffr...(a)gmail.com> wrote: > On Jun 14, 4:49 am, Tony Orlow <t...(a)lightlink.com> wrote: > > > > > Actually, you're right. N should always start with 1 for IFR. I made > > an error of 1 by including 0. If we include 0, then we may say E=N/2+1 > > and O=N/2, because of that extra even element. Apologies. > > Let me see if I've got this right: > > E=N/2+1 and O=N/2 > > so N = E + O = (N/2+1) + (N/2) = N+1 I tried to explain that discrepancy, maybe not well enough. If N+ is the standard countable infinity, which does not include 0, then indeed the evens starting at 0 should include 1 more than those starting at 1, and the odds should remain the same N/2. Hence the difference. Standard omega starts at 1, giving the identity function between element count and value, and the ensuing contradiction regarding the existence of omega.
From: Tony Orlow on 14 Jun 2010 18:01
On Jun 14, 3:10 pm, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > >> In terms of natural density, they probably have a measure of oo? > > David R Tribble wrote: > >> The set of positive natural square roots is not a subset > >> of the naturals, so it does not have a natural density. > > Tony Orlow wrote: > > That is probably correct. The squares have a natural density of 0. If > > natural density is actually restricted to naturals, then there is no > > natural density, but qactually natural density allows for elements > > outside of N, such as the multiples of 1/2 with a density of 2. So, > > your objection is based ona false presumption about a theory that's > > not my invention. My guess is that this set would have an infinite > > natural density if any. > > What false presumption? Natural density is defined only for subsets > of the naturals. You can look it up on Wikipedia if you don't believe > me. > http://en.wikipedia.org/wiki/Natural_density Okay, you're right, I guess. It seems I read slightly more into natural density during my brief lookover than is actually there. It's restricted only to the set of naturals, and not all sets linearly mapped? It's amazing to me that no one ever extended that to include all m*n+b for real m and b. Simply astounding. Okay, IFR does even more than natural density, than I previously thought. Damn I'm good. Tony. |