Uncountability of the Rationals?
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From: MoeBlee on 15 Jun 2010 18:15 On Jun 15, 4:40 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > According to Tribble, there are many questions which TO > can't and won't answer about his theory. One of these > questions (asked IIRC by MoeBlee) is to which of the axioms > of ZFC does TO object? But to me, the answer to this > question is obvious. If a poster disagrees with how the > infinite sets work under ZFC, then they reject the axiom > which guarantees their existence -- and that axiom is, of > course, the Axiom of Infinity. You must have only skimmed the relevant exchanges in this thread on the subject of axioms. MoeBlee
From: Virgil on 15 Jun 2010 18:19 In article <6faffdc1-2c29-4e38-af66-e8e661c720e4(a)z10g2000yqb.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 14, 8:08�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <3b89977b-eb55-4134-803c-2d2b32bd7...(a)g19g2000yqc.googlegroups.com>, > > �Tony Orlow <t...(a)lightlink.com> wrote: > > > > > 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..... > > > > Not necessarily. One can easily have a space without having to have a > > coordinatized space. What bits of geometry are so dependent on > > coordinates that they cannot be done without it? > > Point-set topology. How do you define each unique point? When one starts with a set of points, they are all already there. What need is there of having to replace them with something else? Does TO suppose that there was no geometry prior to de Carte?
From: Tony Orlow on 15 Jun 2010 19:07 On Jun 15, 1:50 am, Brian Chandler <imaginator...(a)despammed.com> wrote: > Tony Orlow wrote: > > 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? > > > 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? > > I can make all sorts of connections, but if I want to persuade someone > else of something, I make those connections explicit. Yes, okay. If you have a matrix x wide and x tall, then you have an x*x grid of values, perhaps with some values repeating, but with expressions unique. > > Look, let's not use "omega" to refer to any of your stuff. It's an > abuse of respected terminology. As I understand it, there's a stage in > which we "declare a unit infinity", so I declare mine to be Tav. OK? Not exactly. Any of the copuntably infinite collection of countable infinities is tied to omega=|N+|, with an algebraic function on |N+|. The countable set has zero measure with respect to the uncountable. In that I agree with Lebesgue. > > I guess that to you an n-dimensional space (let n=2 for simplicity) > consisting of the pairs of naturals > {a, b} is a "square" of side Tav. I further guess that the "number" of > these pairs of naturals {a,b} (a and b being independent naturals) is > Tav^2. > Tav us a bad name, but whether you are talking about omega or a zillion, the equivalence relation holds. > Well, I asked about the set of pairs {p,p}, where (groan) p=p. In > other words typical members of this set are {1, 1}, {72, 72} and so > on. {23, 190} is _not_ a member of this set. Since this set is a > proper subset of the set of _all_ pairs of naturals {a,b}, I believe > its Bigulosity should be smaller. If Big({(p,p)}) = Tav^2, then > Big({(a,b)}) must be more than Tav^2. No that set has exactly one member for each memeber of N+. It's basically the actual diagonal of the set, as long as it is wide, with one element on each row and column, therefore making the height and width equal. > > Kindly explicate, preferably using at least one verb per sentence. I sowwy. > > Brian Chandler > > > Hmmmm..... > > Brian Chandler- Hide quoted text - > > - Show quoted text - Quixoted Text
From: Tony Orlow on 15 Jun 2010 19:16 On Jun 15, 2:00 am, Brian Chandler <imaginator...(a)despammed.com> wrote: > David R Tribble wrote: > > Tony Orlow wrote: > > >> The number of square roots over omega is omega^2. > > > David R Tribble wrote: > > >> 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? > > > Tony Orlow wrote: > > > 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. > > > This is confusing. > > Yes, of course it's confusing, since it's not what you are used to, > and Tony insists on abusing standard terminology for his own wonky > notions. But I really think *you* are being a bit dim here... Well, now, Brian, don't make yourself seem any more so.... > > > Does this also apply to finite sets of natural > > square roots? > > No, of course it doesn't. When you look at a finite set, you can see > the whole set in front of you. Scooby: Rhhurhhr? Great explanation. Of course IFR applies to finite sets. The infinite extension is acheieved through ICI. > > > I'm still confused, because it looks for all the world as if the sets > > N = { 1, 2, 3, ... } > > and > > R = { sqrt(1), sqrt(2), sqrt(3), ... } > > have exactly the same number of members. > > Of course it does to you, because you have grasped the idea of an > infinite set. Tony hasn't. All Tony can do is peer at his imagined > version of what the set looks like, and imagine that the rectangular > (or two-ended, or whatever) "window" through which he is viewing it > expands indefinitely, becoming "Tinfinite", yet remaining a > rectangular window. So when you count the number of naturals, or the > number of roots of naturals, you count from the left edge of the > window to the right edge of the window (even though the window is > "Tinfinite"), and you get a higher count for the roots of naturals > (obviously) because they are closer together. Just because an infinite set is defined by the fact that it can be bijected with a proper subset doesn't mean that this bijection implies the counterintuitive equinumerosity that one would NOT expect from a "correct" theory. > > > The only other explanation I can think of is that you actually > > mean the set: > > T = { x | x*x <= n, for all n in N+ }. > > "For all, schmor all, for every, schmorevery, what is the > difference?" (This is an answer, not a question?) Fascinating! You know, that comes from the same etymological root as "fascism" - "fasces" - a bunch of sticks. Kind of like the Latin (also) root of "religion". They both have to do with fags. > > Brian Chandler- Hide quoted text - > "Bundled with this software may be some third-party offers and opportunities" Tony
From: Tony Orlow on 15 Jun 2010 19:21
On Jun 15, 12:18 am, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > >> The number of square roots over omega is omega^2. > > David R Tribble wrote: > >> 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? > > Tony Orlow wrote: > > 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. > > This is confusing. Does this also apply to finite sets of natural > square roots? Yes, for any finite range of element values, [a,b], and in that case with the use of the floor() and ceiling() functions to account for range limits which are not members of S. If a and b are guaranteed to be members of S mapped from N+ using f, then g(a) and g(b) are both members of N+. > > N_k = { 1, 2, 3, ..., k}, for k in N+. > R_k = { sqrt(1), sqrt(2), sqrt(3), ..., sqrt(k) }, for k in N+. Okay ... R is not the reals, just some other name for set S ... yes? > > You say that bigulosity(N_k) = k (I think). > Is bigulosity(R_k) = k^2? Sort of, yes. In the sense that omega exists as a "zize" for the set N + (bigulosity(N_k) = k) then yes, the number of square roots of naturals is omega^2, and it is proven by ICI that x^2 > x for any x greater than 2, and that includes omega, unless I'm mistaken, since it's greater than 2 last I knew. > > David R Tribble wrote: > >> 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+ } [corrected] > > Tony Orlow wrote: > > Correct. For the positive naturals, N+ > > I'm still confused, because it looks for all the world as if the sets > N = { 1, 2, 3, ... } > and > R = { sqrt(1), sqrt(2), sqrt(3), ... } > have exactly the same number of members. For a given k in N+, > there is a sqrt(k) in R at exactly the same position. (Forget > bijections, I'm applying your BO set ordering and member indexes.) You're still thinking within the bax, David. You know AneN (n>1 -> E(reR: r=sqrt(n)), and that it's a countable set that occurs more often than the naturals, as you move at some measurable speed upward along the number line from that 1. So, you must admit, there is some kind of "moreness" that occurs with the square roots of naturals, and the same measure of "lessness" that occurs with the squares of naturals (of course with a natural density of 0). > > But when you say that R has size omega^2, that implies that > for the k members in N_k, the initial subset of N+, that there > must be k^2 members in the corresponding initial subset R_k > of R. If so, what are those other k^2-k members? Waitaminit. You're not trying to confuse standard set R with the postulated set S, are you, for obfuscational purposes? Lemme see... > > The only other explanation I can think of is that you actually > mean the set: > T = { x | x*x <= n, for all n in N+ }.- Hide quoted text - Why "<="? No I mean the set S = { n^2 | neN+ }, the squares of all positive naturals. Please step back, and consider the finite case. Within the range [a,b], if a and b are both members of S, then they map through g back to N+, with g(b)-g(a)+1 being the size of the finite set, but to account for the possibility that they are not members of S, the floor() and ceiling() (round up and round down) functions are convenient. The set of elements S has size (where monotonically increasing or decreasing one doesn't or does switch a and b, respectively, or uses the absolute value, which is equivalent) as related to the mapping function from N+ to S by f, by its inverse, g: S -> N+. T. |