Uncountability of the Rationals?
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From: David R Tribble on 14 Jun 2010 23:43 David R Tribble writes: >> [...] Tony really needs to put his money where his mouth is and >> just go ahead and write his assumptions down already. He can't >> proceed with any of his other statements about bigulosity, >> ICI, IFR, or whatever until he starts with an axiom or two. >> And he knows it. > Jesse F. Hughes wrote: > It seems to me that he's stated his assumption, but he doesn't get quite > how much it assumes. > > Given any real valued functions f and g, if lim (f - g) > 0, then f(x) > > g(x) where x is any infinite number. > > As a consequence of this, I guess, it follows that f and g are defined > on infinite numbers, though we don't know anything about their values > aside from the fact that f(x) > g(x). "Doesn't quite get it" is right. This formulation opens up several cans of worms, among them being what is an infinite number, what does 'x < y' mean for infinite x and y, how does a real arithmetic function apply this arithmetic to infinite values, and a slew of other deep questions. It's obvious that Tony can't and won't answer any of these questions, and no one here (except Walker) really sees him as capable of going any further in any meaningful sense. So except for deriving some entertainment value and learning a few new things, I don't see the point in indulging him any more. Teaching pigs to sing, and all that.
From: David R Tribble on 15 Jun 2010 00:18 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? N_k = { 1, 2, 3, ..., k}, for k in N+. R_k = { sqrt(1), sqrt(2), sqrt(3), ..., sqrt(k) }, for k in N+. You say that bigulosity(N_k) = k (I think). Is bigulosity(R_k) = k^2? 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.) 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? 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+ }.
From: Brian Chandler on 15 Jun 2010 01:50 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. 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? 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. 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. Kindly explicate, preferably using at least one verb per sentence. Brian Chandler > Hmmmm..... > Brian Chandler
From: Brian Chandler on 15 Jun 2010 02:00 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... > 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. > 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. > 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?) Brian Chandler
From: Aatu Koskensilta on 15 Jun 2010 03:43
Tony Orlow <tony(a)lightlink.com> writes: > On Jun 11, 3:33�pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > >> But >> >> ~ w < w^2 >> >> is just a theorem of ZFC. > > Bull. It's a theorem of the model extension. What is a "model extension"? > It does not follow from the axioms, as the mention nothing about set > size or cardinality. Just as PA proves nothing about primes since they aren't mentioned in any of its axioms? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |