Uncountability of the Rationals?
in [Math]
|
Prev: Collatz conjecture
Next: Beginner-ish question
From: Brian Chandler on 16 Jun 2010 02:12 Tony Orlow wrote: > 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: Excuse me taking your points largely in reverse order... > > 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. Right, so you have now managed to read "set of pairs {p,p}" (call it pp) correctly. That's progress. And you agree that this is the "diagonal" of the "unit infinite square", so B(pp) = Tav. > Tav us a bad name, but whether you are talking about omega or a > zillion, the equivalence relation holds. Why is "Tav" a bad name? In any conversation with you, I am certainly not talking about the "real" omega, because you don't understand what it is. If there is "declaring of unit infinities" to be done, I shall do it in the Hebrew alphabet. (Irrelevant anecdote: I remember Conway using 'Beth', and commenting that the Hebrew alphabet is underused...) > 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'm sure Lebesque is (posthumously) delighted to have your agreement. Personally I have no idea what you are talking about. But working back towards my original question: > > > 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. Sure, a matrix is a matrix. But consider (for example) a two- dimensional vector space over a finite field: say the set of ordered pairs of elements of GF17 (pairs {a, b} where a, b are integers modulo 17). This is a vector space; we could write all 17^2 elements in a grid, but this is not a matrix -- there is no "perhaps with some values repeating". FWIW, I don't know of a definition of "matrix" which forces "expressions" to be unique. But I think this is just another example where you write (much) faster than you think. But anyway: finally my original question... [to David Tribble, I think] > > > > > > 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. Right, well let's avoid the "mat[r]ix" word, and call our "unit infinity" Tav, as usual. Yes, you tell us there are Tav^2 natural roots in the vertical direction, and Tav naturals in the horizontal direction. I'm asking about the set NQ containing values {p,q} such that p^2=q. I think you have probably withdrawn your original claim that there are Tav^3 of these points. But how many are there? I envisage there is some sort of trail of these points, each one above and to the right of the preceding one. If I count them vertically I seem to get Tav^2, horizontally I get Tav. This is what I am asking you to resolve. Brian Chandler
From: Transfer Principle on 16 Jun 2010 02:15 On Jun 15, 2:03 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tony Orlow <t...(a)lightlink.com> writes: > > IF it has an algebraic bijection with N+ it can be compared therewith. > What's "algebraic bijection" mean? I assume TO means a bijection f: N+ -> S such that f is an algebraic function (i.e., a polynomial, rational, radical). According to what TO calls his "Inverse Function Rule," if such a function exists, then the set S has set size (or Bigulosity) of f^-1(zillion). (This is the compositional inverse, not the reciprocal. I actually prefer the notation f_o^-1, with a "o" subscript to denote composition.) The examples given by TO thus far include: S = {2,3,4,5,6,7,8,9,...} f(x) = x+1, Big(S) = f^-1(zillion) = zillion-1 S = {2,4,6,8,10,12,14,16,18,...} f(x) = 2x, Big(S) = f^-1(zillion) = zillion/2 S = {1,4,9,16,25,36,64,81,...} f(x) = x^2, Big(S) = f^-1(zillion) = sqrt(zillion) S = (2,4,8,16,32,64,128,256,512,...} f(x) = 2^x, Big(S) = f^-1(zillion) = lg(zillion)
From: Brian Chandler on 16 Jun 2010 02:25 In a truly extraordinary moment, Transfer Principle wrote: > On Jun 15, 9:18 pm, Brian Chandler <imaginator...(a)despammed.com> > wrote: <snip bit about Conway's ONAG> > So yes, sqrt(omega) is a surreal. But the big difference > between surreals and TO's numbers is that the latter is > supposed to be a sort of set _size_ (Bigulosity). It > makes no sense to state that the set of perfect squares > has the surreal sqrt(omega) as its set size. Have you gone barking mad? Don't you remember the Agreement? You *cannot* say "makes no sense". You *must* say "... mumble mumble, working in a different theory". Janaino? Brian Chandler
From: Jesse F. Hughes on 16 Jun 2010 07:56 Transfer Principle <lwalke3(a)lausd.net> writes: > On Jun 15, 2:03 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Tony Orlow <t...(a)lightlink.com> writes: >> > IF it has an algebraic bijection with N+ it can be compared therewith. >> What's "algebraic bijection" mean? > > I assume TO means a bijection f: N+ -> S such that f is an > algebraic function (i.e., a polynomial, rational, radical). Maybe so, but my question is not really answered. What's an algebraic function? > According to what TO calls his "Inverse Function Rule," if > such a function exists, then the set S has set size (or > Bigulosity) of f^-1(zillion). > > (This is the compositional inverse, not the reciprocal. I > actually prefer the notation f_o^-1, with a "o" subscript > to denote composition.) > > The examples given by TO thus far include: > > S = {2,3,4,5,6,7,8,9,...} > f(x) = x+1, Big(S) = f^-1(zillion) = zillion-1 > > S = {2,4,6,8,10,12,14,16,18,...} > f(x) = 2x, Big(S) = f^-1(zillion) = zillion/2 > > S = {1,4,9,16,25,36,64,81,...} > f(x) = x^2, Big(S) = f^-1(zillion) = sqrt(zillion) > > S = (2,4,8,16,32,64,128,256,512,...} > f(x) = 2^x, Big(S) = f^-1(zillion) = lg(zillion) Notice that you did not mention exponentials or logs as "algebraic functions". So, what counts as an algebraic bijection? (This question is really directed at Tony, not Walker, since Tony is a better authority on what Tony means.) -- "Another factor one has got to look at is the amount of liquidity in the system. In other words, is there enough liquidity to enable markets to be able to correct? And I am told there is enough liquidity in the system to enable markets to correct." -- Guess who.
From: Tony Orlow on 16 Jun 2010 08:16
On Jun 15, 8:19 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <3e8a10bc-7035-492b-9af6-8617c5b2f...(a)f6g2000vbl.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> wrote: > > > 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. > > "Copuntably"? Yeah, "copuntably", meaning, "in a manner which can be kicked by two feet at once." Duh... > > Besides which, for a set A, |A| is its cardinal, not its ordinal, > so |N| = |omega|, but since omega is an rdinal but not a cardinal one > does not have omega = |A|, for any set A whatsoever. "rdinal"? rdninal scmrdinal. Enjoy your cardinality. > > > > > > 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. > > Not with omega! > With ine it does. > > > > > 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+. > > By what bijection? by (x,x) <-> x > > 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. > > But it is its "area" which is relevant. What "area"? It's a series of pairs of two naturals which are identical. > > Consider the cardinalities of Np x Nq, where Nr = {x in N: x <= r} Okay so, you have a Bigulosity of floor(p)*floor(q). And? |