Uncountability of the Rationals?
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From: Virgil on 20 Jun 2010 15:40 In article <f4ac19cf-f3be-44ab-90b7-62cc1727562f(a)u26g2000yqu.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 18, 10:37�pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > On Jun 17, 12:31�pm, Virgil <Vir...(a)home.esc> wrote: > > > > > > > David R Tribble wrote: > > > > > >> So the question becomes, what axiom justifies your logical > > > > > >> leap, applying a property of finite sets to infinite sets without > > > > > >> largest members? > > > > > Tony Orlow wrote: > > > > > > There is no leap. > > > Then there can be no �difference between finite sets and infinite sets, > > > thus everything true of finite sets must be true of infinite sets, thus > > > all infinite sets must be finite since that is true of finite sets. > > > At least following TO's logic. > > > > This is, incidentally, a problem that occurs with many posters > > who want the properties of all finite sets to extend to all > > infinite sets. In general, such posters want more properties > > to extend from finite sets to infinite sets than standard > > theory allows, but they certainly don't want the property of > > _finiteness_ to extend to infinite sets! > > > > So we see how any schema of the form: > > > > (Ax (x finite -> phi(x)) -> Ax phi(x) > > > > fails if we let phi(x) be "x finite" (and, of course, at least > > one infinite set exists). > > > > TO attempts to prevent this by preventing phi from being just > > any function, but instead limiting the schema to _algebraic_ > > functions using a few chosen operations. > > Hi Transfer - > > Actually, I avoid this conundrum by resticting my properties to > statements of inequality among formulaically expressed quantities. By > focusing on inequalities I am establishing quantitative order among > different expressions, and by extending those expressions to the > infinite case, I am concocting a method of distinguishing among > different countable infinities. Whether they are technically algebraic > or not is of little consequence. For the purposes of IFR it is only > required that they be monotonic bijections between N+ and any set. TO allow a monotonic function between N+ and any other set, that other set must be ordered and ORDER ISOMORPHIC TO N+, so that Bigulosity cannot be applied either to unordered sets or to ordered sets whose ordering differs from that of N+. Thus, for example, if one reorders N+ to make odds less than evens then bigulosity does not apply to the reordered set. It cannot even be applied to N+ as an unordered set. Thus bigulosity applies to, at most, only infinite well ordered sets with only one non-successor.
From: Virgil on 20 Jun 2010 15:43 In article <4c9686a9-dd0b-4648-a238-11c37ab78e4e(a)z10g2000yqb.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 20, 8:59�am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > Tony Orlow <t...(a)lightlink.com> writes: > > > On Jun 18, 8:52�am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > >> Tony Orlow <t...(a)lightlink.com> writes: > > >> > On Jun 16, 11:00�pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > >> >> Transfer Principle <lwal...(a)lausd.net> writes: > > >> >> > How about this: > > > > >> >> > An algebraic function is a real-valued function which is the > > >> >> > composition of finitely many real-valued polynomial, radical, > > >> >> > rational, exponential, and logarithmic functions, and whose > > >> >> > inverse (or at least the real-valued branches thereof) is > > >> >> > also the composition of finitely many polynomial, radical, > > >> >> > rational, exponential, and logarithmic functions. > > > > >> >> Yes, that's explicit. > > > > >> >> And, who knows, it might be what Tony meant. �Perhaps he'll say so. > > > > >> > That list probably covers the gamut, at least for now. So, I guess I > > >> > didn't misuse "algebraic" after all. > > > > >> Yes, you did misuse "algebraic". �In my experience, an algebraic > > >> function is one which preserves certain algebraic structure. > > > > >> But no matter. �We'll assume that Walker's definition of "algebraic > > >> bijection" is what you "probably" (probably?) meant. � > > > > >> I guess it will follow that the set P of primes has no size, since there > > >> is no algebraic bijections between P and N+? > > > > >http://primes.utm.edu/howmany.shtml > > > "The Prime Number Theorem: The number of primes not exceeding x is > > > asymptotic to x/log x." > > > So, let's say tav/log(tav). > > > > No, let's not just make things up. > > > > You have said, up 'til now, that the size of a subset S of N+ (I'm still > > not sure what the "+" signifies) depends on having an "algebraic > > bijection" between S and N+. �I don't see any such function between P > > and N+. > > > > Consistency is no vice, Tony. > > > > -- > > Jesse F. Hughes > > > > Baba: Spell checkers are bad. > > Quincy (age 7): C-H-E-K-E-R-S A-R-E B-A-D.- Hide quoted text - > > > > - Show quoted text - > > I didn't say this situation was handled by IFR and ICI. I even stated > that some sets might not have set sizes more specific than cardinality > in some cases. However, given the aymptotic relationship discovered by > others, I would guess that there is justification for quantifying this > set with this formula. Is there a relatively simple inverse to x/log > x? If so, it might be used to estimate the locations of primes, but > probably not. It is easy to see that "Bigulosity", by its very definition, can be applied only to ordered sets which are order isomorphic to Tony's N+, so is, at best, only of extremely limited utility or interest.
From: David R Tribble on 21 Jun 2010 00:27 David R Tribble wrote: >> That is the same fallacious logical jump. N+ is not a finite initial >> segment of N+, so your requirements for finite initial segments >> of N+ (to have an xth element, or for the size to be a member of >> the set) do not apply to N+. >> >> So again, the same question is, what axiom or theorem justifies >> you to state that your property of finite sets also applies to >> infinite sets? > Tony Orlow wrote: > First of all, I never said anything about *finite* initial segments of > N+. You cannot insert that condition into my argument, then claim that > the inconsistency it introduces into my argument is my fault. I stated > that every initial segment of N+ with size x has an xth element whose > value is x, without reference to finiteness. That does not change the fact that that property is only true of finite initial segments of N+. Infinite subsets of N+ do not have a largest element. An initial subset of N+ that is not finite is N+ itself. Surely that much is obvious to you. If not, please provide an example of a non-finite initial subset of N+ different from N+.
From: Transfer Principle on 21 Jun 2010 22:42 On Jun 20, 5:29 am, Tony Orlow <t...(a)lightlink.com> wrote: > On Jun 18, 9:46 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > Thank you. So in that case, we can start with full ZF (or even > > ZFC) as the base theory. The set omega still exists and is the > > cardinality (or ordinality, with aleph_0 being the corresponding > > cardinality) of N+, but now we can assign Bigulosities to sets > > as well. Two sets with the same cardinality can have different > > Bigulosities, but two sets with the same Bigulosities must have > > the same cardinality. > My pleasure. First, I should say, I am not partial to Choice, so I > would like to say, at least tentatively, that Bigulosity should work > with ZF, anyway. Othwerwise, we're good. :) In that case, ZF it is. Let's see whether we can continue our attempt to harmonize the desiderata of TO with the objections of the others. So far, in order to make sure that ICI works for algebraic functions, we had to make all of them primitives. But we know that the more primitives and axioms for those primitives we have, the more likely posters like Hughes will describe the theory as "ad hoc." So let's see whether we can replace all the algebraic primitives with a single primitive. We know that if two sets have the same Bigulosity, then they have the same cardinality, but not vice versa. Thus, the existence of a bijection between two sets is necessary but not sufficient to conclude that their Bigulosities are equal. Thus, let's call a bijection, say, "a strong bijection," or a "TO bijection," if the bijection is sufficient to conclude that the sets have equal Bigulosities. Thus, "strong" is our new primitive (a one-place predicate). So what axioms do we need. Naturally, a "strong" bijection is first and foremost a bijection: Af (f strong -> f bijective) TO has stated that if anything counts as a (strong) bijection, surely the identity function does. After all, we will want equi-Bigulosity to be an equivalence relation, in particular reflexive, so let's include identity functions: Afx ((xe(domain(f)) -> f(x) = x) -> f strong) We want Bigulosity to agree with cardinality for finite sets: Af ((f bijective & domain(f) finite) -> f strong) But we don't want infinite sets to have the same Bigulosity as any of their proper subsets: Af ((range(f) proper subset domain(f) -> ~(f strong)) We want equi-Bigulosity to be symmetric, so let's include inverse functions: Af (f strong -> f_o^-1 strong) We want equi-Bigulosity to be transitive, so let's include compositional functions: Afg ((f strong & g strong) -> gof strong) So we know that the successor function S on N+ is evidently not strong, since it is a bijection between N+ and one of its proper subsets, namely N+\{1}. But what about the restriction of S to the set 2N+ of even naturals, or the set N+\2N+ of odd naturals? The composition of these two restriction functions maps N+\2N+ to one of its proper subsets -- the set of odd naturals greater than 1. So at least one of these restrictions isn't strong. Since TO declares both 2N+ and its complement in N+ to have the same Bigulosity, namely tav/2, we'd like the restriction to the set N+\2N+ to be strong. But in general, how can we tell which restriction, if any, is strong? This is one question that we seek to answer. A related question would be how to incorporate the algebraic functions into our axioms, without introducing them to be more primitives. For example, consider multiplication. We could allow a multiplication primitive, but we can avoid this by taking an idea from cardinality, where multiplication is defined in terms of Cartesian product. So suppose we used Cartesian product in Bigulosity as well, so that the Bigulosity of a Cartesian product equals the product of the Bigulosities of the component sets. But let's go back to a question posed by Hughes. Since TO considers the set {1,4,9,16,25,36,49,64,81,...} of perfect squares to have Bigulosity sqrt(tav), Hughes asks us whether one can prove that: sqrt(tav)sqrt(tav) = tav In other words, we want a strong bijection between N+ and the Cartesian product of {1,4,9,16,25,36,49,64,81,...} with that same set itself. Here's an interesting bijection. Let's start out by mapping the elements on the diagonal to that same natural number: f(n^2,n^2) = n^2 f(1,1) = 1 f(4,4) = 4 f(9,9) = 9 That leaves the ordered pairs off the diagonal to map to the non-perfect squares. Let's try something like this: f(1,4) = 2 f(4,1) = 3 f(1,9) = 5 f(2,9) = 6 f(9,4) = 7 f(9,1) = 8 Here's the pattern: f(m^2,n^2) = (n-1)^2+m, if m<n = m^2-n, if m>n = n^2, if m=n But suppose we defined another function g as follows: g(m,n) = f(m^2,n^2) Then g is a bijection between N+xN+ and N+. But we don't want g to be strong, since otherwise we'd have tav^2 = tav with contradicts IFR. In other words, we want f to be strong but not g. But simplying adding an axiom declaring f to be strong would be "ad hoc." In order to avoid this "ad hoc"-ness, we need to find a way to generalize which bijections are strong and which ones aren't. But this is the hard part -- yet we must do so in order to have any hope of convincing Hughes, Virgil and others that Bigulosity is worth considering.
From: Transfer Principle on 21 Jun 2010 22:53
On Jun 20, 9:29 am, Brian Chandler <imaginator...(a)despammed.com> wrote: > Jesse F. Hughes wrote: > > ...but I'm pretty sure that you haven't > > said anything that implies the conclusion you've just drawn. > Not in what you are talking about Jesse. But the fundamental problem > is obvious: Tony is talking about something else entirely. You're not > sure -- I'm not sure -- what the "+" is in "N+" It refers to the positive naturals, just as "Q+" denotes the positive rationals, "R+" denotes the positive reals, etc. Notice that mathematicians do not agree whether N should be {1,2,3,4,5,6,7,8,9,...} or {0,1,2,3,4,5,6,7,8,9,...}. If the former, then N+ = N and the "+" is redundant. But in either case, N+ = {1,2,3,4,5,6,7,8,9,...} regardless of which convention we're using, which is why we include it. This raises another question. I've noticed that those who are ZFC (Herc-)religionists have a tendency to identify N with omega, and thus they consider 0eN, since 0eomega is uncontroversial (since omega\{0} wouldn't even be an ordinal at all.) But those whom they call using five-letter insults tend to consider ~0eN. In particular, ~0eN to WM, and regardless of whether TO considers 0eN, he clearly prefers working with omega\{0} to omega itself. |