Uncountability of the Rationals?
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From: David R Tribble on 13 Jun 2010 15:56 Jesse F. Hughes wrote: >> I know what the function n |-> n^2 means for infinite values. I'm >> asking you know about the function sqrt. What is sqrt(omega)? What >> properties does it satisfy? > Tony Orlow wrote: > (x>1 -> sqrt(x)<x) ^ (omega>1) -> sqrt(omega)<omega > It's the size of a smaller countably infinite set, the set of squares > of naturals. All this does is state a property of something you call "sqrt(omega)". You left out the parts that a) actually define what sqrt(omega) is, and b) show that it actually exists. Oh, and you also neglected to mention if "x", "1", and "omega" are reals, naturals, ordinals, cardinals, bigulosities, or whatever. (You did state that "sqrt(omega)" is a "set size", but did not state what a "set size" is.) You've been asked this before: Are you saying that sqrt(omega) * sqrt(omega) = omega? If so, what do you mean, exactly, by "omega" and "*"?
From: David R Tribble on 13 Jun 2010 16:17 MoeBlee wrote: >> You simply POSIT that. It's not the case in ordinary mathematics, so >> if you want it, then you need axioms to prove it from. > Tony Orlow wrote: > Consider it a model of the naturals. n is the nth element of N. If the > initial sequence of size n exists, there exists the nth one. > Therefore, if the size is omega, then omega is a member of the set. > However, omega cannot be in the set, because the set contains all, and > only, the finite naturals, and no natural can possibly be the size, > because there is always a natural after it, since the set has no end. > Potential, but not actuality. You need to stop beating around the bush and just go ahead and state your primary axiom: Tony's Axiom. Given property P(x) for all real x, P(w) for any non-finite ordinal w. Everything you state is derived in some way from this assumption. You just need to stop assuming it and state it as an explicit axiom on your part. Now once you have this axiom (suitable rewritten in your inimitable style), you then proceed to built upon in, and demonstrate that it really can be used to build a consistent, coherent theory. And that means starting all of your discussions with "assuming Tony's Axiom, ...". I don't see how you're going to get much further, but that just means you've got your work cut out for you.
From: MoeBlee on 13 Jun 2010 16:23 On Jun 13, 4:02 am, Tony Orlow <t...(a)lightlink.com> wrote: > On Jun 11, 1:43 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > On Jun 11, 8:20 am, Tony Orlow <t...(a)lightlink.com> wrote: > Consider, to begin with, a monotonically increasing function, which of > necessity has a monotonically increasing inverse function. That should > be a good starting point. Yes, I "okay'd" your proposal that far and some more steps along the way too. Notice you're speaking of specific sets w, N and R and orderings on them. I take it that these are the ordinary sets w(the set of natural numbers), N (by your specification, I take it, the set of positive natural numbers) and R (the set of real numbers). If you don't have an axiom of infinity, then you'll have to tell me how you conclude that there are such sets. > > > then that inverse > > > formula may be used to calculate the number of elements in the set > > > within any given value range. > > > "may be used to calculate" (do you mean there is a recursive function, > > or is your notion of 'calculate' something different from Church- > > Turing?) > > No, I mean with a simple function which may be proved inductively to > be greater or less than some other function of the same variable, for > a number greater than any finite number. I mean, if f(x)>g(x) for all > x>a, for some finite a, then the same applies to a positive infinite a > (the size of an infinite set) as long as f(x)-g(x) doesn't have a > limite of 0 as x->oo. Then if you mean something so special by the word 'calculate' then why don't you introduce a new term and define it thus? So what do you have: "If f(x) > g(x) for all x > a for some finite a" You're still talking about a functions from w+ into R? So '>' is being used for both the standard ordering on w+ and on R? "then the same applies to a positive infinite a" Do you mean the following?: then there exists an infinite a such that f(x) > g(x) for all x>a. (1) There is no such x's in the domain if the domain is w+. So we must have the domain of f to be ordinal-greater than w+. And I take it '<' is ordinal-greater here, not cardinal-greater. (2) Anyway, with the domain thus modified, of course there is such an a, and you don't need the antecedent of your statement for that. > > "number of elements" > > Count. Member of N+. "Zize" of a set. Bigulosity. Basta? N+ is the set of all positive natural numbers? Or (I forgot from last time) you mean 'N+' to be w+ (i.e., the wu{w})? So either "count" means a positive natural number, or count means a countable ordinal, or something else? And that's all Biglosity is? > > "value range" > > a<x ^ x<=b = (a,b] I.e, {x | x in the range of f and f in (a b]} I guess? Be careful, just as a matter of keeping variables tidy, previously you were using 'x' for members of the domain. > > > Where a<b, > > > I take it '<' is the standard ordering on R. > > Yes. > > a<b ^ b<c -> a<c > ~(a<b ^ b<a) > ~(a<b v b<a) = (a=b) > > > > and where S is a set mapped > > > from the naturals using f(n)=x, > > > I take it S is a subset of R. > > For the purposes of IFR, yes. I am sure extensions are possible, but I > should establish this first. > > > > > f is the bijection you mentioned earlier? > > Um, yes? I'm doing you a favor by indulging you here. 86 the sarcastic "um"s. > f: N->S is half the bijection, along with g: S->N in the > other direction, S being a subset of R. We don't say "half the bijection" just because the inverse is also a bijection. f is a bijection. And the inverse of f is a bijection. Each is a bijection. There's no reason to say "half the bijection". > > > and where there exists an inverse > > > function g(x)=n such that f(g(x))=g(f(x))=x, > > > I.e., where g is the inverse of f. > > Yes. Okay, so would you now please restate the original statement you made, but this time with the ordinary terminology I just suggested (to which you said 'yes') and with the needed refinements just mentioned? > > > the number of elements in > > > the set S within the range [a,b] is given by floor(g(b))-ceiling(g(a)) > > > +1. Try it on any finite set. Now imagine applying it to the range > > > [0,omega]. > > > As far as I can tell, this is not well defined, since it is RELATIVE > > to f (g being the inverse of f). So what we would have is "number of > > elements of [a b] PER f" is [...]. But, even worse (to the extent I > > understand what you're trying to say) there IS NO bijection between [a > > b] and any segment of w, perforce no bijection between S and a segment > > of w. > > In the words of Scooby Doo, "Rrhuuhrrr?" I made a mistake there that I corrected in a later post. MoeBlee
From: Virgil on 13 Jun 2010 17:45 In article <5fdf516f-a370-4e9e-9e60-29f6caa68fe1(a)k39g2000yqd.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 11, 4:07�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <f13834be-7266-4324-9f8d-37ef78cae...(a)3g2000vbg.googlegroups.com>, > > �Tony Orlow <t...(a)lightlink.com> wrote: > > > > > The inverse function rule. If a set of reals is bijected with a > > > segment of the naturals using an invertible formula, then that inverse > > > formula may be used to calculate the number of elements in the set > > > within any given value range. Where a<b, and where S is a set mapped > > > from the naturals using f(n)=x, and where there exists an inverse > > > function g(x)=n such that f(g(x))=g(f(x))=x, the number of elements in > > > the set S within the range [a,b] is given by floor(g(b))-ceiling(g(a)) > > > +1. Try it on any finite set. Now imagine applying it to the range > > > [0,omega]. > > > > > I hope that is clear enough for you. > > > > A few questions: > > > > What do YOU mean by a "segment of the naturals"? > > Are such "segments" allowed to contain infinitely many naturals? > > Are they allowed to have lacunae? > > They must be contiguous segments of the naturals, sin lacunam, and may > be considered in the infinite case using ICI. > > > > > If they are limited to finite sets without internal gaps, why not > > restrict them to being finite initial segments, which involves no loss > > in generality. > > It does. Any set of contiguous naturals is order isomorphic to an initial set of contiguous naturals (allowing infinite sets in both cases) so there is no loss in generality in restricting to initial sets. > > > > Note than �f(g(x))=g(f(x))=x is nonsense, since f and g here have > > different domains and codomains, so that 'x' would sometimes be both a > > natural and non-natural real simultaneously. > > > > And that is from just a superficial look. > > Very much so, superficial, that is. I already explained that in > another post. Not adequately. Unless your f and g functions both have the same domain and the same image set, your f(g(x))=g(f(x))=x is necessarily false.
From: Virgil on 13 Jun 2010 17:51
In article <b30e5e51-8438-4f0a-93d6-ece7016bf28d(a)g19g2000yqc.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 11, 4:17�pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > Tony Orlow <t...(a)lightlink.com> writes: > > > On Jun 11, 8:47�am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > >> Tony Orlow <t...(a)lightlink.com> writes: > > Your description of ICI only allowed one to conclude that f(w) > g(w), > > given certain conditions. �It did not allow one to conclude that, if f > > is defined for finite values (and some other property?), then there is a > > unique extension of f to infinite values (such that so-and-so is true?). > > By 'w' you mean omega, right? Is that not an infinite value? It is not every infinite value. and if v > w. why must f(v) or g(v) be defined, much less having f(v) > g(v)? > > > > > > Okay, do you disagree with this statement? > > > > > AneN n<omega > > > > Yes. > > I hope you meant you agree... > > > > > > Do you disagree with this one? > > > > > AneN 1<n -> n<n^2 For w being omega or aleph_0, what definition of w^2 makes w < w^2 |