Uncountability of the Rationals?
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From: Tony Orlow on 13 Jun 2010 07:50 On Jun 11, 3:49 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <e84e2ffc-fc2a-4008-8ff6-e9d097aed...(a)g19g2000yqc.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> wrote: > > > > > > > On Jun 10, 3:40 pm, Virgil <Vir...(a)home.esc> wrote: > > No, they are exactly invertible functions, as are used in any > > bijection. IFR is simply an extension of bijection and an extension of > > natural density which derives more distinctions than either of those. > > > > > (1) It can be classically inductively proven that f(x)>g(x) for every > > > > real value greater than some finite x (the base case), and > > > > Does that imply that these are real functions? > > > Yes, they can be any invertible arithmetic functions. Does that pose a > > problem for you? Probably, since I am applying them to infinite > > values. However, that's what infinite case induction is for. > > Real functions do allow infinite quantities either as arguments or > values, so that you are making false statements. > Don't correct me incorrectly. Standard real functions take real value parameters and return a real result. Where they have return an infinite value, they are declare "undefined" or to have some infinite "limit" at that point, and where any parameters are of infinite nature, the modern mathematical procedure is to refer to the "limit" of the function as one or more variables approach oo or -oo. To lie by accusing someone is lying is...lame. > > > > > (2) The difference between the two terms which establishes the > > > > inequality does not have a limit of zero as x approaches infinity, > > > > then > > > > What is the "arithmetic" of the set in which those differences occur? > > > The arithmetic of the set? The difference between f(x) and g(x) is > > f(x)-g(x) for any given natural x. If the limit of f(x)-g(x) tends to > > zero as x tends to oo, then infinite case induction fails. Otherwise > > it holds. > > What if that "difference" were something like (-1)^x (which is quite > possible for the difference between two monotonically increasing > functions)? > Hmmm, an example might be nice. Let's see. We have N mapped to {2,1,4,3,6,5...}. Well, by the axiom of extensionality, that is N. They are same in the infinite case, or any even case. It does not have a limit of 0, so it may be discounted in ICI, but the axiom of extensionality pretty much resolves the issue. > > > > > > (3) If omega is greater than any finite counting number, or c or > > > > aleph_1 (whether they are the same or not), then the same rule also > > > > applies to such large numbers. > > > > > Please let me know which step seems vague. > > > All of them. > > > I hope I clarified some of that for you. > > Still All of them. Whatever that means. Tony
From: Tony Orlow on 13 Jun 2010 07:52 On Jun 11, 3:52 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 11, 2:33 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:> On Jun 11, 2:07 pm, Tony Orlow <t...(a)lightlink.com> wrote: > > do you mean the count of > > > > > {xeS | x e [a b]} > > > ? > > > > is given by floor(g(b))-ceiling(g(a))+1. > > > But this is not well defined, because there are MANY functions f such > > that f is an increasing function from N into R. And for each such f we > > get a different g. > > Wait, I think I'm incorrect. > > Okay for a given S, there is only one increasing function f on w that > gives S as the range. > > But I don't know why you refer to "floor" and "ceiling" the way you > do. The values of g(a) and g(b) are NATURAL NUMBERS. I refer to floor() and celing() to account for one choosing any given range, whether or not its extrema are inverse members of N. > > I think what you mean is > > card({xeS | x e [a b]}) > = > (max(range(g restricted to [a b])) - min(range(g restricted to [a b]))) > +1 > > Yes, that is just a very roundabout way of restating the cardinality > of some finite subset of a closed real interval. > > Now, what exactly is the rest of your proposal? And does it contradict > ZFC or not? > > MoeBlee ICI contraidicts the model of the von Neumann ordinals, and extends natural density through IFR. TOny
From: Tony Orlow on 13 Jun 2010 07:53 On Jun 11, 3:54 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <cd83f80f-337f-40d5-8ae5-764b216c9...(a)j4g2000yqh.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> wrote: > > > On Jun 10, 4:57 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > To have one and to be one are pretty different. Anyway, yes, > > > continence has never seemed your problem; the exact opposite. > > > Don't forget to wipe this week. > > I was always under the impression that it was the responsibility of he > who was incontinent to wipe himself. Perhaps several times a day, not once a week. Is this thing on? (tap tap) TOny
From: Tony Orlow on 13 Jun 2010 07:57 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. I am offering a formula for calulating how many members of a countably infinite set lie within a given value range, not just an "initial" value range [1,x]. > > 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. Tony
From: Tony Orlow on 13 Jun 2010 08:33
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: > >> >> > (1) It can be classically inductively proven that f(x)>g(x) for every > >> >> > real value greater than some finite x (the base case), and > > >> >> Does that imply that these are real functions? > > >> > Yes, they can be any invertible arithmetic functions. Does that pose a > >> > problem for you? Probably, since I am applying them to infinite > >> > values. However, that's what infinite case induction is for. > > >> No, your description of infinite case induction only gave certain > >> sufficient conditions for concluding that f(omega) > g(omega). It does > >> *not* give any reason to believe that a function f defined on natural > >> (or real) numbers can be extended to infinite sets in some unique way. > >> You are presuming that without stating it explicitly. > > > Please reread what I wrote above. I am saying that for IFR the two > > mapping functions are mutual inverses, which means they form a > > bijection. I was saying that your problem with my idea occurs when I > > apply these functions to infinite values, because you don't consider > > them to be defined for nonstandard values. I am telling you that this > > application to the infinite for the purposes of IFR is acheived > > through the application of ICI. Do you understand what I wrote now? > > No. whew! okay... > > 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? > > > 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 > > Okay. Okay. > > > Not so far, right? > > > How's about we talk about the set of all set sizes, or "counts", and > > call it N+? Now, you undoubtedly disagree with this statement: > > > AneN+ 1<n -> n<n^2 > > Of course I disagree if the operations and < relation are the obvious > ones, but this isn't really the meat of my question. It's the crux of the disagreement. > > 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? > (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. > > > > > > It runs completely counter to cardinality, but is a much cleaner > > extension of finite induction than transfinite induction IMHO. Just > > pointing out exactly where our differences lie. There's a concrete > > example. > > >> What you really want, but are not stating, is something like this: > > >> For every (invertible?) function f:R -> R, there is a unique extension > >> f':? -> ? (Ord -> Ord?) such that ... (what?). > > > Exactly!!!! :D You hit the nail on the head. Or, with your head. > > > For every countably infinite set S whose membership is defined by > > monotonically increasing function f:N -> R, which of necessity has > > real inverse function g:seS -> N, the count of elements in the range > > [a,b] is given by floor(g(b))-ceiling(g(a))+1. Do you care to dispute > > whether this works, in the finite case? Please do comment. > > Let's see if I understand. > > Let f:N -> R be monotonically increasing and S = range(f). Let > g:S -> N be the inverse of f. Then The set S does not include all values in the range, and incudes values outside of any finite range. But, continue... > > S n [a,b] = floor(g(b))-ceiling(g(a))+1. Yes, okay 'n' denotes set intersection > > As written, this doesn't make sense, because you didn't restrict a and b > to be elements of S, but dom(g) = S. You have assumed (without > argument) that g has a unique extension to R (satisfying something?). When one says, "g:S -> N", that restricts the domain of g to S. However, a and b need not be in S, but may be any real values, hence the necessity of the floor() and ceiling() functions for arbitrarily chosen intervals. > > But let's look at a case where g is indeed definable on R in a natural > way. Let f:N -> R be defined by f(n) = n^2 and g:R -> N be the sqrt > function. Then, for instance, S is the set of squares of natural > numbers and > > S n [a,b] = floor(sqrt(b)) - ceiling(sqrt(a)) + 1. > > Yes, I think that looks correct, but for the assumption that g is > somehow naturally extended to R. I think that what you have in mind > really is a function f:R -> R with inverse g:R -> R. To see why your > statement doesn't make much sense, let > > f:N -> R > > be defined by f(n) = nth prime. This is an increasing function, and > hence there is an inverse g:Prime -> N, but it makes no sense to ask > what g(b) is unless b is a prime. It certainly makes sense to ask how many prime numbers exist within any given range, but until something new is discovered about primes, the best that can be done with them is to talk about their aymptotic relation to the full set of naturals as one approaches N. There is no algrebraic mapping there. So, IFR doesn't apply. There are various ways to extend g to > R. Here are three: > > g_1(b) = number of primes less than or equal to b. > > g_2(b) = 1 + number of primes less than b > > g_3(b) = g_1(b) if b <= 4 > g_2(b) if b > 4 > > So, as we can see, you have not quite correctly stated the context of > your claim. It will be another thing entirely to tell me what the > claim means when we move to infinite arguments. Please state to me what the function is which maps N to the set of primes, derive the inverse function (it's monotonically increasing, for sure), and then you'll have your answer. It should come out to about 1/logx for the inverse function, at least in the infinite case... TOny |