Uncountability of the Rationals?
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From: MoeBlee on 11 Jun 2010 15:33 On Jun 11, 2:07 pm, Tony Orlow <t...(a)lightlink.com> wrote: > 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? What's "ICI"? > Okay, do you disagree with this statement? > > AneN n<omega What does '<' stand for? If it stands for the ordering cardinal ordering (and below too), then the answer is 'yes'. > Do you disagree with this one? > > AneN 1<n -> n<n^2 Your notation is errant. AneN (1<n -> n<n^2) > Not so far, right? > > How's about we talk about the set of all set sizes, or "counts", Where did such a set come from? In Z set theories, there is no such set. Do you mean the set of all FINITE set sizes. Then, okay, that's w itself again. Or do you mean the set of all COUNTABLE set sizes? Then that's w+. > and > call it N+? Now, you undoubtedly disagree with this statement: So you mean the set of all COUNTABLE set sizes? > AneN+ 1<n -> n<n^2 You mean: Anew+ (1<n -> n<n^2) That's your claim. Or that's what you would LIKE to be the case. So, fine, tell us some axioms from which we can prove it is the case. > 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? I don't see anyting about transfinite induction in the above. All you've done in a LONG WINDED way is point out: Anew (n>1 -> n < n^2) but that in Z set theory ~ w < w^2 (where '<' and '^' stand for the cardinal relation and operation) and that you don't like that result. But you also say you don't dispute ZFC itself but only the way it's "extended" (or whatever you say, and whatever you MEAN by that). But ~ w < w^2 is just a theorem of ZFC. MoeBlee > > 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. What is that meant to convey? > For every countably infinite set S whose membership is defined by > monotonically increasing function f:N -> R, Maybe you mean: For every S that is the range of an increasing function f:N -> R. (And, yes, S is therefore denumerable). > which of necessity has > real inverse function g:seS -> N, Maybe you mean: Let g = the inverse of f. > the count of elements in the range > [a,b] Do you mean the count of [a b] (but there is no such countable "count") or 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. MoeBlee
From: Virgil on 11 Jun 2010 15:49 In article <e84e2ffc-fc2a-4008-8ff6-e9d097aed10a(a)g19g2000yqc.googlegroups.com>, Tony Orlow <tony(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. > > > > > > > > (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)? > > > > > > > > (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.
From: MoeBlee on 11 Jun 2010 15:52 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 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
From: Virgil on 11 Jun 2010 15:54 In article <cd83f80f-337f-40d5-8ae5-764b216c94bc(a)j4g2000yqh.googlegroups.com>, Tony Orlow <tony(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.
From: Virgil on 11 Jun 2010 16:07
In article <f13834be-7266-4324-9f8d-37ef78cae52b(a)3g2000vbg.googlegroups.com>, Tony Orlow <tony(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? 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. 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. |