Uncountability of the Rationals?
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From: Tony Orlow on 14 Jun 2010 08:13 On Jun 13, 3:42 pm, David R Tribble <da...(a)tribble.com> wrote: > Virgil wrote: > >> Real functions do allow infinite quantities either as arguments or > >> values, so that you are making false statements. > > Tony Orlow wrote: > > 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. > > Virgil has the advantage of knowing more math than you, Tony. > You should not pretend to be an expert when you're not. It makes > you look silly and pretentious. It looks like he may also be able to tell who's talking, which gives him an advantage over you. HE'S the one saying real functions can take infinite argument or return infinite values. Look above. I am correcting that misstatement, as you are. You just can't read enough to see who made it. > > A real-valued function, by definition, has a domain from which it > takes its arguments, and a range (or codomain) of values it "returns" > (i.e., maps those arguments to). The function maps each argument > value from the domain to a single result value in the range. > > If the function has no mapping for a particular argument, then the > function is not defined for that argument value, and that value is > not within the domain of the function. > > Likewise, real functions take real arguments, which means that > infinity (+/-oo) cannot be in the domain set of a function, and thus > can never be an argument to a function. > > An example is the reciprocal function, f(x) = 1/x. The domain of f > is R \ {0}, i.e., all real values except 0. The function does not > "return infinity" for argument 0, rather, f is simply does not have > any result for it. It would be more correct to say that f(0) is > "nothing" > or "the empty set". Tell it to Virgil. > > Likewise, the real square root function, s(x) = sqrt(x) for all x >= > 0, > has the domain [0,oo). The function is not defined for any argument > outside its domain, such as -1, meaning that there is no value > for s(-1). It's not an "undefined value" of the function, rather the > function has no value to map that argument to. No kidding. Thanks for the kindergarten lesson. Show it to Virgil. > > It's amazing what you can learn with a little study: > http://en.wikipedia.org/wiki/Domain_of_a_function > http://en.wikipedia.org/wiki/Codomain > http://en.wikipedia.org/wiki/Analytic_function > > -drt Tony
From: Tony Orlow on 14 Jun 2010 08:24 On Jun 13, 3:56 pm, David R Tribble <da...(a)tribble.com> wrote: > 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)". Yes, it's less than omega, Bigulosity-wise due to ICI. > > You left out the parts that > a) actually define what sqrt(omega) is, and Suffice it to say it's countably infinite but less than omega > b) show that it actually exists. It exists as "actually" as omega, which only exists virtually. I've made that clear oodles of times. > > Oh, and you also neglected to mention if "x", "1", and "omega" are > reals, naturals, ordinals, cardinals, bigulosities, or whatever. sqrt is a real function for x>0. "Omega" is the size of N starting at 1, which isn't an absolute number, but an expression of potential infinity. > > (You did state that "sqrt(omega)" is a "set size", but did not state > what a "set size" is.) Bigulosity according to IFR and ICI. > > You've been asked this before: > Are you saying that sqrt(omega) * sqrt(omega) = omega? Yes, if such an operation is desired. For instance, I would say the set all pairs (x,y) where x and y are both squares of naturals would have a Bigulosity of omega. > If so, what do you mean, exactly, by "omega" and "*"? omega is the size of N starting at 1, and * is normal multiplication extended through ICI to establish order among infinities. Tony
From: Tony Orlow on 14 Jun 2010 08:38 On Jun 13, 4:23 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > 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. I'm sorry. It's my turn to cry "quantifier dyslexia". f(x)-g(x) > 0 for x>n|neN lim(x->oo: f(x)-g(x)) >0 Therefore A x>n f(x)>g(x) And since omega>n (and all infinite x>n) Then f(omega)>g(omega) > > (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. No ordinals. They have nothing to do with Bigulosity. > > (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})? N+ is, as others have used it, the set of naturals and hypernaturals, the infinite extension of w. > > So either "count" means a positive natural number, or count means a > countable ordinal, or something else? And that's all Biglosity is? Bigulosity is a better "set size" in the infinite case than cardinality. > > > > "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. > Isn't the value range under consideration a subset of the domain? 'x' is used for all sorts of things. > > > > > > > > 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. Thanks for the favor. > > > 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. Then don't. You understand what I mean. It is not a bijection unless the surjection goes both ways. > > 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". f is a function. If the inverse is a function, you have a 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? Why? So you can do me the favor of "indulging" me? > > > > > 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- Hide quoted text - > Thanx, TOny
From: Jesse F. Hughes on 14 Jun 2010 09:20 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 13, 3:56 pm, David R Tribble <da...(a)tribble.com> wrote: >> 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)". > > Yes, it's less than omega, Bigulosity-wise due to ICI. > >> >> You left out the parts that >> a) actually define what sqrt(omega) is, and > > Suffice it to say it's countably infinite but less than omega You keep dodging the issue. Is sqrt(omega) really the square root of omega? That is, is it a number satisfying sqrt(omega) * sqrt(omega) = omega? -- Jesse F. Hughes "Do not click any hyperlinks that you do not trust. Type them in the Address bar yourself." -- Microsoft gives security advice.
From: Tony Orlow on 14 Jun 2010 09:49
On Jun 13, 6:05 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 13, 4:52 am, Tony Orlow <t...(a)lightlink.com> wrote: > > > > > > > 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 you don't need that. Look at my later formulation below (or, > if I slipped up in it, just make obvious needed correction). > > > > 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 > > Right or not that this is equivalent to what you mean? Well, I am not defining cardinality, first of all. Secondly, one has to determing whether a given value is within the domain of g (not the range) somehow, and this is what is acheieved through the use of floor() and ceiling(). No correction is necessary there. > > It's true in the finite case. > As it was the way I wrote it. > And, if {xeS | x e [a b]} is not finite, then, I'd like to know how > you defined "-". Through ICI by proving inductively that one infinite expression is less than the other. It's an order relation. > > > > 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? > > > ICI contraidicts the model of the von Neumann ordinals, and extends > > natural density through IFR. > > Contradictions are in STATEMENTS, not in models. Models do not consist of statements and definitions? > > As to ICI and IFR, now that we've gotten SOME progress in making those > SOMEwhat more definite, why don't you just state your result and the > proof from your IFR (axiom?) and ICI (axiom?)? > > MoeBlee The result is that there exist many countable infinities distinguishable from each other, both greater than and less than omega (N+). One result is that there are half as many evens as naturals in N +. There are vastly fewer squares, and vastly more square roos of naturals, for instance. IFR makes distinctions down to a difference of a single element, N having one more element than N+ (0, that is). Surely you don't expect me to list all of the countably infinite set of results that arise? Tony |