Uncountability of the Rationals?
in [Math]
|
Prev: Collatz conjecture
Next: Beginner-ish question
From: Jesse F. Hughes on 11 Jun 2010 11:50 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 11, 12:33 am, David R Tribble <da...(a)tribble.com> wrote: >> David R Tribble writes: >> >> So then what is the *general* case for IFR? >> >> Jesse F. Hughes wrote: >> > Could someone remind me what IFR stands for? >> >> I don't remember what Tony's "IFR" stands for, either. Something >> about a "formulaic ratio" or some such. >> >> The funny thing about this is that this does not seem to matter >> in the least. Tony keeps saying that IFR solves all sorts of >> problems and "obfuscations" with Cantor's cardinalities, yet >> he hasn't given a single concrete example of it. > > So, you can't even remember what it stands for or what it means, but > you are sure it's incosistent. You mean that one can't judge IFR unless he knows what the initials stand for? Weird. -- Jesse F. Hughes "Doesn't pay to lie if you aren't good at it." -- Captain Friday, /City of the Dead/ Adventures by Morse radio show
From: Jesse F. Hughes on 11 Jun 2010 12:00 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 10, 5:30 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> David R Tribble <da...(a)tribble.com> writes: >> >> > Tony Orlow wrote: >> >> No, That's IFR in particular which is used on countable sets of reals >> >> mapped from some segment of N. >> >> > So then what is the *general* case for IFR? >> >> Could someone remind me what IFR stands for? >> >> -- >> "I've noticed [...] I routinely have been putting up flawed equations >> with my surrogate factoring work. My take on it is that I have some >> deep fear that the work is too dangerous and am sabotaging myself." >> -- James S. Harris > > 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. What's an invertible formula? > 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. f(g(x)) cannot equal g(f(x)), can it? The functions are f:N -> S and g:S -> N, so f(g(x)) is in S, while g(f(x)) is in N. > Try it on any finite set. Now imagine applying it to the range > [0,omega]. > > I hope that is clear enough for you. I'm not sure. Let's try. Let S = {1/4, 1/16, 1/64}. That's a finite set. Let g:S -> N be defined by g(x) = x^{-1/2} and f:N -> S be defined by f(n) = n^-2. Then f(g(x)) = (x^{-1/2})^{-2} = x and similarly g(f(n)) = n. Now, what do you mean by the number of elements in the set S within the range [a,b]? I think you mean S n [a,b]. Is that right? (I'm confused by your use of the term "range".) Let's try. S n [0,1] = S, of course. Is that given by floor(g(b))-ceiling(g(a)) +1? floor(g(1)) - ceiling(g(0)) + 1 = floor(1^-1/2) - ceiling(0^-1/2) + 1 = 1 - 0 + 1 = 2. Well, that didn't work. How about looking at [1/64,1/4]? floor(g(1/4)) - ceiling(g(1/64)) + 1 = floor(2) - ceiling(8) + 1 = -5 Okay, now I'm sure. I'm sorry, but I really *don't* have any idea what you meant by the above. Can you be a bit clearer about it? -- "So now, The Hammer is here, and with it, the end of days. The world will be destroyed, and then remade, as foretold. You will be lost, with your children, and then there will be others, and one day they will be tested, and will pass, but that is another story." --James S Harris gets a bit excited.
From: David R Tribble on 11 Jun 2010 12:25 Tony Orlow wrote: >> E 0 >> E 1 >> 0<1 >> x<y -> E z: x<z<y >> This leads to some level of infinites between counting numbers. > David R Tribble wrote: >> [...] Tony seems to be missing an axiom or >> two for defining anything beyond 0, 1, and <. Maybe I just missed >> it in the first reading. I'll check again. Maybe I can also find those >> counting numbers and levels of infinities he mentions. > Brian Chandler wrote: > I think you are being *very* slightly unfair. Tony's writing style is > a bit strange, but surely the above can be seen as a definition of UI, > the Unit Interval. > > UI is a set of elements, with a total ordering < such that: > UI includes an element 0 > UI includes an element 1 > If x and y are elements of UI and x < y, then there exists an element > p of UI such that x < p < y. Yes, but he's still not telling us if the elements in his system are reals, naturals, hyperreals, ordinals, or whatever. We can guess that he's talking about the usual reals, but it costs him nothing to actually say so. On the other hand, he may really believe that he can define a complete number system from scratch with so few axioms. He hadn't grasped yet that he has to be more explicit about all his assumptions and "hidden" axioms. For example, he could simply add: x, y, z are members of R or whatever set of elements he's defining his system for. It's telling that he either can't be bothered with such details, or he expects everyone to know what his unstated assumptions are. > My _guess_ is that Tony thinks this gives him something whose > bigulosity he can "declare", despite the fact that there are any > number of sets of different cardinalities even which match this > definition. I believe he also "declares" that the bigulosity of UI is > in some simple relation to the bigulosity of the TNaturals, or > something similar. It looks to me like Tony still thinks that the Peano Axioms, or similar inductive definitions, can lead to "unit infinities". This subtle distinction, btw, appears to be completely lost on Walker. He is correct in goading Tony to take the approach of defining a primitive element (zillion, Big'un, or whatever) and proceed from there. (Indeed, that's exactly what I did with my suprareals.) But I don't think he realizes that Tony believes that infinite numbers are part of the existing naturals (at least as far as I know, last time I checked; he may not be saying this any more). > But this has become sadly repetitive. Reading the old threads is just > as entertaining, and less work. Well, when I stop learning interesting math from these threads, or just plain stop being entertained by them, that's when it will become a waste of time for me. -drt
From: David R Tribble on 11 Jun 2010 12:45 Tony Orlow wrote: > I would like you consider a slightly modified goal: To exactly order > sets according to size, and to distinguish between those that have an > absolute size and those that do not. For me, no countably infinite set > has an absolute size, but only one relative to other countably > infinite sets, especially the standard N, starting at 0. Wouldn't it be easier to say that some chosen countably infinite set has a standard "absolute" size, and then that all other such sets have a size that is some fraction of that size? Thus every set then has an "absolute" size, based on the standard set size. For convenience, you'd want to choose the "largest" countably infinite set for the standard "absolute" set size. This would probably be N, since it contains all of the countable naturals. (Another logical choice would be Z.) Given this, you could then say that every countably infinite set does indeed have an "absolute" set size. Of course I've glossed over all the important details, but that's really your job to provide.
From: Jesse F. Hughes on 11 Jun 2010 12:44
Tony Orlow <tony(a)lightlink.com> writes: > On Jun 11, 12:21 am, David R Tribble <da...(a)tribble.com> wrote: >> Tony Orlow wrote: >> > For, "none >> > shall drive us from the Garden which Cantor has created for us". If it >> > doesn't produce fruit, it's time to plant a new bed, or at least >> > fertilize. >> >> Well, it's done pretty well for the last 100+ years. Have any >> of your ideas produced any results of consequence yet? >> >> I'm still waiting to hear what bigulosity/IFR says about the size >> of the set of natural squares. Any results yet? > > I already did that. According to IFR it's sqrt(omega), which according > to ICI is less than omega. So, it is a(n ordinal) number x such that x * x = omega, is that right? I'm not clever like you, but I'd wager that one can prove no such ordinal exists, when * stands for ordinal multiplication. Does that bother you? As well, one can prove that omega is the least infinite ordinal (where "least" is defined in terms of cardinality, of course). That is, if alpha is an infinite ordinal, then there is an injection from omega to alpha. Is this contrary to your claims? You sometimes say that you don't think omega is the smallest infinite ordinal, but I'm not sure what you mean when you say that. -- "I'm a very well-educated, successful, intelligent person. This is insane to me that I have an armed guard outside my door when I've cooperated with everything other than the whole solitary-confinement- in-Italy thing." --A. Speaker, on the whole T.B.-quarantined thing |