An uncountable countable set
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From: stephen on 30 Oct 2006 23:23 Tony Orlow <tony(a)lightlink.com> wrote: > stephen(a)nomail.com wrote: >> Tony Orlow <tony(a)lightlink.com> wrote: >>> stephen(a)nomail.com wrote: >>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>> stephen(a)nomail.com wrote: >>>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>>> stephen(a)nomail.com wrote: >>>>>> <snip> >>>>>> >>>>>>>> What does that have to do with the sets IN and OUT? IN and OUT are >>>>>>>> the same set. You claimed I was losing the "formulaic relationship" >>>>>>>> between the sets. So I still do not know what you meant by that >>>>>>>> statement. Once again >>>>>>>> IN = { n | -1/(2^(floor(n/10))) < 0 } >>>>>>>> OUT = { n | -1/(2^n) < 0 } >>>>>>>> >>>>>>> I mean the formula relating the number In to the number OUT for any n. >>>>>>> That is given by out(in) = in/10. >>>>>> What number IN? There is one set named IN, and one set named OUT. >>>>>> There is no number IN. I have no idea what you think out(in) is >>>>>> supposed to be. OUT and IN are sets, not functions. >>>>>> >>>>> OH. So, sets don't have sizes which are numbers, at least at particular >>>>> moments. I see.... >>>> If that is what you meant, then you should have said that. >>>> And technically speaking, sets do not have sizes which are numbers, >>>> unless by "size" you mean cardinality, and by "number" you include >>>> transfinite cardinals. >> >>> So, cardinality is the only definition of set size which you will >>> consider.....your loss. >> >> If somebody presents another definition of set size, I will >> consider it. You have not presented such a definition. >> >> > I have presented an approach that works for the majority of infinite > bijections, and explained some of the exceptions. Given that you cannot answer how many Turing machines are, or how many sets of prime numbers there are, or pretty much any set that is not a member of P(N), I do not see how you can claim it works for sets in general. <snip> >>>> In any case, it still does not make any sense. I am not sure >>>> what |IN| is for any n. IN is a single set. There is only >>>> one set, and it does not depend on n. In fact, there isn't >>>> an n specified in the problem. Yes I used the letter n in >>>> the set description, but that does not define an entity named 'n'. >>>> >> >>> There most certainly is an 'n'. The problem describes a repeating >>> process, each repetition of which is indexed with a successive n in n, >>> and during each repetition of which ball n is removed. What do you mean >>> there's no n??? >> >> I am talking about two sets that I have explicitly defined. >> There is no 'n'. I have been very clear about what I am >> talking about. I cannot help it if you cannot get over >> your fixation with balls and vases. >> > Yes, from that perspective, they are the same set. But, there is an n in > the original problem, I am not talking about the original problem. How many times do you have to be told that? <snip> >> So here is one last attempt. The balls and vase problem >> is not a physical problem. > No kidding. And I was going to set up a test bed in the basement.... Who are the one applying physical reasoning to it. > Pretending that it is a physical >> problem is pointless. It is phrased in such a way to suggest >> a physical problem, so that it might confuse your physical >> intuitions, but it is not a physical problem. > Show me where your formulation relates t and n. Where is the schedule > stated in that formulation? It's not. Try again. I never said that was a formulation of the problem. It is however an accurate description of the set of balls that are added to the vase, and the set of balls removed from the vase. But that was not the point of the question. >> >> Even 'finite approximations' of it are not physical problems. >> Consider the case where you start adding balls at 11pm. >> Suppose you had 2000 balls. The last set of 10 balls will >> be added at time at 10^-56 seconds before noon. But that >> is less than Planck time, and is therefore impossible according >> to known physics. How big are these balls? Lets suppose >> they are 1cm in diameter, and moving them into the vase requires >> the ball be moved 1cm. Once you get up to ball 350 or so, >> you will have to move the balls faster than the speed of light. > Blah blah blah. Thanks for the reality lesson. Now take a lesson in realism. More pointless insults. >> >> This is simply not a physical problem. It is an abstract >> problem. It is a mathematical problem. The "physical" part >> is just a distraction. > So is all your blabbering.... More pointless insults. You seem to be incapable of discussing this rationally. >> >> So if instead, someone had just posed this problem >> Let >> IN = { n | -1/2^(floor(n/10)) < 0 } >> OUT = { n | -1/2^n < 0 } >> >> What is | IN - OUT | there would be controversy. >> Note, there are not balls or vases, or times or anything >> in this problem. Just two sets. It would help if you >> bother to stop and think for a second before responding. >> >> > Where are the iterations mentioned there? What iterations? I am not talking about iterations. I am just posing a problem about sets that should be totally uncontroversial. > You're missing the crucial > part of the experiment. By your logic, you could put them in in any > order and remove them in any order, and when you say both processes are > done, nothing's left, but that's BS. It ignores the sequence specified. > This is just a distraction. No, the words "balls" and "vase" are the distraction. If you want to avoid distractions, phrase the question strictly in mathematical terms. >>>>>>>> Given that for every positive integer -1/(2^(floor(n/10))) < 0 >>>>>>>> and -1/(2^n) < 0, both sets are in fact the same set, namely N. >>>>>>>> >>>>>>>> Do you agree, or not? Or is it the case that the >>>>>>>> "formulaic between the sets is lost." >>>>>>>> ? >>>>>>>> >>>>>>>> Stephen >>>>>>> The formulaic relationship is lost in that statement. When you state the >>>>>>> relationship given any n, then the answer is obvious. >>>>>> What relationship? For a given n, -1/(2^(floor(n/10))) < 0 >>>>>> if and only if -1/(2^n) < 0. The two conditions are logically >>>>>> equivalent for positive integers. If n is a member of IN, >
From: Tony Orlow on 30 Oct 2006 23:35 David R Tribble wrote: > Tony Orlow wrote: > David R Tribble wrote: >>> You misunderstand. Your H-riffics are simply finite-length paths >>> (a.k.a. the nodes) of a binary tree. Your definition precludes >>> infinite-length paths as H-riffic numbers. > > Tony Orlow wrote: >>> What part of my definition says that? For the positives: >>> >>> 1 e H >>> x e H -> 2^x e H >>> x e H -> 2^-x e H > > Tony Orlow wrote: >>> Exactly. If you list these H-riffic numbers as a binary tree, each one >>> is a node in the tree along a finite-length path. > > David R Tribble wrote: >>> If you extend your definition and allow the H-riffics to include the >>> infinite paths of your binary tree as well as the finite-length paths >>> (nodes), then, yes, the H2-riffics are uncountable and probably >>> cover the reals. >>> But then the H2-riffics [H-riffics?], like the reals, are not well-ordered. >>> You're back to square one. > > Tony Orlow wrote: >> Yes, you're right, and I admitted that in "Well Ordering the Reals". One >> can linearly order the reals in this way, but eliminating any "countably >> infinite descending sequences" appears to be a matter of infinite >> regression, and I don't see how to prove that it's a *countably* >> infinite regression. If it were, then perhaps it could be considered >> well ordered, but well ordering an uncountably infinite set requires >> predecessor discontinuities as one finds in the limit ordinals. Really, >> despite the Axiom of Choice, I don't see that it's possible to >> explicitly state a well ordering on any uncountable set, besides the >> concoction of ordinals. So, I agree, on that point. >> >> See? I can admit failure, especially when I failed in an experiment that >> didn't guarantee success. :) > > David R Tribble wrote: >>> (If you don't see this, consider the H2-riffic, which we'll call p, as >>> the infinite path in your binary tree, where each successive fork taken >>> is left (2^x) if the next digit in the binary fraction for pi is 0, and >>> the right fork (2^-x) if the next digit is 1. Then we ask, what is the >>> successor to p in the well-ordering? There is no way to know, just >>> as there is no way to name the successor of pi in the usual ordering >>> of the reals.) > > Tony Orlow wrote: >> Oh. You're using "H2-riffics" as the uncountable set. Okay. I get it. >> >> However, you're wrong here. Sorry. There are two successors to every >> H-riffic h, 2^h and 2^-h. So, successors to pi are 2^pi and 1/2^pi. >> Those exist, don't they, as reals? I am sure they (those points on the >> real line) can be calculated (specified) to any arbitrary accuracy. No? > > No, there are no successors in your tree for that number p, > because there is no last L/R branch in the path defining that > H2-number. So there can be no different or next branching for > the "last" node for either successor of p, because there is no last > node in the path. > > I'm talking over your head here, because you obviously do not > follow what I'm saying. > Geeze, David, flatter yourself a little, why don't you? There, have a good time. Okay, done? First, I'd like to thank you for pushing this point. The H-riffics failed as a well ordering, and honestly, my conclusion for the time being is that an explicit well ordering of any uncountable set depends on the kind of predecessor discontinuities which forge paradoxes out of ordinals. I'm really not interested in a lack of infinite descending chains. I'm not shackled by that restriction. But, I AM interested in linear orders, and actual uncountability, and what that means... You are upset because what is clearly an uncountably long string might have a successor. That's understandable. But, allow me to elaborate a tad. I'll break into pieces so you can respond bit by bit: The function 2^x, as a real function, is continuous over R. It goes from an asymptotic y of 0 as x->-oo to an asymptotic y of oo as x->oo, and no discontinuities in between, monotonically increasing from -oo to oo. Right? Since the function is monotonically increasing, one can always find an intermediate value between a and b, which is 2^((log(a)+log(b))/2), that is, f((g(a)+g(b))/2), where f(x)=2^x, f(g(x))=g(f(x)), xeR. Right? Does that not make this set continuous over the range of reals? So, we're left with a little conundrum. Every unique real x has a unique real y such that 2^x=y, and such that -2^x=-y. So each real has two successors. We can't pick a point where that's not true. We can start anywhere (of course, zero's a good spot to start), but in any case, in enumerating this uncountable set, we're going to get to uncountable strings, which have successors. David - If every element in the reals has at least one successor, what does that say to you? TOE knee
From: Virgil on 30 Oct 2006 23:43 In article <4546bfef(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > I have presented an approach that works for the majority of infinite > bijections, and explained some of the exceptions. Not to the satisfaction of anyone except TO, or possibly Ross. > IFR works for all > numeric sets mapped from a common set. But ignores everything that is not naturally ordered like a subset of the reals, which excludes such basics as vector spaces of more than one dimension and most of category theory, among other things, and so TO's toys are useless for any serious mathematical purposes. > N=S^L works for all languages Only when N, S and L are all finite, and then it is trivial. > So, I think Bigulosity is worth > considering. What TO thinks is, in general, not worth considering > > >>> In any case, it still does not make any sense. I am not sure > >>> what |IN| is for any n. IN is a single set. There is only > >>> one set, and it does not depend on n. In fact, there isn't > >>> an n specified in the problem. Yes I used the letter n in > >>> the set description, but that does not define an entity named 'n'. > > I am talking about two sets that I have explicitly defined. > > There is no 'n'. I have been very clear about what I am > > talking about. I cannot help it if you cannot get over > > your fixation with balls and vases. > > > > Yes, from that perspective, they are the same set. So nice of TO to concede what he has been so foolish as to deny. > But, there is an n in > the original problem, which this characterization ignores. Irrelevant to this characterization, but irrelevant anyway. TO will not, or cannot answer: Is there any ball which the original problem requires to be inserted into the vase before noon that the original problem does not require to be removed from that same vase before noon? Until TO can honestly answer that question, he is not dealing with the original problem but only with hobgoblins of his own imagining. > > > > You really are not paying attention, are you? As I said earlier, > > if you are not going to bother to read what I write, you should > > not bother responding. > > > > You should make it pertinent, and respond to the obvious holes in its > relevancy. To is often impertinent, particularly with respect to this vase problem, so he has no right to demand what he is so patently incapable of providing himself. > > > So here is one last attempt. The balls and vase problem > > is not a physical problem. > > No kidding. And I was going to set up a test bed in the basement.... Please do so, and don't bother us again until it is completed. > Show me where your formulation relates t and n. Where is the schedule > stated in that formulation? It's not. Try again. Show me any formulation of that problem that allows a ball inserted before noon to remain until noon. > Blah blah blah. Thanks for the reality lesson. That TO will not listen to those who are wiser than he is only one of his many faults. > Where are the iterations mentioned there? You're missing the crucial > part of the experiment. TO misses the crucial essential part where each ball inserted is also removed before noon. > >> I know you think that logic is valid, as that's what you've been taught, > >> and it sounds nice and clean, but bijections without regard to their > >> formulaic mappings do not provide measure over infinite sets. That's why > >> I keep "babbling" about rates and variables. I'm trying to sprinkle some > >> math in the cauldron. TO is sprinkling way too much eye of bat and toe of frog in his futile efforts to deny the clear requirement that every ball inserted into the vase before noon is removed before noon.
From: Virgil on 30 Oct 2006 23:53 In article <4546c0bb(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > stephen(a)nomail.com wrote: > > Tony Orlow <tony(a)lightlink.com> wrote: > >> stephen(a)nomail.com wrote: > >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>> imaginatorium(a)despammed.com wrote: > >>>>> Tony Orlow wrote: > >>> <snip> > >>> > >>>>>> The formulaic relationship is lost in that statement. When you state > >>>>>> the > >>>>>> relationship given any n, then the answer is obvious. > >>>>> Do "state the relationship given any n"... I mean, what is it, exactly? > >>>>> > >>>> Uh, here it is again. in(n)=10n. out(n)=n. contains(n)=in(n)-out(n)=9n. > >>>> lim(n->oo: contains(n))=oo. Basta cosi? > >>> > >>> What is in(n)? The sets I and everyone but you are talking about are > >>> IN = { n | -1/2^(floor(n/10)) < 0 } > >>> OUT = { n | -1/2^n < 0 } > >>> Noone has ever mentioned or defined in(n) > >>> > >>> What is the definition of in(n)? Is is a set? > >>> > >>> Stephen > >>> > >> out(n) is the number of balls removed upon completion of iteration n, > >> and is equal to n. > > > >> in(n) is the number of balls inserted upon completion of iteration n, > >> and is equal to 10n. > > > > But there are no balls or iterations in the problem I posed. > > So why do you keep talking about balls and iterations? > > Are you really that incapable of participating in a discussion? > > > > Are you incapable of responding to a particular formulation of the > problem? TO is incapable of responding sensibly to the explicit requirement in the original gedankenexperiment that every ball which is to be inserted into the vase before noon is also to be removed from the vase before noon. > Did you start the discussion? What makes you think you can > steer it? We are merely trying to get TO to acknowledge certain requirements of the original gedankenexperiment which he is so carefully ignoring. > > > > The balls and vase problem has absolutley nothing to do > > with physics. Try it out. > > Don't be an idiot. What makes you think my objection has anything to do > with physics? As it has nothing to do with logic or mathematics or the original gedankenexperiment, it would be nice if it had something to do with something, but it appears not to. >It has to do wih properly representing the problem > mathematically without ignoring pertinent details, like the times of > insertion and removal for each ball. You mean like the requirement that each ball to be inserted before non is also to be removed before noon, at least according to the original gedankenexperiment? To seems to keep forgetting that point. > Your example without this > information is irrelevant to the problem. TO's objections. which so deliberately overlook this critical point, are totally irrelevant to the problem. > > So, what's wrong with my logic above, besides the fact that it's not > what YOU are trying to spew? That it deliberately overlooks that point that every ball which is given a time of entry into the vase before noon is also given a time of exist from the vase before noon. That TO's "logic" continually and deliberately omits that requirement, makes it illogical.
From: Virgil on 31 Oct 2006 00:01
In article <4546d27b(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > I'm talking over your head here, because you obviously do not > > follow what I'm saying. > > > > Geeze, David, flatter yourself a little, why don't you? To suppose one to be talking over TO's head is hardly self flattery, except possibly for Ross or Zick or their ilk. |