An uncountable countable set
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From: Tony Orlow on 25 Oct 2006 14:35 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. > > Tony Orlow wrote: >> David, I think I figuredout what your confusion is on this. I presented >> the H-riffics in "Well Ordering the Reals" as a proposed well ordering. >> As it turns out, indeed, it would seem that the well ordered version of >> this set would have to be countable, as you say, much like the >> rationals, or rather, the finite length decimal reals. But, well >> ordering be damned, if infinite-length strings are allowed in the >> H-riffics, it covers the uncountable reals. > > 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. Thank, you, David. I am quite sure they do. :) > > But then the H2-riffics [H-riffics?], like the reals, are not well-ordered. > You're back to square one. 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. :) > > (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.) > 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? Tony
From: Tony Orlow on 25 Oct 2006 14:36 David R Tribble wrote: > Tony Orlow wrote: >>> In order for all the naturals to be removed, one has to actually reach >>> noon, but reaching noon means adding infinite values to the pot. > > Virgil wrote: >>> TO misses again. Every value "added to the pot" is finite. If TO meant >>> infinitely many values, he should have said so. > > Tony Orlow wrote: >> Does anything occur in the vase at noon? If not, then it should have the >> same state as before noon. > > As which state before noon? > The state of non-emptiness that persists continually from t>=-1 until t<0.
From: Tony Orlow on 25 Oct 2006 14:43 David R Tribble wrote: > David R Tribble wrote: >>> I know you don't get this, but go back and read your own definition. >>> Every H-riffic corresponds to a node in an infinite, but countable, >>> binary tree. > > Tony Orlow wrote: >>> No, like the reals, it corresponds to a path in the tree. > > David R Tribble wrote: >>> No, read your own definition again. Each H-riffic is a finite node >>> along a path in a binary tree. > > Tony Orlow wrote: >> Where does it say anything about a node in my definition, or whether >> strings can be infinite? Your baseless declarations about my definitions >> don't fly. > > When you stated that > 1 in H > x in H -> 2^x in H > x in H -> 2^-x in H > > The set H is a countable set. Each x in H corresponds to a node in the > binary tree listing all the x's in H (where each left fork is 2^x and > each right fork is 2^-x from the node of any x). > > In different terms, each x in H is a finite recursion > x = 2^y or 2^-y for some other y in H > where each recursion ends at > y = 1 > > Your definition above does not allow for any infinite-length recursions > or infinite-length paths in the binary tree. > > As I posted previously, if you want to extend your definition to > include infinite-length paths in the tree (which I dubbed the > H2-riffics), you need to define additional numbers using additional > rules. Something akin to the way the irrationals are defined on > top of the rationals (as infinite sequences of rationals) in order to > define the complete set of reals. > > Is that true also of the digital reals? I disagree with the notion that any sequence is countable. In order to prove that the H-riffics really cover the reals I have to use a Cauchy- or Dedekind-like method to prove that any element in the continuum can be specified, even if it requires an infinite specification. But, there is nothing explicit or inherent in my rules that limits such specifications to finite lengths. You are carrying that over from the standard notion of sequences as always countable. I don't adhere to that concept. > Tony Orlow wrote: >>> What makes you think infinite-length strings are excluded? They're not, >>> in either of my riffic number systems. > > David R Tribble wrote: >>> You're confused. Infinite-length fractions are not excluded, >>> obviously. But we're not talking about fractions, we're talking about >>> each H-riffic being a node in the binary tree that lists all of them. >>> Each H-riffic is a node on a finite-length path in the tree. > > Tony Orlow wrote: >> Who the hell said that? Is this your number system now, that you get to >> declare that my H-riffics are nodes in your tree? Get real. > > It's what you did _not_ say that excludes them. There is no way to > produce infinitely recursively-defined H-riffics from your existing > definitions, so you must add another rule or two that allows such > numbers to exist. Which gives you a different set, of course. > Which rules would you recommend that counteract rules that I didn't state? You are applying a rule that says that any sequence is finite. That's not true. Countably infinite sequences exist, as in 1/3 in decimal. They exist here too.
From: Tony Orlow on 25 Oct 2006 14:53 Mike Kelly wrote: > Randy Poe wrote: >> Tony Orlow wrote: >>> Virgil wrote: >>>> In article <453e4a85(a)news2.lightlink.com>, >>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>> >>>>> David Marcus wrote: >>>>>> Tony Orlow wrote: >>>>>>> David Marcus wrote: >>>>>>>> Tony Orlow wrote: >>>>>>>>> David Marcus wrote: >>>>>>>>>> Tony Orlow wrote: >>>>>>>>>>> Your examples of the circle and rectangle are good. Neither has a >>>>>>>>>>> height >>>>>>>>>>> outside of its x range. The height of the circle is 0 at x=-1 and x=1, >>>>>>>>>>> because the circle actually exists there. To ask about its height at >>>>>>>>>>> x=9 >>>>>>>>>>> is like asking how the air quality was on the 85th floor of the World >>>>>>>>>>> Trade Center yesterday. Similarly, it makes little sense to ask what >>>>>>>>>>> happens at noon. There is no vase at noon. >>>>>>>>>> Do you really mean to say that there is no vase at noon or do you mean >>>>>>>>>> to say that the vase is not empty at noon? >>>>>>>>> If noon exists at all, the vase is not empty. All finite naturals will >>>>>>>>> have been removed, but an infinite number of infinitely-numbered balls >>>>>>>>> will remain. >>>>>>>> "If noon exists at all"? How do we decide? >>>>>>>> >>>>>>> We decide on the basis of whether 1/n=0. Is that possible for n in N? >>>>>>> Hmmmm......nope. >>>>>> So, noon doesn't exist. And, there is no vase at noon. I thought you >>>>>> were saying the vase contains an infinite number of balls at noon. >>>>>> >>>>> If the vase exists at noon, then it has an uncountable number of balls >>>>> labeled with infinite values. But, no infinite values are allowed i the >>>>> experiment, so this cannot happen, and noon is excluded. >>>> So did the North Koreans nuke the vase before noon? >>>> >>>> The only relevant issue is whether according to the rules set up in the >>>> problem, is each ball inserted before noon also removed before noon?" >>>> >>>> An affirmative confirms that the vase is empty at noon. >>>> A negative directly violates the conditions of the problem. >>>> >>>> How does TO answer? >>> You can repeat the same inane nonsense 25 more times, if you want. I >>> already answered the question. It's not my problem that you can't >>> understand it. >> Your response requires that the vase contains balls which were >> never, by the stated rules, put in. >> >> You keep saying things like "if the clock runs till noon there are >> balls with infinite numbers on them" even though the rules say there >> are >> no balls with infinite numbers on them. How do you reconcile that? >> >> If I put in balls 1, 2, 3 and stop, can the clock tick till noon >> without >> requiring a 4th ball? >> >> If I specify times for balls 1-1000 only, can the clock till noon >> without >> requiring a 1001-th ball? >> >> How is it, in your world, that when I specify times for all natural >> numbered >> balls, I am required to put in balls that don't have natural numbers? > > The problem is that Tony thinks time is a function of the number of > insertions you've gone through. In order to "get to" any particular > time you have to perform the insertions "up to" that point. He then > thinks that if you want to "get to" noon, you have to have performed > some "infinite" (whatever that means) iterations, where balls without > natural numbers are inserted. That this is obviously not what the > problem statement says doesn't seem to bother him. Nor that it's > absolutely nothing like an intuitive picture of what time is. Time is ultimately irrelevant in this gedanken, but if it is to be considered, the constraints regarding time cannot be ignored. Events occurring in time must occupy at least one moment. > > Obviously, time is an independent variable in this experiment and the > insertion or removal or location of balls is a function of time. That's > what the problem statement says: we have this thing called "time" which > is a real number and it "goes from" before noon to after noon and, at > certain specified times, things happen. There are only > naturally-numbered balls inserted and removed, always before noon. > Every ball is removed before noon. Therefore, the vase is empty. No, you have the concept of the independent variable bent. The number of balls is related to the time by a formula which works in both directions. So, when does the vase become empty? Nothing can occur at noon, as far as ball removals. AT every time before noon, balls are in the vase. So, when does the vase become empty, and how? > > If you follow the sequence of insertions and removals you never "get > to" noon but this doesn't imply that noon is never reached, or that > iterations involving non-naturally numbered balls occur. It just > implies that all insertion and removal is performed before noon. > > Tony won't let himself understand this. He is delusional. His problem. > I won't let myself accept self-contradictory conclusions. There is no moment at which this can occur. The problem is perfectly modeled by a divergent infinite series. No last ball can be removed without there having been an negative number of balls previously. You solution fails.
From: Tony Orlow on 25 Oct 2006 14:54
David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> Tony Orlow wrote: >>>> David Marcus wrote: >>>>> Tony Orlow wrote: >>>>>> stephen(a)nomail.com wrote: >>>>>>> Also, supposing for the sake of argument that there are "infinitely >>>>>>> number balls", if a ball is added at time -1/(2^floor(n/10)), and removed >>>>>>> at time -1/(2^n)), then the balls added at time t=0, are those >>>>>>> where -1/(2^floor(n/10)) = 0. But if -1/(2^floor(n/10)) = 0 >>>>>>> then -1/(2^n) = 0 (making some reasonable assumptions about how arithmetic >>>>>>> on these infinite numbers works), so those balls are also removed at noon and >>>>>>> never spend any time in the vase. >>>>>> Yes, the insertion/removal schedule instantly becomes infinitely fast in >>>>>> a truly uncountable way. The only way to get a handle on it is to >>>>>> explicitly state the level of infinity the iterations are allowed to >>>>>> achieve at noon. When the iterations are restricted to finite values, >>>>>> noon is never reached, but approached as a limit. >>>>> Suppose we only do an insertion or removal at t = 1/n for n a natural >>>>> number. What do you mean by "noon is never reached"? >>>> 1/n>0 >>> Sorry, I meant t = -1/n. So, I assume your answer is that -1/n < 0. >>> >>> But, I don't follow. Translating "-1/n < 0" back into words, I get "all >>> insertions and removals are before noon". However, I asked you what >>> "noon is never reached" means. Are you saying that "noon is never >>> reached" means that "all insertions and removals are before noon"? >> Yes, David. What else happens in this experiment besides insertions and >> removals of naturals at finite times before noon? If the infinite >> sequence of events is actually allowed to continue until t=0, then you >> are talking about events not indexed with natural numbers, so you're not >> talking about the same experiment. If noon is not allowed, and all times >> in the experiment are finitely before noon, well, at none of those times >> does the vase empty, as we all agree. This is why I am asking when this >> occurs. It can't, given the constraints of the problem. > > Does your "Yes" at the beginning of your reply mean that you agree that > "noon is never reached" means that "all insertions and removals are > before noon". By "mean", I mean that that is what the words mean, not > that the two statements are equivalent or deducible from each other. > Yes, every event, every insertion or removal, happens at a specific time before noon. At each of those times, the vase is non-empty. Nothing else occurs, as far as insertions of removals. Is that clear enough? So, when does the vase become empty, and how? |