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From: Tony Orlow on 31 Oct 2006 12:57 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: >>>>>>> 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 relatio
From: David R Tribble on 31 Oct 2006 13:00 [Apologies if this duplicates previous responses] Tony Orlow wrote: > I am beginning to realize just how much trouble the axiom of > extensionality is causing here. That is what you're using, here, no? The > sets are "equal" because they contain the same elements. Yes, the basic definition of set equality, the '=' set operator. > That gives no > measure of how the sets compare at any given point in their production. This makes no sense. Sets are not "produced" or "generated". Sets simply exist. > Sets as sets are considered static and complete. Correct. > However, when talking > about processes of adding and removing elements, the sets are not > static, but changing with each event. Incorrect. If we define set A as containing the elements a, b, and c, then A = {a, b, c}. Period. If we then talk about adding elements d and e to set A, we're not actually changing set A, but describing another set, call it A2, that is the union of A and {d, e}, so A2 = {a, b, c, d, e}. Nothing is ever "added to" a set. Rather, we apply operations (union, intersection, etc.) to existing sets to create new sets. We don't change existing sets. > When speaking about what is in the > set at time t, use a function for that sum on t, assume t is continuous, > and check the limit as t->0. Then you won't run into silly paradoxes and > unicorns. That would be using the wrong nomenclature. We don't talk about what set A contains at any particular time t. We can talk about the sequence of sets A_1, A_2, A_3, etc. We can also talk about the union of the entire sequence of sets as set A, if we like. Or the intersection, or whatever. But every set we're talking about is "static", unchanging, once it (the members it contains) is defined.
From: Tony Orlow on 31 Oct 2006 13:05 imaginatorium(a)despammed.com wrote: > Virgil wrote: >> In article <1162268163.368326.64650(a)m73g2000cwd.googlegroups.com>, >> imaginatorium(a)despammed.com wrote: >> >>> Virgil wrote: >>>> In article <45462ba0(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: >>>>>>>>> stephen(a)nomail.com wrote: >>> <snipola> >>> >>>>>>> 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. >>>> It is the only definition of set size that is known to produce a valid >>>> partial ordering on sets. >>> Huh? I thought cardinality produced a valid *total* ordering on sets. >> The cardinalities are totally ordered, but the sets are not. >> A total order on sets would require that when neither of two sets was >> "larger than" the other that they must be the same set, not merely the >> same size. > > Oh, right. But - and I'm not quite sure how to say this, but the > cardinalities _are_ totally ordered; for any two sets A and B, c(A) < > c(B), or c(A) = c(B), or c(A) > c(B). If you "reduce" the sets by the > cardinality equivalence relation, they are totally ordered. The subset > relation doesn't lead to an equivalence relation, only a partial > ordering: so there is no s(A) = s(B) unless A=B; but for most pairs of > sets, the subset relation simply says nothing at all. (Until His > Master's Voice is heard, telling us something totally arbitrary.) > > Anyway, your claim was clearly wrong, since the subset relation > provides a valid partial ordering on sets. > > Brian Chandler > http://imaginatorium.org > > >>> As I have pointed out when all this started, of course everyone knows >>> that the subset relation produces a perfectly valid *partial* ordering >>> on sets. >>> >>> Brian Chandler >>> http://imaginatorium.org > If that's what a partial ordering vs. a total ordering is, Bigulosity is a partial ordering on sets, not total ordering. Different sets can have the same Bigulosity.
From: stephen on 31 Oct 2006 13:15 David R Tribble <david(a)tribble.com> wrote: > [Apologies if this duplicates previous responses] > Tony Orlow wrote: >> I am beginning to realize just how much trouble the axiom of >> extensionality is causing here. That is what you're using, here, no? The >> sets are "equal" because they contain the same elements. > Yes, the basic definition of set equality, the '=' set operator. >> That gives no >> measure of how the sets compare at any given point in their production. > This makes no sense. Sets are not "produced" or "generated". > Sets simply exist. >> Sets as sets are considered static and complete. > Correct. >> However, when talking >> about processes of adding and removing elements, the sets are not >> static, but changing with each event. > Incorrect. If we define set A as containing the elements a, b, and c, > then A = {a, b, c}. Period. If we then talk about adding elements d > and e to set A, we're not actually changing set A, but describing > another set, call it A2, that is the union of A and {d, e}, so > A2 = {a, b, c, d, e}. > Nothing is ever "added to" a set. Rather, we apply operations (union, > intersection, etc.) to existing sets to create new sets. We don't > change existing sets. Just like when we add 5 to 2 to get 7, we do not change the 5 or 2 to create a 7. Or when you celebrate a birthday, your age changes, but the number that represented your age does not change. A different number is used to represent your age, but the "old" number remains as it always ways. This idea of "changing" sets seems to be at the heart of a lot of people's misconceptions about set theory. Stephen
From: Mike Kelly on 31 Oct 2006 13:24
Tony Orlow wrote: > Mike Kelly wrote: > > Tony Orlow wrote: > >> Mike Kelly wrote: > > <snip> > >>> Now correct me if I'm wrong, but I think you agreed that every > >>> "specific" ball has been removed before noon. And indeed the problem > >>> statement doesn't mention any "non-specific" balls, so it seems that > >>> the vase must be empty. However, you believe that in order to "reach > >>> noon" one must have iterations where "non specific" balls without > >>> natural numbers are inserted into the vase and thus, if the problem > >>> makes sense and "noon" is meaningful, the vase is non-empty at noon. Is > >>> this a fair summary of your position? > >>> > >>> If so, I'd like to make clear that I have no idea in the world why you > >>> hold such a notion. It seems utterly illogical to me and it baffles me > >>> why you hold to it so doggedly. So, I'd like to try and understand why > >>> you think that it is the case. If you can explain it cogently, maybe > >>> I'll be convinced that you make sense. And maybe if you can't explain, > >>> you'll admit that you might be wrong? > >>> > >>> Let's start simply so there is less room for mutual incomprehension. > >>> Let's imagine a new experiment. In this experiment, we have the same > >>> infinite vase and the same infinite set of balls with natural numbers > >>> on them. Let's call the time one minute to noon -1 and noon 0. Note > >>> that time is a real-valued variable that can have any real value. At > >>> time -1/n we insert ball n into the vase. > >>> > >>> My question : what do you think is in the vase at noon? > >>> > >> A countable infinity of balls. > > > > 1) It's not clear to me what you mean by that phrase but I'll assume > > the standard definition. Still, the question remains of which balls you > > think are in the vase? Does every natural number, n, have a ball in the > > vase labelled with that n? > > Conceptually, sure. Yes or no? What is the set of balls in the vase at noon? Which balls are in the vase and which are not? > > 2) How come noon "exists" in this experiment but it didn't exist in the > > original experiment? Or did you give up on claiming noon doesn't > > "exist"? What does that mean, anyway? > > Nothing is allowed to happen at noon in either experiment. Nothing "happens" at noon? I take this to mean that there is no insertion or removal of balls at noon, yes? Well, I agree with that. But what relevence does this have to the statement "noon does not exist"? What does that even *mean*? When you've been saying "noon doesn't exist", you actually mean to say "no insertion or removal of balls occurs at noon"? How about this experiment, does noon "exist" in this experiment : Insert a ball labelled "1" into the vase at one minute to noon. ? >They both end up with countably many balls in the vase at noon. For now, I am going to try to restrict myself to discussing this new experiment, because I want to understand what "noon doesn't exist" is supposed to mean. And, again, your answer is ambiguous. I asked which balls are in the vase at noon, not the cardinality of the set of balls in the vase at noon. I then asked whether "noon exists", not whether anything "happens" at noon. Please try answering the questions people actually ask; it aids in communication. >The experiment's stated sequence logically precludes that the vase become empty. It logically precludes that balls without a finite natural number on them get added to the vase, but that doesn't seem to bother you. Ho hum. <snip more stuff about original experiment> -- mike. |