An uncountable countable set
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From: Tony Orlow on 30 Oct 2006 12:05 David Marcus wrote: > Virgil wrote: >> In article <45439743(a)news2.lightlink.com>, >> Tony Orlow <tony(a)lightlink.com> wrote: >> >>> The purpose being to try to obscure details of the stated problem. >> That does seem to be TO's purpose. >> >> The stated problem was [ with one modification]: >> Given an initially empty vase. >> Given the infinite set of finite natural numbers, staring with 1, and a >> ball with each number marked on it. >> At times in minutes before noon: >> at t = 1/n balls numbered 10*(n -1) +1 to 10*n are inserted into the >> vase and then in that same instant ball n is removed. >> [At noon, a cube is placed in the vase.] >> What is the state of the vase at noon ? > > Seems to me that the cube won't fit in the vase because the vase is full > of balls that have infinite non-standard reals marked on them. > Your vase would have to be uncountably large, and you still might need an infinitesimal cube. ;)
From: Tony Orlow on 30 Oct 2006 12:13 Virgil wrote: > In article <4543b0b3(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > > >> The experiment occurred in [-1,0). Talk of time outside that range is >> irrelevant. Times before that are imaginary, and times after that are >> infinite. Only finite times change anything, so if something changes, >> it's at a finite, negative time. > > Then let us change the experiment to include the insertion into the vase > of a cube at one minute after noon. > > The experiment now ranges over [-1,1]. > > What are the contents of the vase at times in [0,1), TO? > An uncountable number of balls, all infinitely numbered. > >> Why, oh why, do the constraints of the problem have to matter? Why, oh >> why, must we alway mean a continuum when we call something continuous, >> when I want to declare a discontinuity at convenient places like 0? Why >> can't we specify what happens at every moment y between x and z, but >> that something totally different is the case at z, without anything >> changing it? Why, oh why, why don't my labels matter? >> >> I'm not really sure how to answer this, again..... > > TO is confused! Still or again? probably still, as there doesn't seem to > be much time at which he is not.
From: Tony Orlow on 30 Oct 2006 12:16 David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> 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. That gives no >>>> measure of how the sets compare at any given point in their production. >>>> Sets as sets are considered static and complete. However, when talking >>>> about processes of adding and removing elements, the sets are not >>>> static, but changing with each event. 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. >>> There is a lot of stuff in there. Let's go one step at a time. I believe >>> that one thing you are saying is this: >>> >>> |IN\OUT| = 0, but defining IN and OUT and looking at |IN\OUT| is not the >>> correct translation of the balls and vase problem into Mathematics. >>> >>> Do you agree with this statement? >> Yes. > > OK. Since you don't like the |IN\OUT| translation, let's see if we can > take what you wrote, translate it into Mathematics, and get a > translation that you like. > > You say, "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." > > Taking this one step at a time, first we have "use a function for that > sum on t". How about we use the function V defined as follows? > > For n = 1,2,..., let > > A_n = -1/floor((n+9)/10), > R_n = -1/n. > > For n = 1,2,..., define a function B_n by > > B_n(t) = 1 if A_n <= t < R_n, > 0 if t < A_n or t >= R_n. > > Let V(t) = sum_n B_n(t). > > Next you say, "assume t is continuous". Not sure what you mean. Maybe > you mean assume the function is continuous? However, it seems that > either the function we defined (e.g., V) is continuous or it isn't, > i.e., it should be something we deduce, not assume. Let's skip this for > now. I don't think we actually need it. > > Finally, you write, "check the limit as t->0". I would interpret this as > saying that we should evaluate the limit of V(t) as t approaches zero > from the left, i.e., > > lim_{t -> 0-} V(t). > > Do you agree that you are saying that the number of balls in the vase at > noon is lim_{t -> 0-} V(t)? > Find limits of formulas on numbers, not limits of sets. Here's what I said to 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. contains(n) is the number of balls in the vase upon completion of iteration n, and is equal to in(n)-out(n)=9n. n(t) is the number of iterations completed at time t, equal to floor(-1/t). contains(t) is the number of balls in the vase at time t, and is equal to contains(n(t))=contains(floor(-1/t))=9*floor(-1/t). Lim(t->-0: 9*floor(-1/t)))=oo. The sum diverges in the limit. See how that all fits together? Its almost like physics, eh?
From: stephen on 30 Oct 2006 12:22 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. >> >> 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. > There is no spoon. There is no God. There is no mind. There is no n. > No no no no. No. 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. So here is one last attempt. The balls and vase problem is not a physical problem. 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. 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. 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 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. >>>>>> 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, >>>> n is a member of OUT, and vice versa. >>>> >>>> What other relationship do you think there is between >>>> -1/(2^(floor(n/10))) < 0 >>>> and >>>> -1/(2^n) < 0 >>>> ?? >> >>> Like, wow, Man, at, like, each moment, there's, like, 10 going in, and, >>> like, Man, only 1 coming out. Seems kinda weird. There's, like, a rate >>> thing going on.... :D >> >> What rate? There is no rate. There are just two sets >> IN = { n | -1/2^(floor(n/10)) < 0 } >> OUT = { n | -1/2^n < 0 } > 9 balls/iteration. What balls? What iterations? The two sets are defined as IN = { n | -1/2^(floor(n/10)) < 0 } OUT = { n | -1/2^n < 0 } Why do you keep talking about balls and vases? >> >> Why do you keep babbling about rates? We are talking >> about an abstract math problem. > Which involves a process in time which happens at a certain rate at any > given point. No. It involves two sets. There are not rates, just two sets that I have repeatedly defined. >> >> In any case, as Brian pointed out, these two sets can be "constructed" >> at the same rate: >> >> int n=1; >> while (n>=1) >> { >> if (-1/2^(floor(n/10)) < 0) >> IN.add(n); >> if (-1/2^n < 0) >> OUT.add(n); >> } >> >> One element is added to each set each time through the loop. > You forgot to increment n in the loop, so this doesn't produce anything > of interest. If you had included it, first of all, your second 'if' > statement is totally superfluous. Just add n to OUT each time. The first > if statement and resulting addition only adds the same element which is > subsequently removed, not the ten as specified in the problem. Which > vase are you talking about? I am not talking about the balls and vase problem as I have repeatedly said. Why do you bother responding when you clearly do not bother reading? >> >>>> Do you think there exists a positive integer n such that >>>> -1/(
From: Tony Orlow on 30 Oct 2006 12:25
David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> Tony Orlow wrote: >>>> David Marcus wrote: >>>>> Tony Orlow wrote: >>>>>> David Marcus wrote: >>>>>>> Tony Orlow wrote: >>>>>>>> David Marcus wrote: >>>>>>>>> Tony Orlow wrote: >>>>>>>>>> David Marcus wrote: >>>>>>>>>>> Tony Orlow wrote: >>>>>>>>>>>> David Marcus wrote: > >>>>>>>>>>>>> You are mentioning balls and time and a vase. But, what >>>>>>>>>>>>> I'm asking is completely separate from that. I'm just >>>>>>>>>>>>> asking about a math problem. Please just consider the >>>>>>>>>>>>> following mathematical definitions and completely ignore >>>>>>>>>>>>> that they may or may not be relevant/related/similar to >>>>>>>>>>>>> the vase and balls problem: >>>>>>>>>>>>> >>>>>>>>>>>>> -------------------------- >>>>>>>>>>>>> For n = 1,2,..., let >>>>>>>>>>>>> >>>>>>>>>>>>> A_n = -1/floor((n+9)/10), >>>>>>>>>>>>> R_n = -1/n. >>>>>>>>>>>>> >>>>>>>>>>>>> For n = 1,2,..., define a function B_n: R -> R by >>>>>>>>>>>>> >>>>>>>>>>>>> B_n(t) = 1 if A_n <= t < R_n, >>>>>>>>>>>>> 0 if t < A_n or t >= R_n. >>>>>>>>>>>>> >>>>>>>>>>>>> Let V(t) = sum_n B_n(t). >>>>>>>>>>>>> -------------------------- >>>>>>>>>>>>> >>>>>>>>>>>>> Just looking at these definitions of sequences and >>>>>>>>>>>>> functions from R (the real numbers) to R, and assuming >>>>>>>>>>>>> that the sum is defined as it would be in a Freshman >>>>>>>>>>>>> Calculus class, are you saying that V(0) is not equal to >>>>>>>>>>>>> 0? > >>>>>>>>>>>> On the surface, you math appears correct, but that doesn't >>>>>>>>>>>> mend the obvious contradiction in having an event occur in >>>>>>>>>>>> a time continuum without occupying at least one moment. It >>>>>>>>>>>> doesn't explain how a divergent sum converges to 0. >>>>>>>>>>>> Basically, what you prove, if V(0)=0, is that all finite >>>>>>>>>>>> naturals are removed by noon. I never disagreed with that. >>>>>>>>>>>> However, to actually reach noon requires infinite >>>>>>>>>>>> naturals. Sure, if V is defined as the sum of all finite >>>>>>>>>>>> balls, V(0)=0. But, I've already said that, several times, >>>>>>>>>>>> haven't I? Isn't that an answer to your question? > >>>>>>>>>>> I think it is an answer. Just to be sure, please confirm >>>>>>>>>>> that you agree that, with the definitions above, V(0) = 0. >>>>>>>>>>> Is that correct? > >>>>>>>>>> Sure, all finite balls are gone at noon. > >>>>>>>>> Please note that there are no balls or time in the above >>>>>>>>> mathematics problem. However, I'll take your "Sure" as >>>>>>>>> agreement that V(0) = 0. > >>>>>>>> Okay. > >>>>>>>>> Let me ask you a question about this mathematics problem. >>>>>>>>> Please answer without using the words "balls", "vase", >>>>>>>>> "time", or "noon" (since these words do not occur in the >>>>>>>>> problem). > >>>>>>>> I'll try. > >>>>>>>>> First some discussion: For each n, B_n(0) = 0 and B_n is >>>>>>>>> continuous at zero. > >>>>>>>> What??? How do you conclude that anything besides time is >>>>>>>> continuous at 0, where yo have an ordinal discontinuity???? >>>>>>>> Please explain. > >>>>>>> I thought we agreed above to not use the word "time" in >>>>>>> discussing this mathematics problem? > >>>>>> If that's what you want, then why don't you remove 't' from all >>>>>> of your equations? > >>>>> It is just a letter. It stands for a real number. Would you >>>>> prefer "x"? I'll switch to "x". > >>>> It is still related to n in such a way that x<0. > >>>>>>> As for your question, let's look at B_2 (the argument is >>>>>>> similar for the other B_n). >>>>>>> >>>>>>> B_2(t) = 1 if A_2 <= t < R_2, >>>>>>> 0 if t < A_2 or t >= R_2. >>>>>>> >>>>>>> Now, A_2 = -1 and R_2 = -1/2. So, >>>>>>> >>>>>>> B_2(t) = 1 if -1 <= t < -1/2, >>>>>>> 0 if t < -1 or t >= -1/2. >>>>>>> >>>>>>> In particular, B_2(t) = 0 for t >= -1/2. So, the value of B_2 >>>>>>> at zero is zero and the limit as we approach zero is zero. So, >>>>>>> B_2 is continuous at zero. > >>>>>> Oh. For each ball, nothing is happening at 0 and B_n(0)=0. >>>>>> That's for each finite ball that one can specify. > >>>>> I thought we agreed to not use the word "ball" in discussing this >>>>> mathematics problem? Do you want me to change the letter "B" to a >>>>> different letter, too? > >>>> Call it an element or a ball. I don't care. It doesn't matter. > >>>>>> However, lim(t->0: sum(B_n| B_n(t)=1))=oo. Why do you >>>>>> conveniently forget that fact? > >>>>> Your notation is nonstandard, so I'm not sure what you mean. Do >>>>> you mean to write >>>>> >>>>> lim_{x -> 0-} sum_n B_n(x) = oo ? >>>>> >>>>> If so, I don't understand why you think I've forgotten this fact. >>>>> If you look in my previous post (or below), you will see that I >>>>> wrote, "Now, V is the sum of the B_n. As t approaches zero from >>>>> the left, V(t) grows without bound. In fact, given any large >>>>> number M, there is an e < 0 such that for e < t < 0, V(t) > M." > >>>> Then don't you see a contradiction in the limit at that point >>>> being oo, the value being 0, and there being no event to cause >>>> that change? I do. > >>>>>>>>> In fact, for a given n, there is an e < 0 such that B_n(t) = >>>>>>>>> 0 for e < t <= 0. > >>>>>>>> There is no e<0 such that e<t and B_n(t)=0. That's simply false. > >>>>>>> Let's look at B_2 again. We can take e = -1/2. Then B_2(t) = 0 >>>>>>> for e < t <= 0. Similarly, for any other given B_n, we can find >>>>>>> an e that does what I wrote. > >>>>>> Yes, okay, I misread that. Sorry. For each ball B_n that's true. >>>>>> For the sum of balls n such that B_n(t)=1, it diverges as t->0. > >>>>>>>>> In other words, B_n is not changing near zero. > >>>>>>>> Infinitely more quickly but not. That's logical. And wrong. > >>>>>>> Not sure what you mean. > >>>>>> The sum increases without bound. > >>>>>>>>> Now, V is the sum of the B_n. As t approaches zero from the >>>>>>>>> left, V(t) grows without bound. In fact, given any large >>>>>>>>> number M, there is an e < 0 such that for e < t < 0, V(t) > >>>>>>>>> M. We also have that V(0) = 0 (as you agreed). >>>>>>>>> >>>>>>>>> Now the question: How do you explain the fact that V(t) goes >>>>>>>>> from being very large for t a little less than zero to being >>>>>>>>> zero when t equals zero |