An uncountable countable set
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From: Tony Orlow on 20 Oct 2006 12:45 David R Tribble wrote: > Tony Orlow wrote: >>> The sequence of events consists of adding 10 and removing 1, an infinite >>> number of times. In other words, it's an infinite series of (+10-1). > > David Marcus wrote: >>> Sorry, but I don't quite understand. When you stated the problem in >>> English, it ended with a question mark. But, your statement in >>> Mathematics does not end with a question mark. If it is a >>> problem/question, I think it should end with a question mark. Please >>> give the statement of the problem in Mathematics. > > Tony Orlow wrote: >> What is sum(n=1->oo: 9)? > > Because 9 = 10 - 1, it obviously must be true that > sum{n=1 to oo} 9 > is equal to > [sum{n=1 to oo} 10] - [sum{n=1 to oo} 1] > > Did I get it right, Tony? > If those three oo's are exactly the same, yes. The fact that the insertions and removals alternate and are therefore exactly 1-1 makes that the case. Tony
From: Tony Orlow on 20 Oct 2006 12:50 David R Tribble wrote: > Tony Orlow wrote: >>> What is sum(n=1->oo: 9)? > > Alan Morgan wrote: >>> I think you actually mean, what is 10-1+10-1+10-1.... >>> >>> It was recognized long before Cantor that there isn't a simple answer to >>> that question. > > Tony Orlow wrote: >> There is if you prohibit rearranging the terms to change the relative >> frequencies of the two terms. Group all you like without rearranging. >> This series is (+10-1)+(10-1)+(10-1)+... > > So you're saying that > s = (10 - 1) + (10 - 1) + (10 - 1) + ... > is just > s = 9 + 9 + 9 + ... Yep. > > But surely then > s = (10 - 1) + (10 - 1) + (10 - 1) + ... > is exactly the same as > t = 10 + (-1 + 10) + (-1 + 10) + ... > (notice the "relative frequencies" of the 10's and 1's are the same), > which is just > t = 10 + 9 + 9 + ... > > So since the terms of s and t have the same relative frequencies, > we conclude that s = t. Right, Tony? > If you do that, then you don't have all the same terms repeated, but an odd one out. Yes, you produced a difference of 1, which is an insignificant iota compared to the infinite sum. No, don't rearrange them, and yes, group them into identical groups if possible. Each iteration yields a net addition of 9 balls, and there are an infinite number of such iterations. Tony
From: Tony Orlow on 20 Oct 2006 12:58 Virgil wrote: > In article <4535884c(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: > >>> One can separate the reals into everything before 0 as one set and 0 >>> and everything after it as the other. Does TO claim time is less >>> seperable? >>> >> Linear time? no. > > Does TO claim that anything in the statement of the problem justifies an > assumption of non-linear time. > Yes, the fact that each event comes twice as fast as the last. >> Yeah, noon doesn't exist in the description of the problem. It's like >> saying, "Everyone on Earth has 3 children which survive, for four >> generations, and then half the population of the planet dies. This >> happens an infinite number of times. What happens when there is no more >> planet?" The question is itself a non-sequitur. > > One can claim that the vase problem cannot occur in any physical sense, > but if one accepts the statement of the problem, the only conclusion > which does not require assumptions not inherent in the problem itself is > that at noon the vase is empty. > Incorrect. > >>> One certainly starts with more balls. At what time do more balls become >>> less balls? And why? >> When the net addition of nine balls overtakes the mere subtraction of one. > > Non sequitur. > Then you'll say I never answered the question.... >> When ALL balls are added, and then balls are ONLY removed, to say that >> gives the same result as repeatedly adding more balls than you remove, >> that's what's idiotic, to borrow your obnoxious term. > > When one starts with all infinitely many balls in the vase and then > balls are removed on the original schedule, there will be infinitely > many in the vase at all times from that group insertion up to but not > including noon. Yes, that is correct. You get a cookie. >>>> You really >>>> don't understand the implications of the Zeno machine, do you? > > As I am not using one, that is irrelevant. So, now you're doing it in linear time? Let me know when you're done.... >>> I do not understand how having more balls in the vase for longer times >>> can produce less balls in the urn at any time. >> There is so much you fail to understand, or succeed in misunderstanding, >> that I don't even know where to begin with you. If you can't grasp the >> logic here, I really don't see any hope for you. > > I can grasp logic well enough, but from TO I have not seen any. You have to step out of the cave to see it in the light, Virgil. They're only birds......
From: Tony Orlow on 20 Oct 2006 13:02 Virgil wrote: > In article <453589db(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <45343e6f(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: >>>>>>>>>> David Marcus wrote: >>>>>>>>>>> Tony Orlow wrote: >>>>>>>>>>>> David Marcus wrote: >>>>>>>>>>>>> How about this problem: Start with an empty vase. Add a ball to a >>>>>>>>>>>>> vase >>>>>>>>>>>>> at time 5. Remove it at time 6. How many balls are in the vase at >>>>>>>>>>>>> time >>>>>>>>>>>>> 10? >>>>>>>>>>>>> >>>>>>>>>>>>> Is this a nonsensical question? >>>>>>>>>>>> Not if that's all that happens. However, that doesn't relate to >>>>>>>>>>>> the >>>>>>>>>>>> ruse >>>>>>>>>>>> in the vase problem under discussion. So, what's your point? >>>>>>>>>>> Is this a reasonable translation into Mathematics of the above >>>>>>>>>>> problem? >>>>>>>>>> I gave you the translation, to the last iteration of which you did >>>>>>>>>> not >>>>>>>>>> respond. >>>>>>>>>>> "Let 1 signify that the ball is in the vase. Let 0 signify that it >>>>>>>>>>> is >>>>>>>>>>> not. Let A(t) signify the location of the ball at time t. The >>>>>>>>>>> number >>>>>>>>>>> of >>>>>>>>>>> 'balls in the vase' at time t is A(t). Let >>>>>>>>>>> >>>>>>>>>>> A(t) = 1 if 5 < t < 6; 0 otherwise. >>>>>>>>>>> >>>>>>>>>>> What is A(10)?" >>>>>>>>>> Think in terms on n, rather than t, and you'll slap yourself awake. >>>>>>>>> Sorry, but perhaps I wasn't clear. I stated a problem above in >>>>>>>>> English >>>>>>>>> with one ball and you agreed it was a sensible problem. Then I asked >>>>>>>>> if >>>>>>>>> the translation above is a reasonable translation of the one-ball >>>>>>>>> problem into Mathematics. If you gave your translation of the >>>>>>>>> one-ball >>>>>>>>> problem, I missed it. Regardless, my question is whether the >>>>>>>>> translation >>>>>>>>> above is acceptable. So, is the translation above for the one-ball >>>>>>>>> problem reasonable/acceptable? >>>>>>>> Yes, for that particular ball, you have described its state over time. >>>>>>>> According to your rule, A(10)=0, since 10>6>5. Do go on. >>>>>>> >>>>>>> OK. Let's try one in reverse. First the Mathematics: >>>>>>> >>>>>>> >>>>>>> Let B_1(t) = 1 if 5 < t < 7, >>>>>>> 0 if t < 5 or t > 7, >>>>>>> undefined otherwise. >>>>>>> >>>>>>> Let B_2(t) = 1 if 6 < t < 8, >>>>>>> 0 if t < 6 or t > 8, >>>>>>> undefined otherwise. >>>>>>> >>>>>>> Let V(t) = B_1(t) + B_2(t). What is V(9)? >>>>>>> >>>>>>> >>>>>>> Now, how would you translate this into English ("balls", "vases", >>>>>>> "time")? >>>>>> That's not an infinite sequence, so it really has no bearing on the vase >>>>>> problem as stated. I understand the simplistic logic with which you draw >>>>>> your conclusions. Do you understand how it conflicts with other >>>>>> simplistic logic? It's the difference between focusing on time vs. >>>>>> iterations. Iteration-wise, it never can empty. There's something wrong >>>>>> with your time-wise logic which has everything to do with the Zeno >>>>>> machine and the indistinguishability of iterations outside of N. For >>>>>> every n e N, iteration n results in 9n balls left over, a nonzero >>>>>> number. If all steps are indexed by n in N, then this result holds for >>>>>> the entire sequence. Within your experiment, with ball numbers all in N, >>>>>> you never reach noon, and at every moment for the minute before it the >>>>>> vase is nonempty. >>>>> I didn't say it was an infinite sequence nor did I say it had a "bearing >>>>> on the vase problem as stated". However, I asked you a question. I don't >>>>> believe you answered my question. So, let me try again: >>>>> >>>>> How would you translate the mathematical problem I wrote above into >>>>> English ("balls", "vases", "time")? >>>>> >>>> We could say we insert one ball, then another, then remove one, then the >>>> other, and how many balls are in the vase after that? 0. That's the >>>> sequence of events and the result. >>> Now merely repeat similarly one insertion and one removal for each n in >>> N to get the original vase problem. >> Oh, did you forget the order of events, where ten are added as each is >> removed? That didn't occur in the irrelevant gedankenette that David >> offered. > > As the only important part of that is that each nth ball, for n in N, > is inserted before being removed and removed before noon. > > Absolutely ANY arrangement of insertions and removals satisfying those > constraints will leave the vase empty at noon. 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.
From: Tony Orlow on 20 Oct 2006 13:19
imaginatorium(a)despammed.com wrote: > Tony Orlow wrote: >> cbrown(a)cbrownsystems.com wrote: >>> Tony Orlow wrote: >>>> cbrown(a)cbrownsystems.com wrote: >>>>> Tony Orlow wrote: >>>>>> cbrown(a)cbrownsystems.com wrote: >>>>>>> Tony Orlow wrote: >>>>>>>> cbrown(a)cbrownsystems.com wrote: >>>>>>>>> Tony Orlow wrote: >>>>>>>>>> cbrown(a)cbrownsystems.com wrote: >>>>>>>>>>> Tony Orlow wrote: >>>>>>>>>>>> cbrown(a)cbrownsystems.com wrote: >>>>>>> <snip> >>> <snipitty-snip> >>> >>>>> Do you accept the above statements, or do you still claim that there is >>>>> /no/ valid proof that ball 15 is not in the vase at t=0? >>>>> >>>> 15 is a specific finite number for which we can state its times of entry >>>> and exit. >>> Agreed, >>> >>>> At its time of exit, balls 16 through 150 reside in the vase. >>> Agreed. >>> >>>> For every finite n in N, upon its removal, 9n balls remain. >>> "upon its removal" = "at the time of ball n's removal"; Agreed. >>> >>>> For every n >>>> e N, there is a finite nonzero number of balls in the vase. >>> "For every n e N, there is a finite non-zero number of balls in the >>> vase at t = -1/n". Agreed. >>> >>>> Every >>>> iteration in the sequence is indexed with an n in N. >>> "Balls are only added or removed at a time t = -1/n for some natual n." >>> Agreed. >>> >>>> Therefore, nowhere >>>> in the sequence... >>> ..., i.e, at no time t such that t = -1/n for some natural n, ... >>> >>>> is there anything other than a finite nonzero number of >>>> balls in the vase. >>> Agreed. >>> >>>> Now, where, specifically, in the fallacy in that argument? >>>> >>> Well, what do you state is the conclusion of this argument? >> You have agreed with everything so far. At every point before noon balls >> remain. You claim nothing changes at noon. Is there something between >> noon and "before noon", when **those balls** disappeared? > > Tony, which balls does "those balls" refer to here? Your supposed entire set of naturally-numbered balls. You have a diverging sum of them at every time before noon, and then at noon the sum is 0. So, this subtraction, or whatever, has to have occurred sometime between noon, and "before noon". Do you have such a moment in between x and everything before x? > > You mention *any* time before noon, and we can work out what balls are > in the vase, and we can also work out exactly when those balls [refers > to the balls earlier in this sentence] will all have been removed, and > we know "in advance" that that time of removal will be before noon. So > at *no* time before noon will there be any balls in the vase with any > chance of lingering after noon. At every time before noon there are a growing number of balls in the vase. The only way to actually remove all naturally numbered balls from the vase is to actually reach noon, in which case you have extended the experiment and added infinitely-numbered balls to the vase. All naturally numbered balls will be gone at that point, but the vase will be far from empty. > > Hmm. So your "those balls" must have been introduced into the vase in > this mysterious zone "between before noon and noon". But, see, in > mathematics, it's quite clear there is no such zone - here's a proof. If there is no such zone, and nothing changes AT noon, then the state must be the same as it was immediately BEFORE noon. If at every time before noon there are balls in the vase, then there are still balls in the vase, because nothing happened, at noon. If something DID happen at noon, then it involved infinitely-numbered balls, and the vase has an uncountable number of balls. > > Let B = { t : t is a time, and t is before noon } // the set of all > times before noon > Let N = {noon} // the singleton set of noon > > I suggest that if there is a time _between_ before noon and noon, it > must be a member of the following set: > > Let Z = { t : t is a time, t is after b for all b in B, t is before n > for all n in N } > > Do you agree? > > Would you like to prove that Z is the empty set, just as a little > exercise? > > Brian Chandler > http://imaginatorium.org > Being obnoxious just kinda makes you look dumb, when you agree that the idea of "between before noon and noon" is stupid, but fail to see that it's a direct consequence of your conclusion regarding the vase. |