An uncountable countable set
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From: Virgil on 27 Oct 2006 18:56 In article <454229fa(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > RLG wrote: > > "Tony Orlow" <tony(a)lightlink.com> wrote in message > > news:4540f9d0(a)news2.lightlink.com... > >> This is very simple. Everything that occurs is either an addition of ten > >> balls or a removal of 1, and occurs a finite amount of time before noon. > >> At the time of each event, balls remain. At noon, no balls are inserted or > >> removed. The vase can only become empty through the removal of balls, so > >> if no balls are removed, the vase cannot become empty at noon. It was not > >> empty before noon, therefore it is not empty at noon. Nothing can happen > >> at noon, since that would involve a ball n such that 1/n=0. > > > > > > Tony, I think your confusion results from imagining the balls without any > > labels. In this case at 1 minute before noon 10 balls are inserted into > > the > > vase, at 1/2 minute before noon 9 balls are inserted into the vase, at 1/4 > > minute before noon, 9 more balls are inserted into the vase and, in > > general, > > at (1/2)^n minutes before noon 9 balls are inserted into the vase. So you > > are saying that the number of vase balls at noon is: > > > > > > 10 + 9 + 9 + 9 + 9 + 9 + ... = Infinite. > > > > Yes, but I don't consider that confusion. If the problem is solvable > without the labels, then the labels don't matter. But the problem is NOT "solvable" without labels if that means being able to determine what is left in the vase at noon. > > > > > Or, since one ball is removed each time ten more are added, we should > > write: > > > > > > 10 + (10-1) + (10-1) + (10-1) + (10-1) + ... = Infinite. > > > > > > Now, this divergent series is conditionally convergent. That means we can > > make the sum equal any value we like depending on how the terms are > > arranged. So if we choose 0 for the sum that is perfectly valid: > > > > > > 10 + (10-1) + (10-1) + (10-1) + (10-1) + ... = 0. > > > > No, we went through this in another thread. The only way to get a sum of > 0 is by rearranging the terms and grouping so you have ten -1's for > every +10. But, the sequence of events is specified NOT to be in that > order. No ball can be removed without having ten inserted immediately > before. And no ball can be inserted without being removed before noon. > So, despite the silly games that mathematicians may play with > "conditionally convergent" series, none of that applies to the ball and > vase problem as stated. Does that sound confused to you? No more so that TO's usual level of total confusion. > > > > > In this case there are no balls in the vase at noon. Without labels on the > > balls there is no criterion by which to select what the sum should be and > > the end state of the supertask is undefined. As I noted in an earlier > > post, > > if some of the balls are labeled with numbers that are not naturals, for > > example transfinite ordinal numbers, we can choose "Infinite" for the sum > > if > > the circumstances require it. > > > >. You cannot remove a ball without adding ten > more. And you also cannot insert a ball without removing it before noon.
From: Lester Zick on 27 Oct 2006 18:56 On 27 Oct 2006 14:31:46 -0700, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: >Lester Zick wrote: >> According to MoeBlee's recent lectures on the subject of exhaustive >> mathematical definitions one cannot simply define IN and OUT, one must >> use a placeholder such as IN(x) and OUT(x) to establish the domain of >> discourse. > >Please stop mangling what I've said and then representing your mangled >interpretations as if they are what I said. I will if you'll just stop doubletalking, Moe. ~v~~
From: Virgil on 27 Oct 2006 18:58 In article <45422a2f(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > imaginatorium(a)despammed.com wrote: > > David Marcus wrote: > >> imaginatorium(a)despammed.com wrote: > >>> Virgil wrote: > >>>> In article <45417528$1(a)news2.lightlink.com>, > >>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>> <snip> > >>> > >>>>> For what it's worth, and I know this doesn't add a lot of credibility to > >>>>> Ross in your eyes, coming from me, but I think Ross has a genuine > >>>>> intuition that isn't far off with respect to what's controversial in > >>>>> modern math. Sure, he gets repetitive and I don't agree with everything > >>>>> he says, but his cryptic "Well order the reals", which I actually > >>>>> haven't seen too much of lately, is a direct reference to his EF > >>>>> (Equivalence Function, yes?) between the naturals and the reals in > >>>>> [0,1). The reals viewed as discrete infinitesimals map to the > >>>>> hypernaturals, anyway, and his EF is a special case of my IFR. So, to > >>>>> answer your question, I think Ross makes some sense. But, of course, > >>>>> coming from me, that probably doesn't mean much. :) > >>>> Coming from TO it damns Ross. > >>> Even by your standards, Virgil, this is egregiously silly. TO skips the > >>> basic exposition in Robinson's book, but finds a sentence he likes. So > >>> this "damns" Robinson's non-standard analysis, does it? > >> Virgil said "Ross", not "Robinson", I believe. > > > > Yes, of course. But Virgil's implication is that "TO says person P is > > right about something" implies P is wrong. This may, contingently, be > > true about Ross, but the argument could equally be applied to Robinson, > > in which case the conclusion is obviously not true. > > > > Brian Chandler > > http://imaginatorium.org > > > > And, what about those rare occasions when I agree with Virgil? Uh oh. It actually has happened that TO and I agree on something. It does not happen boringly often but it does happen.
From: Virgil on 27 Oct 2006 19:00 In article <45423853(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> stephen(a)nomail.com wrote: > >>> What are you talking about? I defined two sets. There are no > >>> balls or vases. There are simply the two sets > >>> > >>> IN = { n | -1/(2^floor(n/10)) < 0 } > >>> OUT = { n | -1/(2^n) < 0 } > >> For each n e N, IN(n)=10*OUT(n). > > > > Stephen defined sets IN and OUT. He didn't define sets "IN(n)" and "OUT > > (n)". So, you seem to be answering a question he didn't ask. Given > > Stephen's definitions of IN and OUT, is IN = OUT? > > > > Yes, all elements are the same n, which are finite n. There is a simple > bijection. But, as in all infinite bijections, the formulaic > relationship between the sets is lost. What never existed cannor be lost.
From: Virgil on 27 Oct 2006 19:05
In article <454238b9(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > So, noon doesn't exist in this case either? > > > > Nothing happens at noon, and as long as there is no claim that anything > happens at noon, then there is no problem. By noon, it is all over and every ball has been removed. >Before noon there was an > unboundedly large but finite number of balls. At noon, it is the same. Let's see! The first numbered ball is removed before noon. If the nth ball is removed before noon, then so is the n+1st ball. So that in any system consistent with ZF or NBG, EVERY naturally numbered ball (which is all the balls allowed to be inserted by the gedankenexperiment) are removed. |