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From: Tony Orlow on 29 Sep 2006 14:57 cbrown(a)cbrownsystems.com wrote: > Tony Orlow wrote: >> cbrown(a)cbrownsystems.com wrote: >>> Tony Orlow wrote: > <snip> > >>>> Like I said, there were >>>> terms in my infinitesimal sections of moving staircase which differed by >>>> a sub-infinitesimal from those in the original staircase. So, they could >>>> be considered to be two infinitesimally different objects in the limit. >>> Here's a thing that confuses me about your use of the term "limit". >>> >>> In the usual sense of the term, every subsequence of a sequence that >>> has as its limit say, X, /also/ has a limit of X. >>> >>> For example, the sequence (1, 1/2, 1/2, 1/3, ..., 1/n, ...) usually is >>> considered to have a limit of 0. And the subsequence (1/2, 1/4, 1/6, >>> ..., 1/(2*n), ...) which is a subsequence of the former sequence has >>> the same limit, 0. >>> >>> But the way you seem to evaluate a limit, the sequence of staircases >>> with step lengths (1, 1/2, 1/3, ..., 1/n, ...) is a staircase with >>> steps size 1/B, where B is unit infinity; but the sequence of >>> staircases with step lengths (1/2, 1/4, 1/6, ..., 1/(2*n), ...), which >>> is a subsequence of the first sequence, would seem to have as its limit >>> a staircase with steps of size 1/(2*B). >>> >>> Unless steps of size 1/B are the same as steps of size 1/(2*B), I don't >>> see how that can be possible. >>> >>> Cheers - Chas >>> >> It's possible because no distinction is currently made between countable >> infinities, even to the point where a set dense in the reals like the >> rationals is considered equal to a set sparse in the reals like the >> naturals. Where there is no parametric understanding of infinity, >> infinity is just infinity, and 0 is just 0. > > Uh, OK. I assume that you somehow resolve this lack of "parametric > understanding" in /your/ interpretation of T-numbers. Well, yes. That's the whole point, but I'm not sure which particular numbers you mean. Maybe the T-riffics, which are center-indefinite digital numbers? > >> Where there is a formulaic >> comparison of infinite sets as n->oo, the distinction can be made. The >> fact that you have steps of size 1/n as opposed to steps of size 1/(2*n) >> is a reflection of the fact that the first set has twice the density on >> the real line as the first. As a proper superset, it SHOULD be larger. >> So, it's quite possible to make sense of my position, with a modicum of >> effort. > > Well, let me ask you this: > > Suppose we have the original sequence of staircases, with step lengths > (1, 1/2, 1/3, ..., 1/n, ...). Let S be the T-limit staircase; you claim > that it has step sizes 1/B, where B is unit infinity. Well, where B is some infinite number of iterations for both staircase and diagonal. We can call it the unit infinity if you want, especially since it covers a space of 1 unit. :) > > Now, suppose we just forget about the very first staircase, but > otherwise continue normally. Now we have step lengths (1/2, 1/3, 1/4, > ..., 1/(n+1), ...). Do you claim that the T-limit staircase has step > size 1/(B+1)? Do you propose that 1/B is or is not equal to 1/(B+1)? Oh, I see where this is going. I'm not saying that every problem breaks down into B steps. Smaller and larger infinities can be expressed with B formulaically. I am simply saying that, for infinite n, as for finite n, the ratio between 1/n and 1/n remains 1 (slope of the diagonal). The ratio between 0 and 1/n is 0 or infinite, if 1/n>0, and these are the slopes of the segments comprising the staircase. You might conclude that, in the limit, 1/n=0, and the staircase breaks down to segments of (0,0). In that case, your measure is lost. What I would say is that the sequence starting at n=2 has 1 less element than that starting at n=1, if they have the same upper limit. > > I mean, suppose you actually started constructing staircases, starting > with side length 1; I then enter the room when you start to construct > the second staircase with length 1/2. I copy everything that you do > from thereon, just as you do it: you make a staircase of length 1/2, > and at the same time, so do I. You make a staircase with side length > 1/3, and at the same time, so do I. And so on. > > But somehow, we end up with different results "in the infinite case"? > > Cheers - Chas > No, that grows out of an assumption that I think Big'un is some limit to how many iterations one could possibly have, or something. When I read what you wrote I went, "huh?? no." It seemed to be a mixture of standard transfinite thinking and what I was saying, kind of like a cross between a platypus and a tiger. In any case n, whether finite or infinite, the staircase has length 2, as the sum of the lengths of the segments. In the moving staircase, removing terms of order n^2, we have the same limit curve as well. I liked it, anyway. Tony
From: Tony Orlow on 29 Sep 2006 15:00 Han de Bruijn wrote: > stephen(a)nomail.com wrote: > >> In TO-matics, it is also possible to end up with >> an empty vase by simply adding balls. According to TO-matics >> >> ..1111111111 = 1 + 1 + 1 + 1 + ... >> >> and ..1111111111 + 1 = 0 >> >> So if you just keep on adding balls one at a time, >> at some point, the number of balls becomes zero. >> You have to add just the right number of balls. It is not >> clear what that number is, but it is clear that it >> exists in TO-matics. > > Commonly known with digital computers as "overflow" ? > If TO-matics is an idealization of overflow, then it _is_ consistent > anyway. Sad for you :-( > > Han de Bruijn > Haha. True. It is a matter of overflow, for positive registers. That's why it's wrong here - an error. For signed 2's complement, it's not considered overflow, but interestingly, 0 isn't the only number which is its own additive inverse. In addition to 000...000, it is also true of 100...000. That's supposedly the largest negative, but it is its own negative. Hmmm... 100...000 is really the representation of +/- oo in signed binary. Have a nice day! Tony
From: Tony Orlow on 29 Sep 2006 15:03 Han de Bruijn wrote: > Tony Orlow wrote: > >> Your axiom system is a farse. > > I'd rather think it is a farce. > > Han de Bruijn > Ooops, yes, but it might as well be in Farsi. Tony
From: Virgil on 29 Sep 2006 15:11 In article <ce018$451cd047$82a1e228$14108(a)news1.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Virgil wrote: > > > In article <b1fd9$451b85a3$82a1e228$11085(a)news1.tudelft.nl>, > > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > > > >>Virgil wrote: > >> > >>>In article <d12a9$451b74ad$82a1e228$6053(a)news1.tudelft.nl>, > >>> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>> > >>>>Randy Poe wrote, about the Balls in a Vase problem: > >>>> > >>>>>It definitely empties, since every ball you put in is > >>>>>later taken out. > >>>> > >>>>And _that_ individual calls himself a physicist? > >>> > >>>Does Han claim that there is any ball put in that is not taken out? > >> > >>Nonsense question. Noon doesn't exist in this problem. > > > > What has noon to do with my question? > > > > According to my recollection of the problem, there was a specific time > > (prior to noon) at which any ball which had been put into the vase was > > taken out. > > > > Does Han dispute my recollection? > > Maybe we're not talking about the same version of the SuperTask. > So: where can I find this recollection of yours? > > Han de Bruijn The problem as I recall it was this: Given an infinite set of balls numbered with the infinite set of naturals and an "infinitely large" initially empty vase, and a positive time interval in seconds, t, and a small positive time interval in seconds, epsilon ( much smaller than t/2). (1) At time t before noon balls 1 through 10 are put into the vase and at time t - epsilon before noon ball 1 is removed. (2) At time t/2 before noon balls 11 through 20 are put into the vase and at time (t - epsilon)/2 before noon ball 2 is removed. .... (n) At time t/2^(n-1) before noon balls 10*(n-1)+1 through 10*n are put in the vase and at time (t-epsilon)/2^(n-1) before noon, ball n is removed. .... The question is what will be the contents of the vase at or after noon. Of course, there can be no physical analog of this thought experiment, but in mathematics the question has a clear answer.
From: Virgil on 29 Sep 2006 15:20
In article <97d49$451cd1e0$82a1e228$15818(a)news1.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Virgil wrote: > > > In article <1159437157.109258.97400(a)b28g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > >>Virgil schrieb: > >> > >>>In article <1159186907.615747.304410(a)h48g2000cwc.googlegroups.com>, > >>> mueckenh(a)rz.fh-augsburg.de wrote: > >>> > >>>>1/3 is a number, properly defined, for instance, by the pair of numbers > >>>>1,3 or 2,6 or 3,9 etc. But 0.333... is not properly defined because you > >>>>cannot index all positions, you cannot distinguish the positions of > >>>>this number from those with finite sequences (and you cannot > >>>>distinguish them from other infinte sequences which could exist, if one > >>>>could exist). > >>> > >>>Def: 0.333... = lim_{n -> oo} Sum_{k = 1..n} 1/3^n > >> > >>Definitions (even correct definitions unlike this one) don't guarantee > >>existence (I used above "to be properly defined" but I meant "to > >>exist"). Example: The set of all sets is defined but is not existing. > > > > Def, corected: 0.333... = lim_{n -> oo} Sum_{k = 1..n} 3/10^k > > > > And it exists because there is a real number (actually a ratonal number) > > L = 1/3 such that for every epsilon greater than 0, there is a largest > > n such that | L - sum_{k+1..n}| >= epsilon. > > Shouldn't that be "such that | L - sum_{k = 1..n}| <= epsilon" ? And why > that "largest n" instead of just "n" ? There are a number of equivalent variations on the definition of the limit of a series. This is Note that if there is a largest n , say n_0, for which | L - sum_{k+1..n_0}| >= epsilon then for n > n_0, | L - sum_{k+1..n}| < epsilon so the result is the same as for the other definition. |