An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: David R Tribble on 25 Sep 2006 19:32 [I'm a bit late in this subthread, but no matter...] Tony Orlow wrote: > If omega is the successor to the set of all finite naturals, ... It's not. It's a limit ordinal. > ... it is > greater than all finite naturals, as any successor is greater then all > those that precede it. It is certainly a positive number, if it is a > count or size of anything. If it is a positive natural greater then 1, ... No, it can't be a natural, because of what you just said, that's it larger than all naturals. A natural can't be larger than itself, so omega can't be a natural. > ... then its reciprocal is a real in (0,1). Nope. If w > x for all real x, then 1/w < 1/x for all real x as well. Which means that 1/w cannot be a real. If 1/w was a real, then 1/(1/w) = w would also be a real, because 1/x is real for all real x > 0. But then w could not be greater than all reals (or all naturals), since it would have to be a real that is greater than itself. > To say that some count which is > greater than any finite count does not obey this general rule is a > kludge, like all the transfinite "arithmetic". Your "count" is not a well-defined term. Do you mean real, ordinal, cardinal, or something else?
From: Tony Orlow on 25 Sep 2006 19:50 cbrown(a)cbrownsystems.com wrote: > Tony Orlow wrote: > > <snip> > >> I did give another curve with the same "Tlimit" as the staircase in the >> limit, which produced an interesting result, giving weight to the notion >> that an infinitesimal is something distinct from 0, whose square is not >> distinct from 0. > > Suppose we let B represent Big'un; then B*1/B = 1, where 1/B is an > infinitesimal. Then what you are saying means > > 1/B = 1/B > 1*1/B = 1/B > (B*1/B)*1/B = 1/B > B*(1/B*1/B) = 1/B > B*(1/B^2) = 1/B > > Since 1/B is infinitesimal, its square is not distinct from 0; so... > > B*(0) = 1/B > 0 = 1/B > > So 1/B is identical to 0. Where is my error? > > Cheers - Chas > Hi Chas - I haven't been online lately, so please excuse the delay. You haven't made any error, except in interpretation of my position. I said that I got an interesting result from applying a segment-sequence definition of your staircase-and-diagonal comparison, which gave "weight" to the notion of an infinitesimal being defined in such a way. There are a number of ways infinitesimals may be considered. Mine supported the notion of a "nilpotent" infinitesimal in the sense that, even at the first-order infinitesimal level, where there is a unit infinitesimal, squares of such units would be negligible, being infinitely smaller than the unit. That general idea seems reasonable for any infinitely small quantity. Then the moving staircase which I defined is the same as your staircase in the limit, with the same descriptive sequence of infinitesimal offset pairs. :) Tony
From: Tony Orlow on 25 Sep 2006 20:02 Dik T. Winter wrote: > In article <eet4ru$6ku$1(a)news.msu.edu> stephen(a)nomail.com writes: > > cbrown(a)cbrownsystems.com wrote: > > > Tony Orlow wrote: > > >> I did give another curve with the same "Tlimit" as the staircase in the > > >> limit, which produced an interesting result, giving weight to the notion > > >> that an infinitesimal is something distinct from 0, whose square is not > > >> distinct from 0. > > > > > Suppose we let B represent Big'un; then B*1/B = 1, where 1/B is an > > > infinitesimal. Then what you are saying means > > > > > 1/B = 1/B > > > 1*1/B = 1/B > > > (B*1/B)*1/B = 1/B > > > B*(1/B*1/B) = 1/B > > > B*(1/B^2) = 1/B > > > > > Since 1/B is infinitesimal, its square is not distinct from 0; so... > > > > > B*(0) = 1/B > > > 0 = 1/B > > > > > So 1/B is identical to 0. Where is my error? > > > > Assuming that Tony's definition of infinitesimal has anything > > to do with any standard definition of infinitesimal. :) > > It has with one of the definitions. Have a look at "synthetic differential > geometry". Lavendhomme, Kock, amongst others. They use infinitesimals > that are nil-potent. But, of course, they deny the law of the excluded > middle in that system. I am not sure that the law of the excluded middle holds in all cases, particularly where probabilistic/arithmetic logic is concerned. In my naivete, am not sure how that relates to the notion of a nilpotent infinitesimal. But, there is sense in the notion that, expressing a measure in a formula using some particular infinitesimal unit, squares of that unit would be negligible, since they are infinitesimal relative to that unit and would therefore have no measure (at least for a finite number of terms). Alternatively, those terms can be considered nonzero, but sub-infinitesimal, and analysis can continue through any number of successive infinitesimal levels. I'd say they are two equally valid alternatives. :) Tony
From: Tony Orlow on 25 Sep 2006 20:08 Dik T. Winter wrote: > In article <eeu7fn$ti$2(a)news.msu.edu> stephen(a)nomail.com writes: > > Dik T. Winter <Dik.Winter(a)cwi.nl> wrote: > > > > > Tony Orlow wrote: > > > > >> I did give another curve with the same "Tlimit" as the staircase > > > > >> in the limit, which produced an interesting result, giving weight > > > > >> to the notion that an infinitesimal is something distinct from 0, > > > > >> whose square is not distinct from 0. > > Note what Tony writes above. > > > > It has with one of the definitions. Have a look at "synthetic differential > > > geometry". Lavendhomme, Kock, amongst others. They use infinitesimals > > > that are nil-potent. But, of course, they deny the law of the excluded > > > middle in that system. > > > > From the context, I gather that nil-potent infinitesimals > > are infinitesimals whose square equals 0? I could not find > > a clear definition on the web, but that interpretation seems > > consistent with what I did find. > > That is it. I believe Anders Kock has quite some information online. > > > I do not think Tony's infinitesimals are nil-potent. > > But see what Tony did write! > > > Tony's > > infinitesimals are just really small numbers whose inverses are > > infinite that otherwise behave just like ordinary real numbers. > > I think that Tony's idea of infinitesimals is actually more > > in line with Robision's infinitesimals which do not have > > nilpotent infinitesimals, if my understanding is correct. > > Indeed, Robinson's infinitesimals are not nil-potent. But if Tony > states that his infinitesimals are nil-potent, they are closer to > the infinitesimals of Kock. Off-hand, I do not remember whether > inversion of these infinitesimals does work. Thank you for the references, Dik. I'll bookmark and read them over. My position is that infinitesimals can be viewed as nilpotent for purposes of measure, but that the same notions can be applied recursively so that there is no end to the formulaic smallness that is considered, in which case we have inverses for the full spectrum of definable infinities. And, that's good. :) I simply said this example gave weight, or lent credibility, to the idea of the nilpotent infinitesimal. That's because it was a measure problem. :) Tony
From: Tony Orlow on 25 Sep 2006 20:28
Virgil wrote: > In article <451149ef(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > > >> Consider the equally spaced staircase from (0,0) to (1,1), as the number >> of steps increases from 1 without bound. Is it the same as the diagonal >> line? Inductively we can prove that the length of the staircase is 2 at >> every step. Does it really suddenly become sqrt(2) in the infinite case? > > There is no "infintie case", there is only a limit case. > Then noon never comes and the vase is never empty, and you have just agreed entirely with Han's entire point. Congratulations. > > If the cases for a finite number of steps are sets of points, so is the > limit case. Correct. There is no measure. > > If the finite cases are sets of segments with specific directions > determined by their endpoints, the limit case will only contain pairs of > identical points which do not determine any direction at all, and so is > ill defined. Incorrect. First of all, each segment was defined by a xy-offset pair, starting at point (0,0). Each such pair denotes, through these relative coordinates, a length and a direction. While the pairs of coordinates in the set of points have negligible difference in location relative to (0,0), the length and direction of each and every segment pair in the sequences remain constantly different between the two, and the sum of the staircase in the limit is indeed 2 and not sqrt(2). It's not surprising that you can't measure a curve with points. They have no measure. :) Tony |