An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Han de Bruijn on 21 Sep 2006 03:11 Mike Kelly wrote: > Han.deBruijn(a)DTO.TUDelft.NL wrote: > >>Mike Kelly wrote: >> >>>The limit of a sequence need not have a property that each of its >>>elements has. >> >>The limit of the sequence 1,1,1,1,1,1,1,1, ... ,1, ... is 1. > > True. How does that refute anything I said? > > We have a sequence of sets.. > > {0}, {0,1}, {0,1,2}, {0,1,2,3} .... > > and for each we can define a uniform probability distribution to choose > one element of the set. The limit of this sequence of sets is N. > > We construct a sequence of corresponding "Kolmogorov sums", the > summations of the elementary probabilities. > > 1/1, 2/2, 3/3, 4/4, ... = > 1, 1, 1, 1... > > The limit of this sequence of real numbers is 1. > > Now you want to conclude that there is a uniform distribution on N. > However, you offer no justification for this. No. I just want to conclude that this is a sufficient base for inferring that the probability of a natural being divisible by three is 1/3 . Han de Bruijn
From: Mike Kelly on 21 Sep 2006 03:22 Han de Bruijn wrote: > Mike Kelly wrote: > > > Han.deBruijn(a)DTO.TUDelft.NL wrote: > > > >>Mike Kelly wrote: > >> > >>>The limit of a sequence need not have a property that each of its > >>>elements has. > >> > >>The limit of the sequence 1,1,1,1,1,1,1,1, ... ,1, ... is 1. > > > > True. How does that refute anything I said? > > > > We have a sequence of sets.. > > > > {0}, {0,1}, {0,1,2}, {0,1,2,3} .... > > > > and for each we can define a uniform probability distribution to choose > > one element of the set. The limit of this sequence of sets is N. > > > > We construct a sequence of corresponding "Kolmogorov sums", the > > summations of the elementary probabilities. > > > > 1/1, 2/2, 3/3, 4/4, ... = > > 1, 1, 1, 1... > > > > The limit of this sequence of real numbers is 1. > > > > Now you want to conclude that there is a uniform distribution on N. > > However, you offer no justification for this. > > No. I just want to conclude that this is a sufficient base for inferring > that the probability of a natural being divisible by three is 1/3 . It isn't. -- mike.
From: Han de Bruijn on 21 Sep 2006 03:24 Virgil wrote: > In article <3a6c6$4510f00a$82a1e228$27505(a)news2.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>David R Tribble wrote: >> >>>mueckenh wrote: >>> >>>>>Nothing has changed. There is no complete set of natural numbers. Any >>>>>set that can be established is a finite set. Hence, the probability to >>>>>select a number divisible by 3 is 1/3 or very very close to 1/3. >>> >>>Virgil wrote: >>> >>>>That presumes that the allegedly finite set of naturals that can be >>>>constructed is nearly uniform with respect to divisibility by 3 at >>>>least, and probably by other numbers as well. What is the justification >>>>for this assumption? >> >>Wolfgang says litteraly: "_or_ very very close to 1/3". > > Which requires "_nearly uniform_ with respect to divisibility by 3". Sorry, Virgil. I don't get it. Han de Bruijn
From: Han de Bruijn on 21 Sep 2006 03:29 stephen(a)nomail.com wrote: > Assuming that Tony's definition of infinitesimal has anything > to do with any standard definition of infinitesimal. :) What is "any standard definition of infinitesimal" ? Han de Bruijn
From: Han de Bruijn on 21 Sep 2006 06:08
Tony Orlow wrote: > Han de Bruijn wrote: > >> Tony Orlow wrote: >> >>> Han, if I prove inductively, say, that 2^x>2*x for all x>2, do you >>> find it objectionable to say that this also applies to any infinite >>> value, if such a thing existed, given that any infinite value would >>> be greater than any finite value, and therefore greater than 2? >> >> Give me one reason, Tony, why I would find such a theorem interesting >> in the first place. I'd prefer the ultimate terseness in mathematics, >> especially if it comes to infinities. > > Okay, but it's not that particular theorem that is of interest, but the > system of theorems this extension of induction makes possible. If we can > say that such inductive arguments hold in the infinite case, and within > any range up to n one set has size 2n (multiples of 1/2) and the other > x^2 (logs base 2 of naturals), then proving that n>2 -> n^2>2n proves > the second to be bigger than the first, for infinite n, since they are > larger than 2. This gives us a nice exact way of comparing infinite sets > over an infinite range, free from "limit ordinals" and other nonsense. > This allows us to distinguish between the sizes of the naturals vs. the > evens, and even detect the addition or removal of a single element from > an infinite set. > > Hope that made sense. Thought I've made my point. When I said that your infinities are not worse than their standard alephs, then I didn't say that I find your infinities "better" :-( Sorry ... Han de Bruijn |