countability of real numbers? What am I missing?
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From: Chip Eastham on 18 May 2010 10:58 On May 18, 10:36 am, Saijanai <saija...(a)gmail.com> wrote: > On May 18, 7:29 am, Torsten Hennig <Torsten.Hen...(a)umsicht.fhg.de> > wrote: > > > > > > On May 18, 6:43 am, Torsten Hennig > > > <Torsten.Hen...(a)umsicht.fhg.de> > > > wrote: > > > > > OK, all you set/number theorists, what is wrong > > > with > > > > > this binary > > > > > fraction sequence. I assert it lists all real > > > numbers > > > > > [0,1] (allowing > > > > > for duplicates): > > > > > {0.0, 0.1}, > > > > > {0.00, 0.01, 0.10, 0.11}, > > > > > {0.000, 0.001, 0.010, 0.011, 0.100, 0.101, 0.110, > > > > > 0.111}, > > > > > ... > > > > > I only see rational numbers in your enumeration ; > > > > where are the irrational ones ? > > > > > Best wishes > > > > Torsten. > > > > The sequence is never-ending, so you can follow an > > > arbitrarily long > > > trail from level to level to create a Cauchy-Sequence > > > that corresponds > > > to any real, including the irrationals. My assertion > > > is that you don't > > > miss anything in the ordering, because the initial > > > squence-ordering > > > isn't numerical, just lexical. > > > > I'm not sure if that's cheating or not. I suspect it > > > is, but I'm only > > > auditing elementary level number theory/set > > > theory/analysis lectures > > > online right now and I'm no doubt missing something > > > (or just don't > > > understand the lectures I've already seen in the > > > first place). > > > > L. > > > The point is that it's not enough that you can form > > a Cauchy sequence from the numbers of your > > enumeration to reach every real number in [0;1]. > > An enumeration (x_n) of the reals in [0;1] would mean: > > for a specified real number x in [0;1] you must be able to name the index K for which x_K = x. > > > Best wishes > > Torsten. > > OK, I understand what you're saying, but is it a valid issue? I'm not > claiming to be able to give the index or a specific real only that my > enumeration doesn't miss any. > > L. Your use of "doesn't miss any" seems disingenuous. Either every specific real number in [0,1] is in the list you gave or there's some real number in [0,1] that is not in the list. As discussed up- thread, countability is not about coming close. Your list misses almost everything in [0,1], as apparently you only hit the finite (terminating) binary fractions. As pointed out, you don't get to even such rational numbers as 1/3 no matter how far out the list is taken, because 1/3 has no terminating binary representation. regards, chip
From: William Hughes on 18 May 2010 11:17 On May 18, 10:27 am, Saijanai <saija...(a)gmail.com> wrote: > OK, all you set/number theorists, what is wrong with this binary > fraction sequence. I assert it lists all real numbers [0,1] (allowing > for duplicates): > > {0.0, 0.1}, > {0.00, 0.01, 0.10, 0.11}, > {0.000, 0.001, 0.010, 0.011, 0.100, 0.101, 0.110, 0.111}, > ... Without actually looking at your list, Karnac says you only have finite decimal expansions in your list. Karnac is right again. It is well known that the finite sequences are countable. If you want to list an infinite decimal then approximating it is not good enough. It has to be in the list. Cantor's argument will not work with finite decimals, because the anti-diagonal produced is not a finite decimal. - William Hughes
From: Aage Andersen on 18 May 2010 11:17 "Saijanai" > OK, all you set/number theorists, what is wrong with this binary > fraction sequence. I assert it lists all real numbers [0,1] (allowing > for duplicates): > {0.0, 0.1}, > {0.00, 0.01, 0.10, 0.11}, > {0.000, 0.001, 0.010, 0.011, 0.100, 0.101, 0.110, 0.111}, > ... I don't see 0.111111111111111111111111111.......... in your list. Aage
From: W^3 on 18 May 2010 12:25 In article <6b027f1d-3ba8-4f9f-9a12-7fb28e08fafd(a)y18g2000prn.googlegroups.com>, Saijanai <saijanai(a)gmail.com> wrote: > OK, all you set/number theorists, what is wrong with this binary > fraction sequence. I assert it lists all real numbers [0,1] (allowing > for duplicates): > {0.0, 0.1}, > {0.00, 0.01, 0.10, 0.11}, > {0.000, 0.001, 0.010, 0.011, 0.100, 0.101, 0.110, 0.111}, > ... Your list doesn't include x = .010101010.... Yes, from your list you can find a sequence converging to x. And this might be interesting to you. But it's irrelevant to the point at hand: Yours is not a list of all real numbers because x is a real number and it is nowhere in your list.
From: James Burns on 18 May 2010 12:42
Saijanai wrote: > On May 18, 7:29 am, Torsten Hennig > <Torsten.Hen...(a)umsicht.fhg.de> wrote: >>The point is that it's not enough that you can form >>a Cauchy sequence from the numbers of your >>enumeration to reach every real number in [0;1]. >>An enumeration (x_n) of the reals in [0;1] would mean: >>for a specified real number x in [0;1] you must be able >>to name the index K for which x_K = x. > OK, I understand what you're saying, but is it a valid > issue? I'm not claiming to be able to give the index > or a specific real only that my enumeration doesn't > miss any. The question rests upon what you (or we) mean by "doesn't miss any". The complete binary expansion for 1/3 doesn't show up on your list anywhere, although matches to the initial piece of it, for arbitrarily long initial pieces, do show up. If that counts as "showing up", then, yes, every real "shows up". This is well known, but it is not the sense used in comparing the sizes of sets. If two sets have the same cardinality, then there is a bijection between them. You will have natural numbers (your index) corresponding to all those approximations to 1/3, but no natural number corresponding to 1/3 itself. We know that there is no such natural number, because every real in your list has the form k/2^m, for some naturals k and m. Do you accept different size infinities, such as the more general Cantor's proof shows? That is a much simpler proof, with less room for wiggling around to look for a counter-example. For any set S, and any function f: S -> P(S), from S to the powerset of S, f is not a bijection, therefore S and P(S) do not have the same cardinality. Proof: Let Y = the subset of S {x| x not in f(x)}. There is no y in S such that f(y) = Y. Notice that you can biject binary sequences of {0,1} with the subsets of the naturals, .10101... <-> {1,3,5,...} If you take care of cases like 0.1000... = 0.0111... (messy but quite doable), you will have shown that the reals are a larger infinity than the naturals. Jim Burns |