countability of real numbers? What am I missing?
in [Math]
|
Prev: Computation of definite integral of Sin(x)/x and more
Next: Supernova as the redshift distance candle-basis Chapt10, Fiberglass/Prism & Eclipse-test Experiments #94; ATOM TOTALITY
From: Tonico on 18 May 2010 09:59 On May 18, 4:41 pm, Saijanai <saija...(a)gmail.com> wrote: > On May 18, 6:34 am, Tonico <Tonic...(a)yahoo.com> wrote: > > > > > > > On May 18, 4:27 pm, 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}, > > > ... > > > Hey, wonderful! Now, just for the sake of fun, please do tell us what > > binary fraction from the ones you list represents...I dunno...say, the > > number 1/sqrt(2)? Or the numer 1/e? > > > You know what? Forget the above: you wrote: "I assert it lists all > > real numbers in [0,1]...", so what about a little proof for your > > assertion? > > > Tonio > > The proof should be self-evident in the same way that listing the > Naturals as 1, 2, 3, 4,... obviously doesn't miss any. > > What I'm claiming (and I'm not claiming I'm correct, only that I don't > see the problem) is that the usual issues like Cantor's > Diagonalization don't refute it. This is almost certainly because I > don't understand Cantor's Diagonalization, but regardless, I don't see > where the problem is. Problems? For one, and from what you wrote, I can see only rational numbers, and even that of a very special form. For example, the rational number 0.10101010....(continuing in this fashion) doesn't appear in your list according to the model you wrote... Tonio > > My own guess is that I'm conflating lexical ordering with numerical > ordering but I'm not sure if that is relevant. An order is an order > and a countable sequence is a countable sequence, regardless of how > you arrive at them. > > L.-
From: Saijanai on 18 May 2010 10:26 On May 18, 6:59 am, Tonico <Tonic...(a)yahoo.com> wrote: > On May 18, 4:41 pm, Saijanai <saija...(a)gmail.com> wrote: > > > > > > > On May 18, 6:34 am, Tonico <Tonic...(a)yahoo.com> wrote: > > > > On May 18, 4:27 pm, 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}, > > > > ... > > > > Hey, wonderful! Now, just for the sake of fun, please do tell us what > > > binary fraction from the ones you list represents...I dunno...say, the > > > number 1/sqrt(2)? Or the numer 1/e? > > > > You know what? Forget the above: you wrote: "I assert it lists all > > > real numbers in [0,1]...", so what about a little proof for your > > > assertion? > > > > Tonio > > > The proof should be self-evident in the same way that listing the > > Naturals as 1, 2, 3, 4,... obviously doesn't miss any. > > > What I'm claiming (and I'm not claiming I'm correct, only that I don't > > see the problem) is that the usual issues like Cantor's > > Diagonalization don't refute it. This is almost certainly because I > > don't understand Cantor's Diagonalization, but regardless, I don't see > > where the problem is. > > Problems? For one, and from what you wrote, I can see only rational > numbers, and even that of a very special form. For example, the > rational number 0.10101010....(continuing in this fashion) doesn't > appear in your list according to the model you wrote... > My assertion is that the sequence doesn't miss any real and by extension, you can get arbitrarily close to any real by going out far enough. The relevant issue is countability, at least to me. This gives a countable sequence of numbers that doesn't miss any reals. The fact that the ordering isn't numerical is not relevant to the claim that the sequence is ordered and doesn't miss numbers. Again, I think its an issue with conflating different definitions of ordering, but the fact that a specific number, repeating or non-repeating, doesn't appear in some finite sublist isn't relevant to the claim that no number is "missed". I'm side-stepping the issues, not countering them. The more interesting question is: does the sidestep make the question meaningless or is there some deeper thingie going on. Lawson
From: Torsten Hennig on 18 May 2010 06:29 > 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.
From: Saijanai on 18 May 2010 10:36 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.
From: Chip Eastham on 18 May 2010 10:50
On May 18, 10:26 am, Saijanai <saija...(a)gmail.com> wrote: > On May 18, 6:59 am, Tonico <Tonic...(a)yahoo.com> wrote: > > > > > On May 18, 4:41 pm, Saijanai <saija...(a)gmail.com> wrote: > > > > On May 18, 6:34 am, Tonico <Tonic...(a)yahoo.com> wrote: > > > > > On May 18, 4:27 pm, 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}, > > > > > ... > > > > > Hey, wonderful! Now, just for the sake of fun, please do tell us what > > > > binary fraction from the ones you list represents...I dunno...say, the > > > > number 1/sqrt(2)? Or the numer 1/e? > > > > > You know what? Forget the above: you wrote: "I assert it lists all > > > > real numbers in [0,1]...", so what about a little proof for your > > > > assertion? > > > > > Tonio > > > > The proof should be self-evident in the same way that listing the > > > Naturals as 1, 2, 3, 4,... obviously doesn't miss any. > > > > What I'm claiming (and I'm not claiming I'm correct, only that I don't > > > see the problem) is that the usual issues like Cantor's > > > Diagonalization don't refute it. This is almost certainly because I > > > don't understand Cantor's Diagonalization, but regardless, I don't see > > > where the problem is. > > > Problems? For one, and from what you wrote, I can see only rational > > numbers, and even that of a very special form. For example, the > > rational number 0.10101010....(continuing in this fashion) doesn't > > appear in your list according to the model you wrote... > > My assertion is that the sequence doesn't miss any real and by > extension, you can get arbitrarily close to any real by going out far > enough. The relevant issue is countability, at least to me. This gives > a countable sequence of numbers that doesn't miss any reals. The fact > that the ordering isn't numerical is not relevant to the claim that > the sequence is ordered and doesn't miss numbers. Again, I think its > an issue with conflating different definitions of ordering, but the > fact that a specific number, repeating or non-repeating, doesn't > appear in some finite sublist isn't relevant to the claim that no > number is "missed". > > I'm side-stepping the issues, not countering them. The more > interesting question is: does the sidestep make the question > meaningless or is there some deeper thingie going on. > > Lawson While you can give a countable sequence of numbers that come "arbitrarily close" to any real number in [0,1], this is a weaker result than actually listing _all_ real numbers in [0,1], which is what it would mean for the real numbers to be countable. What you describe with "arbitrarily close" is that the real numbers contain a countable dense subset (in the sense of the metric topology), which is well-known. (For historical reasons this property is termed "separable" by topologists.) Again, it does not conflict with the equally well-known theorem that the real numbers are uncountable. regards, chip |