Cantor Confusion
in [Math]
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From: Virgil on 21 May 2007 02:03 In article <1179696860.625351.146190(a)u36g2000prd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 20 Mai, 17:57, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > WM says... > > > > >> > It is. Set theory is simply biased. Consider the list > > > > >> > 0.666... > > >> > 0.3666... > > >> > 0.33666... > > >> > 0.333666... > > >> > ... > > > > >> > If the diagonal number is defined by "replace 6 by 3", then we have > > >> > two answers none of which can be preferred by logic, but the second of > > >> > which is suppressed by convention. > > > > The diagonal number is 0.333... which is not on the list. > > The diagonal number cannot have more 3's than every list number has > 3's. Even if that were so, which it is not, the diagonal MUST have fewer 6's that any list element has, as every list element has at least one, but the diagonal has none. > If the diagonal number is complete in the sense that at every > position indexed by a natural number there is a 3, then there must be > a complete sequence of 3's among the list entries too. Each list member has MORE 6's than 3's, so at what point does WM claim MORE becomes LESS? > Otherwise the > list is not complete, i.e., there is always a 6 at some natural index. There always is a 6 at SOME natural index, since there are always more 6's than 3's. There are never more than finitely many 3's so there can always be at least as many 6's plus one extra 6. > If the list is not complete however, the diagonal number cannot be > complete either. But in any case the "diagonal" does not have a 6 but every listed number does. > And the saying that at *every* position there is a > digit b_n =/= a_n,n is nonsense. For every listed number there is some position at which it has a digit 6. In the diagonal number there is no digit position at which has a 6. > > > > >For the entries E(n) of the list we find > > >lim[n-->oo] (E(n) - 0.333...) = 0. > > >It is the same case as lim[n-->oo] (1 - 0.999...9 with n 9's) = 0. > > > > That's true. In this particular case, the limit of the sequence > > is equal to the diagonal of the sequence. So what? > > Why only in this particular case? Is lim[i --> oo] (b_i - a_i) * 10^-i > = 0 correct only in one special case? Of what relevance is lim[i --> oo] (b_i - a_i) * 10^-i = 0 ? The Cantor theorem says nothing about limits. > > >If every initial segment of the diagonal number is represented by the > > >initial segment of an entry of the list, then the full diagonal number > > >is represented by an entry of the list. > > > > That's false. > > No, that's correct. It is the only valid interpretation of the notion > infinity. If it says that 0.333..., with nothing but digits of 3 in it must have a digit of 6 in it, then WM is spouting nonsense again. WM's MathUnRealism fails again.
From: Virgil on 21 May 2007 03:31 In article <1179696860.625351.146190(a)u36g2000prd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > So the problem of replacing 0 by 9 is not existing. But we have in the > limit j --> oo: D_j = 0. Without this limit there are no real numbers > existing but at most sequences of digits. One cannot have the sequences of digits without having the numbers that they represent. Except possibly in WM's MathUnRealism.
From: WM on 21 May 2007 07:29 On 21 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179653443.812676.189...(a)u30g2000hsc.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 19 Mai, 04:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > Another question about chapter 10. Do you understand what a normal > > > > > number is? I think not. Off-hand I do not know whether there are > > > > > normal numbers that are normal with respect to all bases (although > > > > > it is expected that pi is one). > > > > > > > > Such numbers are called absolutely normal. But to know that is neither > > > > required for the readers of my book in order to understand > > > > MatheRealism nor would it be useful to expand the number of pages and > > > > the price of the book by a large factor. > > > > > > So you prefer to talk nonsense? > > > > I consider the difference between absolutely normal and weakly normal > > not important with respect to the topic of my book, in particular > > since even such experts as you seem to have no clue about that. > > Weakly normal is *not* a common definition, but it is Borel's terminology Borel wrote in French. > for what is now called "simply normal". Please do not conclude from your lacking knowledge on that of others. Weakly normal is a common definition. See fort instance. http://eom.springer.de/N/n067560.htm Or do you think that "weakly normal" is not common because it is missing in Wikipedia? I would estimate Springer somewhat higher than Wikipedia. > > > > > There are different notions (for instance weakly normal numbers and > > > > absolutely normal numbers). Of course normal numbers can be > > > > constructed, one of the simplest cases is the rational number > > > > 0.12012012... with respect to base 3, > > > > > > That number is not normal to base 3. > > > > That number is weakly normal, namely normal to base 3. > > It is *not* normal to base 3. It is "simply normal to base 3", or in > Borel's terminology "weakly normal to base 3". You cannot omit the > base. I did not. I said with respect to base 3. > > If you don't > > know about the definition of normal numbers you should first inform > > you. Online for instance > >http://eom.springer.de/N/n067560.htm > > Read what is written there, A number is normal to a particular base if > *all* n-digit sequences are equi-probable. > > > > > but as there must be included > > > > also normally distributed frequencies of 10^100-tuples and larger > > > > tuples most normal numbers cannot be constructed. > > > > > > Do you know about the Chapernowne numbers? But be also aware that the > > > Copeland-Erdos number is normal to base 10. A quote: > > > "While Borel proved the normality of almost all numbers with respect > > > to Lebesgue measure, with the exception of a number of special classes > > > of constants, the only numbers known to be normal (in certain bases) > > > are artificially constructed ones such as the Champernowne constant > > > and the Copeland-Erdos constant." > > > > Another quote: "The weakly-normal number (to base 10) > > 0.01234567890123456789... is of course rational." > >http://eom.springer.de/N/n067560.htm > > Yes, so what? The Champerowne constants and the Copeland-Erdos constants > are *not* rational. Read just below your quote, where they give a > Champerowne constant. So what? I never said that every however normal number must be rational. But 0.01234567890123456789... (weakly normal to base 10) and 0,012012012... (weakly normal to base 3) *are* rational. Regards WM
From: WM on 21 May 2007 07:44 On 21 Mai, 04:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179654356.461792.242...(a)n59g2000hsh.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 19 Mai, 04:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > According to current mathematics, pi is well defined. Even according > > > > to MatheRealism pi is well defined (as an idea). > > > > > > Your distinction between "number" and "idea" is just terminology, and not > > > more than that. > > > > Its is more. You cannot answer the question whether the numbers P = > > [pi*10^10^100] and P' = P with the last digit replaced by 3 nsatisfy P > > < P'. > > Yes, so what? Your distinction is just terminology, and not more than that. It is by far more than a difference in terminology if one will never or always be able to answer a question. > > > > > > It is > > > > > indeed easy to show that there is an injection from that set to the > > > > > set of natural numbers. Consider all finite sentences over some > > > > > alphabet (let's say the 26 latin letters plus a space). Each such > > > > > sentence can be considered as a base-27 number, so we have an > > > > > injection. > > > > > > > > Claim of injection is correct. Claim of bijection is wrong. (pi for > > > > instance is defined by many different definitions). > > > > > > Do you not know the theorem that if there is an injection of some set S to > > > a countable set T that S is also countable? > > > > I use it for the paths and nodes of the tree. But you keep on asking > > for a bijection. The injection has already been shown. > > No. You do *not* give an injection from paths to nodes. What node does > the path 0.0101010101... inject to? How do you define the injection? I define it in the same way as you define the injection for definable numbers, namely by a catalogue. Map, for instance, node no. 17 on the path 0.010101... Obviously, the catalogue can contain all nodes. So let's wait whether it also can contain all paths. > > > > > An injection is also possible for the set of all paths into the set of > > > > all nodes. (There are two nodes per path.) > > > > > > *Give* that injection. > > > > Map every node onto the path which leaves it to the left-hand side. > > I would think that that is *not* an injection from paths to nodes. Moreover, > at each node there are many paths that leave it on the left-hand side, so it > is not even an injection. Here is another mapping: Map the node on one of the many paths, call it A. The others will get their nodes when they separate from A. > > > > > > > > What node is bijected with the branch-off of 0.101010101010...? > > > > > > > > For an injection you can choose whatever node you want. > > > > > > Wrong. For an injection it is needed that two paths do not map to the same > > > node, so you have to be careful in your mapping. You simply refuse to give > > > an injection because you are not able to give one. > > > > Map every node onto the path which leaves it to the left-hand side. > > Which of the paths that leaves it on the left-hand side must I chose? Choose the path bunch. If the paths which you assume to be present in the bunch will become separated then choose the due nodes by which this happens. > > > > > > The number of paths is the same from the root node, because every path > > > > > starts at the root. Or are you suggesting that there are paths *not* > > > > > starting at the root? > > > > > > > > Every path starts at the root node. But in order to count the paths, > > > > they must be distinguishable, i.e. separated. > > > > > > Makes no sense. > > > > How would you count inseparated paths? > > You are trying to do the counting. When I count I find at every node > uncountably many paths. You do not find any path. You find at most path bunches. > > > > > Every bunch starts at the root node. But in order to count the > > > > bunches, they must be distinguishable, i.e. thy must be separated > > > > bunches. The number of separated bunches is doubled at every level. > > > > > > You were talking about bunches going in and out of nodes. What you are > > > doing is counting edges, not bunches, and the number of edges is countable. > > > > The number of paths cannot be larger than the number of edges. > > Why not? Because every path contains more edges than it has in common with any other path. Regards, WM
From: Daryl McCullough on 21 May 2007 07:47
WM says... >On 20 Mai, 17:57, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> WM says... >> >> >> > It is. Set theory is simply biased. Consider the list >> >> >> > 0.666... >> >> > 0.3666... >> >> > 0.33666... >> >> > 0.333666... >> >> > ... >> >> >> > If the diagonal number is defined by "replace 6 by 3", then we have >> >> > two answers none of which can be preferred by logic, but the second of >> >> > which is suppressed by convention. >> >> The diagonal number is 0.333... which is not on the list. > >The diagonal number cannot have more 3's than every list number has >3's. Of course it can. With your rule for the diagonal: replace 3 by 6, then if we start with the following list: 0.6000 0.06000 0.006000 0.0006000 0.00006000 ... then the diagonal will be the same: 0.333... There is one "3" for each number in the list. >If the diagonal number is complete in the sense that at every >position indexed by a natural number there is a 3, then there must be >a complete sequence of 3's among the list entries too. No, that's false. It's *provably* false. For your particular example, what is true is this (letting r_j be the jth number in the list) 1. For each natural number n > 0, there exists an index j such that r_j agrees with the diagonal number in the first n decimal places. 2. For each index j, there exists a natural number n > 0 such that r_j does *not* agree with the diagonal number in the first n decimal places. Statement 1. says that the diagonal number can be approximated arbitrariy well. Statement 2. says that the diagonal number does not occur on the list. Both are true. These two statements have the logical forms: 1. forall n, exists j, Phi(r_j, d, n) 2. forall j, exists n, ~Phi(r_j, d, n) where Phi(r_j, d, n) is the statement "r_j agrees with the diagonal number d in the first n decimal places". >Otherwise the list is not complete, i.e., there is always a 6 at >some natural index. The definition of the diagonal number d is the following: (letting r[n] mean the nth decimal place of the real r) d[n] = 9 - r_n[n] The definition of the numbers r_j is this: r_j[n] = 6, if j=n = 3, otherwise It follows that r_n[n] = 6 Putting those together, we have d[n] = 9 - 6 = 3 So every digit of the diagonal number is 3. >If the list is not complete however, the diagonal number cannot be >complete either. The list is "complete" in the sense that for every natural number j, there is a corresponding real number r_j. Similarly, the diagonal number is complete in the sense that for every natural number j, there is a corresponding digit d[j]. >And the saying that at *every* position there is a >digit b_n =/= a_n,n is nonsense. It's provably true. (I'm using d, rather than b, and I'm using r_n instead a_n). d[n] = 3 r_n[n] = 6 Clearly d[n] is never equal to r_n[n]. >> >For the entries E(n) of the list we find >> >lim[n-->oo] (E(n) - 0.333...) = 0. >> >It is the same case as lim[n-->oo] (1 - 0.999...9 with n 9's) = 0. >> >> That's true. In this particular case, the limit of the sequence >> is equal to the diagonal of the sequence. So what? > >Why only in this particular case? Is lim[i --> oo] (b_i - a_i) * 10^-i >= 0 correct only in one special case? I say in this case because not every sequence has a limit. For example, the sequence 0.600... 0.330... 0.6660... 0.33300... 0.666600... 0.3333000... has no limit. But it still has a diagonal, namely 0.363636... >> >If every initial segment of the diagonal number is represented by the >> >initial segment of an entry of the list, then the full diagonal number >> >is represented by an entry of the list. >> >> That's false. > >No, that's correct. It's provably false. The diagonal number in your case satisfies d[n] = 3 Every element has r_n[n] = 6. Clearly they are not the same, so d is not equal to r_n. >It is the only valid interpretation of the notion infinity. I don't know what that's supposed to mean, but you seem to be confusing three different concepts: 1. The number d appears on the list r_j. For this to be true, it must be the case that Exists j, Forall n, d[n] = r_j[n]. That is *false*. Provably false. 2. The number d is the limit of the numbers r_j. For that to be true, it must be the case that Forall n > 0, Exists j, Forall m < n, d[m] = r_j[m]. That's true in your case. >> To say that the number r appears on the list r_0, r_1, ... >> is to say that there is some natural number j such that r = r_j. If >> we let D_j = |r - r_j|, then the criterion for r appearing on the list >> is that >> >> exists j such that D_j = 0 > >This criterion is false. If D_j is not zero, then that means that the diagonal number is not equal to r_j. >In case lim [i --> oo] (10 - 9) * 10^-i = 0 we see it clearly. What do limits have to do with anything? There are two different questions: 1. Does the diagonal number d appear on the list? The answer is no. 2. Is the diagonal number the limit of the numbers on the list? The answer for your case is yes. >> In the case we are talking about, that is false. If r_0 = 0.6666...., >> r_1 = .3666..., etc and r = 0.333..., then >> >> |r-r_0| = .333... >> |r-r_1| = .0333... >> ... >> >> in general, >> >> |r - r_j| = 0.333... * 10^{-j} >> >> So we have >> >> forall j, D_j > 0 > >Forall n e N we have 0 < 1 - 0.999...9 (with n 9's). Yes, that's true. 1 does not appear on the list 0.9 0.99 0.999 0.9999 0.99999 etc. and 0.33333 does not appear on the list 0.6666... 0.36666 0.336666 0.3336666 etc. >> So r does not appear on the list. > >So the problem of replacing 0 by 9 is not existing. What are you talking about? Of course the number 0.9999... (all 9s) exists. And it is equal to 1. But it is not on the list 0.9 0.99 0.999 0.9999 0.99999 etc. >But we have in the limit j --> oo: D_j = 0. Yes. That's true, but irrelevant. For the diagonal to be on the list, it must be that D_j = 0 for some finite j. It's not good enough that limit of D_j = 0. -- Daryl McCullough Ithaca, NY |