Cantor Confusion
in [Math]
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From: Dik T. Winter on 20 May 2007 22:10 In article <1179652465.304840.3760(a)e65g2000hsc.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 19 Mai, 04:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > The set is finitely defined. Not all lucky numbers can get > > > > > > > finitely defined. > > > > > > > > > > > > In general a recursive definition is not considered a finite > > > > > > definition. > > > > > > > > > > Every definition which ends after finitely many words is a finite > > > > > definition. > > > > > > > > Ah, so you disagree with common mathematical terminology. > > > > > > No. A finite definition means a definition by a finite number of > > > words. Every other definition is nonsense. I agree with the common > > > mathematical definition which implies that there are only finitely > > > many definitions. > > > > Which implication? Again contradicting the axiom of infinity? > > Pardon, I meant "countably many definitions". This is implied by the > finity of every definition. If there were infinite definitions, then > there were uncountably many definitions. Yup. So the question remains: "you disagree with common mathematical terminology?" But whatever, you can not apply the diagonal argument to "finitely defined numbers". You can supply a list of finite definitions, but not all of them define a number. And for the diagonal argument a list of numbers is needed. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 20 May 2007 22:28 In article <1179653443.812676.189260(a)u30g2000hsc.googlegroups.com> WM <mueckenh(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 for what is now called "simply normal". > > > 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. > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 20 May 2007 22:37 In article <1179654356.461792.242730(a)n59g2000hsh.googlegroups.com> WM <mueckenh(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 > > > > 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? > > > 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. > > > > > > 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? > > > > 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. > > > 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? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 20 May 2007 22:58 In article <1179655065.969901.59820(a)k79g2000hse.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > So you refuse to post an answer in the forum where I asked the question, > > giving proper references ... I have given you reasons *why* I do not > > access that group. That you ignore those reasons just shows arrogance. > > We have seen this same behaviour earlier by James Harris. Are you going > > to mimick him? I have *no* idea what you mean with "this set F of > > functions". > > You defined the functions: > (1) If two functions f and g are not equal, there is a smallest n such > that f(n) != g(n). > (2) We define f < g if f(n) < g(n), and f > g if f(n) > g(n). This is > a complete ordering on that set of functions. I still distrust transitivity of intercession. There is indeed a bijection between that set of functions and the reals, but as far as I know there is no order-preserving bijection. So if the rationals intercede the reals, it is unproven that their images also intercede the images if the irrationals. > > > 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. > > > > But, again, that is *not* the diagonal proof of Cantor. And even with > > that notation you write nonsense. "Replace 6 by 3" yields the sequence > > 0.33333..., which is not in the list. > > For the entries E(n) of the list we find lim[n-->oo] (E(n) - 0.333...) > = 0. Yes? I thought we were talking about a list, not about the limit of the elements of the list (which in general is not an element of the list). > > Both are true, if you replace (2) by: > > 2) Every initial segment of the diagonal number is represented by the > > initial segment of an entry of the list. > > To wit: > > 0.333666... > > does *not* represent > > 0.333, > > or you have a very strange interpretation of the word "represent". > > If very 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. Why? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 20 May 2007 23:00
In article <1179663816.282116.232880(a)h2g2000hsg.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > But, again, that is *not* the diagonal proof of Cantor. > > The following wm-proof certainly even in your opinion belongs to the > diagonal proofs considered by Cantor: > > 0) mmm... > 1) wmmm... > 2) wwmmm... > 3) wwwmmm... > 4) wwwwmmm... > ... .......... > > And if the list can be considered as a completed entity, then there > must be all natural numbers in the first column. And there must be a > line with all natural indexes mapped on w's, i.e., no w must be > missing (as would be the case if one m was present). Why? Show a proof. You are again assuming that there is a last natural number. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |