Cantor Confusion
in [Math]
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From: Dik T. Winter on 18 May 2007 22:49 In article <1179427830.763838.270380(a)w5g2000hsg.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 17 Mai, 02:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179346617.111313.267...(a)o5g2000hsb.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 16 Mai, 04:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > Each of the numbers is finitely definable, because each is a halting > > > > Turing machine, and Turing machines are finitely definable. It is > > > > the set that is not finitely definable. > > > > > > If each number is finitely defined, then there are only finitely many > > > numbers, because there are only finitely many Turing machines. Of > > > course, this finite set then is finitely defined too. > > > > Back again to how you started. There are only finitely many natural numbers... > > That is not mathematics. > > That is the obvious basis which can be proven by mathematics. It is most ridiculous to state that there are only finitely many Turing machines. Can you prove that, using mathematics? > > > > Now you come up again with a new term. 'Well defined'. What is > > > > 'well defined'? > > > > > > A number is well defined if it is defined by a finite number of words > > > such that, in principle (i.e., given an infinite amount ressources) > > > the Cauchy epsilon can be made arbitrary small. > > > > So pi is well defined as the limit of the circumference of the inscribed > > n-gon? The sequence of those circumferences *is* a Cauchy sequence... > > 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. > > > I showed that there are countable sets which cannot be bijected with > > > the set of natural numbers, for instance the set of finitely definable > > > numbers. > > > > But that set *can* be bijected with the set of natural numbers. > > Wrong. Just assertion? > > 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? > 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. > > > > 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. > Obviously for > every number represented in the tree there is a node. And there is no > diagonal construction possible in the tree. So pray, *give* an injection. > > > I claim that there are no more branching-offs than odes and that there > > > are no more real numbers represented in the tree than are branching- > > > offs. There is no path ever finished, but it is only branching off > > > from other paths in infinity. But the number of paths separated from > > > other paths cannot surpass the number of branching-offs. > > > > 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. > > > One bunch goes in, because the two which come out have not yet been > > > separated when the paths which they consist of, went in. > > > > So it is your opinion that bunches that start of the root nevertheless do > > not come in at some node but only come out? > > 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. > > 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? > > But if a number is normal with respect to some base that does > > *not* mean that the digits are unpredictable. Nor does unpredictability > > of digits mean that a number is 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. > 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." So, as I said: if a number is normal with respect to some base that does *not* mean that the digits are unpredictable. Nor does unpredictability of digits mean that a number is normal. But, of course, most normal numbers cannot be constructed, as most numbers cannot be constructed, and most numbers are normal. (Now try to interprete the three occurrences of the word most.) -- 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 18 May 2007 22:50 In article <1179427901.710765.103690(a)n59g2000hsh.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 17 Mai, 02:24, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179346353.430169.24...(a)y80g2000hsf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 16 Mai, 04:01, "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? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 20 May 2007 05:14 On 19 Mai, 04:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179427901.710765.103...(a)n59g2000hsh.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 17 Mai, 02:24, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1179346353.430169.24...(a)y80g2000hsf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > > On 16 Mai, 04:01, "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. Regards, WM
From: WM on 20 May 2007 05:30 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. > > > > But if a number is normal with respect to some base that does > > > *not* mean that the digits are unpredictable. Nor does unpredictability > > > of digits mean that a number is 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. 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 > > > 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 Regards, WM
From: WM on 20 May 2007 05:45
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'. > > > > > I showed that there are countable sets which cannot be bijected with > > > > the set of natural numbers, for instance the set of finitely definable > > > > numbers. > > > > > > But that set *can* be bijected with the set of natural numbers. > > > > Wrong. > > Just assertion? Define the bijection. > > > > 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. > > > 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. > > > > > > 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. > > > Obviously for > > every number represented in the tree there is a node. And there is no > > diagonal construction possible in the tree. > > So pray, *give* an injection. > > > > > I claim that there are no more branching-offs than odes and that there > > > > are no more real numbers represented in the tree than are branching- > > > > offs. There is no path ever finished, but it is only branching off > > > > from other paths in infinity. But the number of paths separated from > > > > other paths cannot surpass the number of branching-offs. > > > > > > 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? > > > > > One bunch goes in, because the two which come out have not yet been > > > > separated when the paths which they consist of, went in. > > > > > > So it is your opinion that bunches that start of the root nevertheless do > > > not come in at some node but only come out? > > > > 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. Regards, WM |