Cantor Confusion
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From: Han de Bruijn on 12 Oct 2006 07:26 mueckenh(a)rz.fh-augsburg.de wrote in response to Virgil: > For the vase problem with the number n(t) of balls in the vase after t > transactions we can find always a positive eps such that for t > t_0: > 1/n(t) < eps, hence n(t) larger than an arbitrary positive number. > > Therefore, your assumption of lim {t-->oo} n(t) = 0 is absurd. Precisely! Virgil doesn't know how to handle limits. Han de Bruijn
From: mueckenh on 12 Oct 2006 07:34 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > > > 0.1 > > > > > 0.11 > > > > > 0.111 > > > > > ... > > > > > > > > > > > That is correct. But every element of the natural numbers is finite. > > > > Hence every element covers its predecessors. If 0.111... is covered by > > > > "the whole list", then it is covered by one element. That, however, is > > > > excuded. > > > > > > > > > > Since no one has claimed that '0.111... is covered by "the whole > > > list"', I fail > > > to see the relevence of a sentence that starts out > > > 'If 0.111... is covered by "the whole list"'. > > > > If every digit position is well defined, then 0.111... is covered "up > > to every position" by the list numbers, which are simply the natural > > indizes. I claim that covering "up to every" implies covering "every". > > > Quantifier dyslexia. Quantifier magic may apply and may be useful at several occasions. But to state that in a unary representation of natural numbers the union of "up to every" and "every" have different meaning is easily disproved. > The fact that > > for every digit position N, there exists a natural number, M, > such that M covers 0.111... to position N > > does not imply > > there exist a natural number M such that for every digit > position N, M covers 0.111... to position N In case of linear sets we have a third statement which is true without doubt: 3) Every set of unary numbers which covers 0.111... to a finite position N can be replaced by a single unary number. This holds for every finite position N. If 0.111... has only finite positions, then (3) holds for every position. As 0.111... does not consist of mre than every position, it holds for the whole number 0.111.... Regards, WM
From: William Hughes on 12 Oct 2006 07:34 mueckenh(a)rz.fh-augsburg.de wrote: > It is not > contradictory to say that in a finite set of numbers there need not be > a largest. There is no such thing as a number with arbitrary size (i.e. a number that has the property that its size is arbitrary). Take a finite set B. By definition, B must have a finite number of elements. This number does not have arbitrary size. Each element has a size This size is not arbitrary. So we have a finite number (which does not have arbitrary size) of elements, each of which does not have arbitrary size. Thus B has a largest element. - William Hughes.
From: mueckenh on 12 Oct 2006 07:43 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > It is not > > contradictory to say that in a finite set of numbers there need not be > > a largest. > > There is no such thing as a number with arbitrary size > (i.e. a number that has the property that its size is > arbitrary). The number I will write down here has that property, before I write it down. 7 Now it has no longer this property. The prime number to be discovered next has the property to be unknown. - Until it is discovered. > Take a finite set B. > > By definition, B must have a finite > number of elements. This > number does not have arbitrary size. > > Each element has a size This size is not > arbitrary. > > So we have a finite number (which does not > have arbitrary size) > of elements, each of which does not > have arbitrary size. > > Thus B has a largest element. What is the largest number of the set of numbers mentioned in the New York Times in 2007? You see, your proof is rubbish. B will have a largest element. And the set of all numbers ever used in the universe in eternity also will have a largest element. But it has not yet. Therefore you think the set of natural numbers were infinitely large. It is not. It is only your positon which, compared to the universe, is very, very tiny. Regards, WM Regards, WM
From: William Hughes on 12 Oct 2006 08:09
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > > > 0.1 > > > > > > 0.11 > > > > > > 0.111 > > > > > > ... > > > > > > > > > > > > > > That is correct. But every element of the natural numbers is finite. > > > > > Hence every element covers its predecessors. If 0.111... is covered by > > > > > "the whole list", then it is covered by one element. That, however, is > > > > > excuded. > > > > > > > > > > > > > Since no one has claimed that '0.111... is covered by "the whole > > > > list"', I fail > > > > to see the relevence of a sentence that starts out > > > > 'If 0.111... is covered by "the whole list"'. > > > > > > If every digit position is well defined, then 0.111... is covered "up > > > to every position" by the list numbers, which are simply the natural > > > indizes. I claim that covering "up to every" implies covering "every". > > > > > Quantifier dyslexia. > > Quantifier magic may apply and may be useful at several occasions. But > to state that in a unary representation of natural numbers the union of > "up to every" and "every" have different meaning is easily disproved. > > > The fact that > > > > for every digit position N, there exists a natural number, M, > > such that M covers 0.111... to position N > > > > does not imply > > > > there exist a natural number M such that for every digit > > position N, M covers 0.111... to position N > > In case of linear sets we have a third statement which is true without > doubt: > > 3) Every set of unary numbers which covers 0.111... to a finite > position N can be replaced by a single unary number. > > This holds for every finite position N. If 0.111... has only finite > positions, then (3) holds for every position. As 0.111... does not > consist of mre than every position, it holds for the whole number > 0.111.... > So we have for every digit position N, there exists a set of unary numbers which covers 0.111... to position N and for every digit position N, if there exists a set of unary numbers which covers 0.111... to position N, there exists a single unary number, M, such that M covers 0.111... to position N this implies for every digit position N, there exists a single unary number, M, such that M covers 0.111... to position N this does not imply there exists a single unary number M such that for every digit position N, M covers 0.111... to position N - William Hughes |