Cantor Confusion
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From: Albrecht on 12 Oct 2006 03:20 Dik T. Winter schrieb: > In article <1160551520.221069.224390(a)m73g2000cwd.googlegroups.com> "Albrecht" <albstorz(a)gmx.de> writes: > > David Marcus schrieb: > ... > > > I don't follow. How do you know that the procedure that you gave > > > actually "defines/constructs" a natural number d? It seems that you keep > > > adding more and more digits to the number that you are constructing. > > > > What is the difference to the diagonal argument by Cantor? > > That a (to the right after a decimal point) infinite string of decimal > digits defines a real number, but that a (to the left) infinite string > of decimal digits does not define a natural number. I claim that the diagonal number of Russell isn't infinite (to the left) since it is finite different from any natural number by construction. In respect to the diagonal number of Cantor, you have to proof if the sentence "any infinite string of decimal digits (to the right after a decimal point) defines a real number" is a correct sentence. Or it is just an arbitrary axiom? Best regards Albrecht S. Storz
From: mueckenh on 12 Oct 2006 04:01 Dik T. Winter schrieb: > In article <1160551520.221069.224390(a)m73g2000cwd.googlegroups.com> "Albrecht" <albstorz(a)gmx.de> writes: > > David Marcus schrieb: > ... > > > I don't follow. How do you know that the procedure that you gave > > > actually "defines/constructs" a natural number d? It seems that you keep > > > adding more and more digits to the number that you are constructing. > > > > What is the difference to the diagonal argument by Cantor? > > That a (to the right after a decimal point) infinite string of decimal > digits defines a real number, but that a (to the left) infinite string > of decimal digits does not define a natural number. And why is this so? Because an infinite string of digits is not at all defined. Only by the factors 10^(-n) this is veiled. The due digits become more and more unimportant because their contributions to the number size are pulled down by the increasing exponents. But this has been forgotten by Cantor whose diagonal proof attaches the same weight to every digit. That is obviously wrong. Or it could also be applied to the left of the decimal point, constructing an infinite natural number and showing that every list of natural numbers is incomplete. Regards, WM
From: mueckenh on 12 Oct 2006 05:54 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". Regards, WM
From: mueckenh on 12 Oct 2006 06:03 Tony Orlow schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > >> mueckenh(a)rz.fh-augsburg.de wrote: > >> > >> [...] > >> > >>> There is no largest natural! There is a finite set of > >>> arbitrarily large naturals. The size of the numbers is unbounded. > >>> > >> I can only conclude you have knocked youself out. > > > > Try the following gedanken-experiment to become accustomed with it: > > a) How many different natural numbers can you store using a maximum of > > 100 bits? > > b) What is the largest natural number you can store with a maximum of > > 100 bits? > > > > Regards, WM > > > > Answer to a) less than 100. > > Answer to b) unknown, depends on representation. > > > > No, if they are binary digits, or bits, then there are 2^100 unique > strings possible with 100 digits. Be careful. I did not mean to ask how many numbers the set containes, from which you may choose, but how many numbers you can really realize and store at a given time. (With three decimal digits you can, in principle, realize all numbers between 0 and 999, but you can realize at most three numbers at a given time.) Regards, WM
From: mueckenh on 12 Oct 2006 06:09
William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > [...] > > > > > > > There is no largest natural! There is a finite set of > > > > arbitrarily large naturals. The size of the numbers is unbounded. > > > > > > > > > > I can only conclude you have knocked youself out. > > > > Try the following gedanken-experiment to become accustomed with it: > > a) How many different natural numbers can you store using a maximum of > > 100 bits? > > b) What is the largest natural number you can store with a maximum of > > 100 bits? > > > > c) Once you have decided on a representation, what is the largest > number you can store with a maximum of 100 bits? That depends on the representation. In unary representation it is 100. > > You are confusing "unknown" with "arbitrary". A natural number > X may have unknown size. It cannot have arbitrary size. The largest number you construct today may have arbitrary size. Perhaps it is 777? > > if you pick an even natural number you end up with an > even natural number. If you pick an arbitrarially large natural > number, you do not end up with an arbitrarially large natural > number. My purpose was to explain to you why your unreflected assumption is wrong that all numbers up to a given one must exist. It is not contradictory to say that in a finite set of numbers there need not be a largest. It seems that this false assumption is one of the basic reasons for set theory. Regards, WM |