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From: William Hughes on 12 Oct 2006 11:20 Han de Bruijn wrote: > Dik T. Winter wrote: > > > In article <1160646886.830639.308620(a)c28g2000cwb.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > 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". > > > > Yes, you claim. Without proof. You state it is true for each finite > > sequence, so it is also true for the infinite sequence. That conclusion > > is simply wrong. > > That conclusion is simply right. And yours is wrong. Completed infinity > does not exist. So _each_ finite sequence "means" the infinite sequence. > If you insist that '_each_ finite sequence "means" the infinite sequence' you have the empty conclusion that if each finite sequence 0.1,0.11, 0.111, ... is covered by a list of numbers then each finite sequence 0.1,0.11, 0.111, ... is covered by a list of numbers. You do not have, there exists one element of the list of numbers which covers 0.1,0.11, 0.111, ... . - William Hughes
From: William Hughes on 12 Oct 2006 12:02 William Hughes wrote: > Han de Bruijn wrote: > > Dik T. Winter wrote: > > > > > In article <1160646886.830639.308620(a)c28g2000cwb.googlegroups.com> > > > mueckenh(a)rz.fh-augsburg.de writes: > > > ... > > > > 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". > > > > > > Yes, you claim. Without proof. You state it is true for each finite > > > sequence, so it is also true for the infinite sequence. That conclusion > > > is simply wrong. > > > > That conclusion is simply right. And yours is wrong. Completed infinity > > does not exist. So _each_ finite sequence "means" the infinite sequence. > > > > If you insist that '_each_ finite sequence "means" the infinite > sequence' > you have the empty conclusion that if each finite sequence 0.1,0.11, > 0.111, ... > is covered by a list of numbers then each finite sequence > 0.1,0.11, 0.111, ... is covered by a list of numbers. > You do not have, there exists one element of the list of numbers > which covers 0.1,0.11, 0.111, ... . To ellaborate: To get the result you want (there is a single number which covers all finite sequences) it is not enough to deny completed infinity. You must also insist that 0.111... has an end (i.e you must also deny potential infinity). You cannot say that the 0.111... has an end but it is arbitrary because "arbitrary" is not a property ends can have. - William Hughes > > - William Hughes
From: stephen on 12 Oct 2006 12:18 Han de Bruijn <Han.deBruijn(a)dto.tudelft.nl> wrote: > Dik T. Winter wrote: >> In article <1160647755.398538.36170(a)b28g2000cwb.googlegroups.com> >> mueckenh(a)rz.fh-augsburg.de writes: >> ... >> > It is not >> > contradictory to say that in a finite set of numbers there need not be >> > a largest. >> >> It contradicts the definition of "finite set". But I know that you are >> not interested in definitions. > Set Theory is simply not very useful. The main problem being that finite > sets in your axiom system are STATIC. They can not grow. Which is quite > contrary to common sense. (I wouldn't imagine the situation that a table > in a database would have to be redefined, every time when a new row has > to be inserted, updated or deleted ...) > Han de Bruijn I suppose you object to the fact that when you add 1 to a number you have to write down an entirely different number? When I add 1 to 7, 7 does not change. It remains the same. Of course this is not true on a computer. If I have a set of bits that represent a 7, and then I add 1, then set of bits will now represent an 8, and there is no guarantee that there is any representation of 7 anywhere. That does not mean 7 ceases to exist. Your idea seems to be at the heart of lots of peoples confusion about set theory. Sets are static, just as numbers are. When we work with them, we often have dynamic variables that take on the values of different sets and numbers, but the sets and numbers themselves do not change. Stephen
From: William Hughes on 12 Oct 2006 12:59 Tony Orlow wrote: > William Hughes wrote: > > Tony Orlow wrote: > >> Dik T. Winter wrote: > >>> 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. > >> It defines something. What do you call that? If the value up to and > >> including every digit is finite, how can the string represetn anything > >> but a finite value? > >> > > > > Because there are two types or strings. Strings that end and strings > > that don't end. Only strings that end represent finite values. > > > > -William Hughes > > > > And what about countably infinite strings which cannot achieve actually > infinite values? If the strings do not end they do not represent finite values. You have a hole between what most people call finite and what you call infinite. You argue that since the strings cannot represent infinite things they must represent finite things. This does not follow. You need at least three catagories, finite, unbounded and infinite. Countably infinite strings have unbounded length and do not represent finite values. [ You have four types of ordered sets of integers. -sets that end with a finite integer -sets that do not end but do not contain an infinite integer -sets that end with an infinite integer -sets that do not end and do contain an infinite integer You have never really come to terms with the second type of set. (You have never really come to terms with the fourth type of set, but this type of set has rarely been discussed). ] -William Hughes I agree that "countably infinite string which cannot actually achieve infinite values" are a problem with your theory. Your theory, your problem.
From: mueckenh on 12 Oct 2006 13:45
William Hughes schrieb: > 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 Why shouldn't it? If every digit position of 0.111... is a finite position then exactly this is implied. Your reluctance to accept it shows only that you do not understand how an infinite set can consist of finite numbers. In fact, nobody can understand it, because it is impossible. Regards, WM |