Cantor Confusion
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From: Virgil on 13 Oct 2006 00:02 In article <452ecfab(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > William Hughes wrote: > > 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. > > Please prove this in specific mathematical terms, without using the > circular argument involving the supposed infinity of omega. Actually endless strings are acceptable as representing finite numbers provided they have only a finite number of non-zero digits, and thus a most significant digit. Such limited strings can be truncated to finite strings without changing their values, so count as finite strings. > > > > > You have a hole between what most people call finite and > > what you call infinite. > > Uh, no, you do. > > You argue that since the strings > > cannot represent infinite things they must represent finite > > things. This does not follow. > > So, they can represent a third kind of number, which is not finite, nor > infinite? They could fail to represent anything at all except themselves. > > Countably infinite strings have unbounded length and > > do not represent finite values. > > > > *Because* they are unbounded? Does "unbounded" MEAN *infinite*? The set of all finite strings has no finite bound on the lengths of the finite strings it contains, but each finite string is bounded.
From: Virgil on 13 Oct 2006 00:04 In article <452ed045(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > But Wolfgang, surely that consideration does not impact, say, the set of > reals in (0,1], which are all finite, yet whose number is infinite. It > is not a requirement that a set of all finite values be finite. TO blows hot and cold with the same breath. He has argued today that a set of all finite values must be finite but now argues the reverse.
From: mueckenh on 13 Oct 2006 03:13 William Hughes schrieb: > > > > > > > > > 0.1 > > > > > > > > > 0.11 > > > > > > > > > 0.111 > > > > > > > > > ... > > > > 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? > > Because > > for every digit position N, > there exists a single unary number, M, > such that M covers 0.111... to position N > > can be true even if M depends on N. The statement > > there exists a single unary number M such that for every digit > position N, M covers 0.111... to position N > > cannot be true if M depends on N. Therefore the first does > not imply the second. I did not claim that the first statement implied the second. I did not claim that one letter depends on the other. I claim that the second statement is true for every finite N. And that is obvious. Take just M = N. Hence every finite position of 0.111... will be covered by one single number M of the list. This statement does not refer in any respect to the number of digit positions. All finite digit positions are covered by one finite number M. If 0.111... has only finite digit positions, then the whole number is covered. > >If every digit position of 0.111... is a finite > > position then exactly this is implied. > > No, the fact that every digit position of 0.111... is finite does not > mean that M does not depend on N, and that is what we need. Why should we need that? Why should the number of finite positions be of any influence, if we know that the theorem is true for any finite position? You think we cannot name the M, so it cannot exist? You know we cannot construct a well-order of the reals. But nevertheless it does exist, according to Zermelo's proof. And here, there *is* such an M, for all finite positions, according to a very simple proof, independent of whether or not we can name it. The problem is that you believe in an actually infinite series of finite numbers. This belief leads to intermingling digit position and number of digits just according to the needs and, after all, it leads to such self-contradictive statements as we have seen here with "the vase". > What we need is not that "every digit position of 0.111... is finite" > but "there are a finite number of digit positions in 0.111..." > These two statements are not the same. Whether they are the same or not is a question which I will not discuss in this context. But if they are not the same, then the second one (or its negation) is at least completely irrelevant with respect to the problem in question. This depends only on a particular digit position and not at all on the observation that there may be further positions. Regards, WM
From: Virgil on 13 Oct 2006 03:40 In article <1160723585.727441.311990(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > > > > > > 0.1 > > > > > > > > > > 0.11 > > > > > > > > > > 0.111 > > > > > > > > > > ... > > > > > > 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? > > > > Because > > > > for every digit position N, > > there exists a single unary number, M, > > such that M covers 0.111... to position N > > > > can be true even if M depends on N. The statement > > > > there exists a single unary number M such that for every digit > > position N, M covers 0.111... to position N > > > > cannot be true if M depends on N. Therefore the first does > > not imply the second. > > I did not claim that the first statement implied the second. I did not > claim that one letter depends on the other. I claim that the second > statement is true for every finite N. And that is obvious. Take just M > = N. > Hence every finite position of 0.111... will be covered by one single > number M of the list. That essentially claims that there is a largest member of a list that has no largest member. >This statement does not refer in any respect to > the number of digit positions. All finite digit positions are covered > by one finite number M. What about position M + 1? > > > > No, the fact that every digit position of 0.111... is finite does not > > mean that M does not depend on N, and that is what we need. > > Why should we need that? Because for every M there is an N = M+1 which M does not cover. > You think we cannot name the M, so it cannot exist? You know we cannot > construct a well-order of the reals. But nevertheless it does exist, > according to Zermelo's proof. And here, there *is* such an M, for all > finite positions, according to a very simple proof, independent of > whether or not we can name it. There is no such thing in ZF or NBG or any other standard set theory, and I very much doubt that there is any such thing in any constructionist theory. > > The problem is that you believe in an actually infinite series of > finite numbers. This belief leads to intermingling digit position and > number of digits just according to the needs and, after all, it leads > to such self-contradictive statements as we have seen here with "the > vase". There is nothing self contradictory in "the vase" but there are contradictions between the axiom systems on which "the vase" is based and the assumptions of those who oppose those axiom systems. > > > What we need is not that "every digit position of 0.111... is finite" > > but "there are a finite number of digit positions in 0.111..." > > These two statements are not the same. > > Whether they are the same or not is a question which I will not discuss > in this context. Because discussing it would reveal the flaws in your arguments. > But if they are not the same, then the second one (or > its negation) is at least completely irrelevant with respect to the > problem in question. Not in the opinion of anyone else besides "Mueckenh".
From: Han de Bruijn on 13 Oct 2006 04:11
William Hughes wrote: > 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. How about: I'm not interested in "the end". I don't know where the end is. And I don't care as well. As long as the end is somewhere where it causes a uncertainity which is acceptable for my purpose. 0.11111111111111111111111111111111111111111111111 = 1 (binary) Okay. 0.000000000000000000000000000000000000000000000001 : error accepted. Han de Bruijn |