Cantor Confusion
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From: jpalecek on 19 Oct 2006 16:33 mueckenh(a)rz.fh-augsburg.de napsal: > jpalecek(a)web.de schrieb: > > > > The set of constructible numbers is countable. Any diagonal number is a > > > constructed and hence constructible number. > > > > No. Definition (from MathWorld): Constructible number: A number which > > can be represented by a finite number of additions, subtractions, > > multiplications, divisions, and finite square root extractions of > > integers. > > > > How do you represent the diagonal number (which is sort of a limit of a > > series) > > via FINITE number of +,-,*,/,sqrt( ) ? > > Which digit should not be constructible by a finite number of > operations? No digit, but the number has to be constructed (according to MathWorld definition). So you must do something like diag=x^2+sqrt(y+z/x) > > > > > Every list of reals can be shown incomplete in exactly the same way as > > > every list of contructible reals can be shown incomplete. > > > > No. > > A constructible number is a number which can be constructed. Definition > obtained from Fraenkel, Abraham A., Levy, Azriel: "Abstract Set > Theory" (1976), p. 54: "Why, then, the restriction to the digits 1 and > 2 in our proof? Just to kill the prejudice, found in some treatments of > the proof, as if the method were purely existential, i.e. as if the > proof, while showing that there exist decimals belonging to C but not > to C0, did not allow to construct such decimals." > > Definition (by me): A number which can be constructed like pi, sqrt(2) > or the diagonal of a list is that what I call constructible. If you > dislike that name, you may call these numbers oomflyties. Anyhow that > set is countable. And that set cannt be listed. Therefore the diagonal > proof shows that a set of countable numbers is uncountable. I like the name, but your "definition" isn't a definition. Please restate your definition in the language of ZF, and note, that there are no "formulas" in the language of ZF. It seems that your definition is as vague as "interesting" numbers in a paradox of first uninteresting number. Regards, JP
From: jpalecek on 19 Oct 2006 16:51 mueckenh(a)rz.fh-augsburg.de napsal: > jpalecek(a)web.de schrieb: > > > > > > > > > 0. > > > > > > > /a \ > > > > > > > 0 1 > > > > > > > /b \c / \ > > > > > > > 0 1 0 1 > > > > > > > ..................... > > > > > > > > > > > An edge is related to a set of path. If the paths, belonging to this > > > set, split in two different subsets, then the edge related to the > > > complete set is divided and half of that edge is related to each of the > > > two subsets. If it were important, which parts of the edges were > > > related, then we could denote this by "edge a splits into a_1 and a_2". > > > But because it is completely irrelevant which part of an edge is > > > related to which subset, we need not denote the fractions of the edges. > > > > Sorry, but your "proof" doesn't work. Imagine an infinite path in the > > tree. Which is the edge it inherits as a whole? Whenever you give me > > that edge, I can tell you're lying because if a path inherits an edge > > as a whole, it means that the path terminates by that edge. > > How should I be able to name the last term of a sum which has no last Indeed, INDEED. But that means, that there is no term 1 in the series, too. Moreover, it means that there is no positive term in the series. (Because 1 would be counted for the last edge, which does not exist, etc.). > term? But while we cannot name any individual edge we can prove: No > path splits into two paths without the supply of two new edges, one > edge for each path. This implies there cannot be less edges than paths. No. This holds for the finite case only. > Or the other way round: Assume there were more paths than edges, then > at least two paths could no be distinguished. (A path can be > distinguished from every other path by at least one edge.) Which doesn't mean anything. See the marriage theorem (or Hall's theorem) for the conditions you would have to fulfill to have an injective mapping of paths to their edges). And note, you cannot use marriage theorem to prove existence of any injective mapping of paths to edges). > > This is > > impossible for infinite paths. > > Of course that is impossible. Therefore the sum 1 + 1/2 + 1/4 + ... is This series is irrelevant. > a infinite sum. But nevertheless your argument covers only half of the > story. Whenever you give me two infinite paths, I can name an edge > which belongs to only one of them. Again, this doesn't buy you anything. See marriage theorem. > > The same argument applies to other terms > > in the sum. (That edge is inherited by an infinite path by 1/1024! > > Ok, but that means that the path terminates 10 levels lower). This > > means that infinite path inherit zero edges in your proof. > > Then the series 1 + 1/2 + 1/4 + ... contains zeros? No, that means that series is a wrong one for your problem. The correct one is 0+0+0+... > The distance between any two edges of one path is infinite? The distance of any edge from the end of an infinite path is infinite. Regards, JP
From: david petry on 19 Oct 2006 18:01 Dik T. Winter wrote: > In article <1161080567.156919.211680(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > And there is no complete list of computable numbers. But they are > > countable. Hence the diagonal argument does not prove anything. > > You are, again, missing the essential information. There *is* such a > list, but it is not computable. I suggest that you are missing a key point: it all depends on what the definition of *is* is.
From: Dik T. Winter on 19 Oct 2006 21:24 In article <eh89qg$jdf$1(a)mailhub227.itcs.purdue.edu> Dave Seaman <dseaman(a)no.such.host> writes: > On Thu, 19 Oct 2006 14:42:32 GMT, Dik T. Winter wrote: .... > > I have checked in the dictionary, and indeed, just like in Dutch there are > > also in German two verbs with the meaning 'count': z?hlen and abz?hlen. > > The first in general meaning obtaining the total number of objects you are > > counting, while the second is more or less the *process* of counting (one > > of the possible meanings is 'to count down' at the launching of a rocket). > > So also when you have obtained the number of objects you are likely to use > > 'gez?hlt' and not 'abgez?hlt'. > > I won't dispute the meaning of the verb, but it seems very strange to me > that Cantor could make such a fundamental mistake. Yes, I found it also very strange. That is why I did remark on it. The only thing I can imagine is that he first started to use the common meaning and later switched over to a fairly non-standard meaning. On the other hand, at that time he is still struggling with 'potential infinite' and 'actual infinite'. (These two play a major role between philosophers, but also between German mathematicians.) > Am I right that "sets of first cardinality" are the finite sets, "sets of > second cardinality" are the denumerable sets, and so on? I have not > found an explanation of that terminology, although "numbers of the second > class" are clearly explained. No. The first cardinality is aleph-0. The second cardinality is aleph-1. Finite sets just have there size as cardinality without an ordinal attached to it. So when he states 'sets of the first cardinality can only be "counted" with ordinals of the second class' he is nearly correct. Hou would have been correct if he had omitted the worde 'only'. > > > I read that to mean that the correspondence goes like this: > > > > 0 -> w > > > 1 -> w+1 > > > 2 -> w+2 > > > I think not. .... > I don't follow you here. I was the one describing the table, and > according to my intended description, it is definitely the case that > every number of the second number class appears in the right-hand column. Yes, that was your table, and I understand what you did intend, but that is *not* what Cantor wrote. What you did write is indeed correct. But Cantor (in his statement) was not counting sets, he was counting elements of a set with cardinality aleph-0. And when counting elements of such sets not all ordinals of a particular class will come along. > What German word would you use to describe the contents of the above > table? Notice that the table does not have an end. Abz?hlen. But now I see what you mean, and I also see how you come to your interpretation. First: w?hrend die Mengen erster M?chtigkeit nur durch (mit Hilfe von) Zahlen der zweiten Zahlenklasse abgez?hlt werden k?nnen might indeed mean that you are counting sets rather than elements of sets. That statement is ambiguous. But the successor: die Abz?hlung bei Mengen zweiter M?chtigkeit leaves a lot of doubt. If there was 'von' rather than 'bei', I would agree with your interpretation. But I think it is only a native German speaker who could explain precisely what was written. -- 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 19 Oct 2006 21:30
In article <1161295304.522422.233980(a)b28g2000cwb.googlegroups.com> "david petry" <david_lawrence_petry(a)yahoo.com> writes: > > Dik T. Winter wrote: > > In article <1161080567.156919.211680(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > And there is no complete list of computable numbers. But they are > > > countable. Hence the diagonal argument does not prove anything. > > > > You are, again, missing the essential information. There *is* such a > > list, but it is not computable. > > I suggest that you are missing a key point: it all depends on what the > definition of *is* is. Pray provide me with a mathematical definition. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |