Cantor Confusion
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From: David Marcus on 11 Nov 2006 12:56 mueckenh(a)rz.fh-augsburg.de wrote: > Daniel Grubb schrieb: > > > >> >The diagonal of actually infinite length omega cannot exist without the > > >> >existence of a line of actually infinite length omega. > > >> > > >> Why not? Do you have a proof of this claim? > > > > >The diagonal of a matrix is defined as consisting of elements of this > > >matrix. For a diagonal longer than every line (or every column) this > > >is impossible. > > > > Let me restate the situation: > > > > We have the 'matrix' > > > > 1 > > 2 3 > > 4 5 6 > > 7 8 9 10 > > etc > > > > and you argue that the length of the diagonal cannot be > > infinite. > > > > Of course, the first problem is that this is not a square > > matrix. If you want to make it into a square matrix, you can > > make it into this matrix: > > > > 1 0 0 0 0 0 .... > > 2 3 0 0 0 0 ... > > 4 5 6 0 0 0 ... > > .... > > .... > > > > In which case, it is a square matrix and every line > > has infinite length. So the diagonal, which has infinite > > length, has the same length as every row and every column. > > There is no line without trailing zeros. The diagonal has no zeros. That is correct. Were you agreeing with Daniel or disagreeing with him? > > An alternative is to look at the incomplete, triangular array > > that we started with and notice that in triagular arrays the > > length of the diagonal is always at least as large as the > > length of each row. In the situation where there are finitely > > many rows, the length of the diagonal is the same as the number > > of rows. > > > > The sticking point is whether the length of the diagonal is > > the same as the length of some row even when there are infinitely > > many rows. You seem to be claiming the answer is yes. > > As we have not the least idea of what "actual infinity" could be, we > can do nothing but extrapolate from the finite domain. Using precise definitions and rigorous logic. -- David Marcus
From: David Marcus on 11 Nov 2006 13:02 mueckenh(a)rz.fh-augsburg.de wrote: > > David Marcus schrieb: > > > If you think that your "mathematical proof" is valid and you prove > > something that is absurd, this only shows that your mathematical > > reasoning leads to absurdities. Everyone who uses mathematical reasoning > > consistent with ZFC cannot prove your absurdity. So, if you learn ZFC, > > you will stop proving absurdities. > > Only then the absurdities would begin. But I would not notice. > > > > > We need no experts in trees. This tree does nothing else but represent > > > the real numbers of an interval. Here it is in a form which can be > > > understod by ever mathamatician. > > > > > > Consider a binary tree which has (no finite paths but only) infinite > > > paths representing the real numbers between 0 and 1 as binary strings. > > > The edges (like a, b, and c below) connect the nodes, i.e., the binary > > > digits 0 or 1. > > > > > > > > > > > > 0. > > > > > > /a \ > > > > > > 0 1 > > > > > > /b \c / \ > > > > > > 0 1 0 1 > > > > > > .......................... > > > > > > > > > > > > The set of edges is countable, because we can enumerate them. Now we > > > set up a relation between paths and edges. Relate edge a to all paths > > > which begin with 0.0. Relate edge b to all paths which begin with 0.00 > > > and relate edge c to all paths which begin with 0.01. Half of edge a is > > > inherited by all paths which begin with 0.00, the other half of edge a > > > is inherited by all paths which begin with 0.01. > > > > Half of the paths include edge a. One quarter of the paths include edge > > b. > > > > Suppose we have a finite tree of height n. Then there are 2^n paths. > > Consider a path (e1,e2,...,en). Edge e^j is contained in 2^-j of the > > paths. We can define a real-value function on paths by g(e1,e2,...,en) = > > sum_{j=1}^n 2^-j = 1 - 2^-n. Not sure what to do next... > > > > > Continuing in this > > > manner in infinity, we see by the infinite recursion > > > > > > f(n+1) = 1 + f(n)/2 > > > > What is f(n)? > > f(n) is the number of edges related to the initial segment of one path > which has passed through the first n edges. Are you saying that f(1) is the number of edges "related" to the paths that contain edge a? What is the value of f(1)? -- David Marcus
From: David Marcus on 11 Nov 2006 13:10 mueckenh(a)rz.fh-augsburg.de wrote: > > David Marcus schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Therefore we can compare the diagonal with every line. We find that the > > > diagonal is longer than every line. > > > > 1 > > 1 1 > > 1 1 1 > > 1 1 1 1 > > ... > > > > The length of the diagonal is clearly the same as the length of the > > first column. And, the first column "goes on forever", while each line > > does not "go on forever". Therefore, for each line L, the length of the > > diagonal is longer than the length of line L. Hence, "the diagonal is > > longer than every line." > > > > What is wrong with my reasoning? > > Nothing. Wrong is only the assumption that "goes on forever" could be > considered a finished infinity, i.e., could be denoted by a fixed > cardinal number being larger than any natural number. I'm not sure what you mean. In particular, I'm not sure what "considered", "finished" (or "finished infinity"), "denoted", and "fixed" mean here. > If you look at the diagonal, you always see that it is only there where > lines are. Therefore you will always see that it is of finite length. > "It goes on forever" does not mean actually infinite length. The > lengths of the lines also increase from line to line forever. > Nevertheless, an infinite length will never be reached. To me, "infinite length" just means "goes on forever". To you, the two phrases have different meanings, it seems. So, what do you mean by the phrase "infinite length"? -- David Marcus
From: William Hughes on 11 Nov 2006 13:25 mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > ... apparently, he thought that if a set was actually infinite, it should > > have an actually infinite element, or somesuch. > > It should have an integer number counting its elements. Of course this > number had to appear in the sequence of numbers if it was a number. But > where? Obviously after all natural numbers. But then, what is the > number of 1,2,3,...,omega ? omega+1 > It cannot be omega + 1 yes it can > because, according to his equation above, > 2,3,4, ....,1 has already the number omega + 1. Yes, there is more that one representation of omega+1. This is like saying 2/4 cannot be one half, because one half already has the representation 1/2. [Note that I cannot tell which ordinal is represented by a set just by looking at the elements of the set. For example, just by using the finite natural numbers I can get: omega 1,2,3,4... omega+1 2,3,4 ...,1 omega+2 3,4,5 ...,1,2 2*omega 1,3,5... 2,4,6... Indeed I can represent any countable ordinal.] [Note further that if we insist that a representation of an ordinal be an initial sequence of ordinals then 2,3,4, ....,1 does not represent an ordinal. If we insist that a representation of an ordinal be an initial segment of ordinals then the representation of an ordinal by a sequence of ordinals is unique.] > Therefore omega cannot appear in the sequence. > > So we have the irrevocable dilemma: > 1) omega is the number of countable many numbers like 1,2,3,... or > 7,8,9,... and as such a single term in the sequence following the > counted terms like 1,2,3,...,omega or 7,8,9,...,omega > 2) omega + 1 = 2,3,4,...,1 If we allow representations like the above, then omega+1 does not have a unique representaion > and therefore omega cannot not be a term in > the sequence It can however be a term in another sequence representing omega+1. There is no single sequence which represents omega+1. > but omega is the first part of it, omega = 2,3,4,... > > This dilemma is what I have been trying to explain for years now. > The explanation is that if we do not insist that an ordinal be represented by an initial seqment of ordinals, then there is no unique representation of an ordinal. - William Hughes
From: David Marcus on 11 Nov 2006 13:27
mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > > > It is now to be shown how one is lead to the definition of the new > > > > > numbers. And this definition, given by Cantor and translated by myself > > > > > is given above. There is no point to draw but only to understand > > > > > Cantor's *definition* (or not). > > > > > > > > Okay, so for you there is no point to draw about your quote other than > > > > understanding Cantor's own writings. And, I'll add that in this > > > > particular instance, Cantor, as you translated, is not in conflict with > > > > the theorem of current set theory that omega is a limit ordinal and the > > > > first oridinal that is greater than all natural numbers. > > > > > > That is what I said. omega is a limit. In modern set theory there are > > > limits. > > > > Moe said that "omega is a limit ordinal". He did not say that "omega is > > a limit". The two statements are not the same. > > Moe said "Cantor, as you translated, is not in conflict with the > theorem of current set theory". Cantor said "omega can be understood as > a limit". Cantor said "understood". That is not the same as saying "is". Regardless, Cantor's papers are very old and predate the modern formulations. > > Do you really believe the > > two statements are the same? Or, are you trolling? > > A bottle of beer is not a beer bottle. But both notions have to do with > beer and with bottles. Why is a limit ordinal called so if it is not > the limit of some set of ordinals? Because the notions of limit ordinal and of limit are related to intuitive ideas that are similar enough that people decided to use similar terminology to name them. However, you can't deduce anything from the name. You must use the precise, rigorous definition. Thinking that you can use the name is a common fallacy. The fact remains that a limit ordinal is not the limit of a set of ordinals. A limit ordinal is an ordinal other than zero that is not a successor ordinal. This definition is given by both Halmos and Kunen in their books on set theory. > > is rude to use a common word with a personal meaning and not point this > > out. If someone asks you for a "definition", etiquette and honesty > > requires you to say, "I'm sorry, but I do not know what you mean." > > In mathematics it is practical to give names to various particular > properties and objects, i.e., to define new properties. Mathematics > without definitions would be possible, but exceedingly clumsy. Yes, but the problem is that you fail to state what definitions you are using. > > The phrases "actually existing" and "cannot exist" are not defined. > Eetiquette and honesty requires you to say, "I'm sorry, but I do not > know what the definitions of these words are. And then you should > attach a list of words you know. It can't be too long. So I will look > whether there are words which could be used to explain "actually > existing" and "cannot exist". Amusing. I (and others) have repeatedly asked you what you mean. I have suggested that you use the terminology in Halmos's Naive Set Theory or in Kunen's Set Theory. If you don't like these books, then pick a different one. However, Cantor's papers are much too old to be used for a discussion today. -- David Marcus |