Cantor Confusion
in [Math]
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From: Virgil on 14 May 2007 19:29 In article <1179170680.845959.226840(a)n59g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 14 Mai, 17:15, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > Given any path p in a CIBT. > > > > For each n in N, Let p_n be a path that has in common with p > > > > the first n nodes of p, but no others. There is always such a path, > > > > any of the ones branching the other way at level n. > > > > Since every node of p is at some finite level n in N, the infinite > > > > set {p_n: n in N} of paths "covers" all nodes of p. > > > > > > Is that simple enough for you to grasp, WM? > > > > > Now remove all the path p_1, > > > > and you are left with infinitely many paths > > > > >and remove the path p_2, > > > > and you are left with infinitely many paths > > > > > and remove .... > > > > and you are left with infinitely many paths... > > more than half of which are useless for our purpose. If you remove "half" of an infinite set, you have as many left as you started with. And while they may appear useless individually, they are so useful collectively that they conclusively prove WM totally wrong!! Which some of us regard as quite useful. > And, therefore, it cannot be remedied by infinity. Even > the help of a pink elephant is not sufficient, to take the unnecessary > paths for necessary paths and the insufficient paths for sufficient > paths. That sounds like the last bleatings of someone who knows he is wrong but is desperate to deny it.
From: Dik T. Winter on 15 May 2007 21:55 In article <1178795907.957410.94020(a)o5g2000hsb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 10 Mai, 03:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > So in your opinion there are only finitely many paths in the > > > > infinite tree. > > > > > > I did not say that. > > > > Quote: "Only from those few you can ask for". I can ask only for finitely > > many paths. > > That is correct. But that does not mean that I said that there are > only finitely many paths in the tree, as you implied. Why not? I can ask only for finitely many paths, and (if I interprete your thinking correctly) path which I can not ask for do not exist. > > Not at all. I ask the single question: "are all other paths different > > from a particular path?" > > That is circular, because by "other" you imply that path p can be > distinguished by a node from all the others. No, that is not implied at all. Rather, for *each* other path there is a particular node where they differ. > But we know, that this > not possible. Hence there is another path with p for all the time. There is not. For each path p' != p there is a node where they differ. > If > you state > > 1) p is not alone at any node. > 2) There is no single path p' with p at all nodes. Right. > Then you should show at least two (or more) paths p' which are with p > such that one of them is always with p but not always the same. Prove > that more than one further path p' is required. And this makes no sense. If there is a path always with p, it is equal to p, and so is the same path. > You always state that this was true, but you cannot prove it. Is that > mathematics? What do you think of the following proof: p1 = {n11, n12, ...} and p2 = {n21, n22, ...}. If all of n1i = n2i the paths are the same. If they are not the same, there is an i where they are different. > > In the same way: "what is sum{i = 1..n} i" a single > > question. You want to split the first question into infinitely many > > questions about all paths. In that case you should also split the second > > question into infinitely many questions about all n. Using your logic > > the split questions can only be answered for finitely many cases, so the > > formula "sum{i = 1..n} i = (n + 1) * n / 2" is only valid for those cases > > that you did ask. > > It is valid for those cases I did prove. I did not prove it (and one > cannot prove it) for "all natural numbers which obey this formula", > because all natural numbers which obey this formula do not obey this > formula. This sentence makes no sense. I do not ask it for "all natural numbers which obey this formula", I ask it for "all natural numbers". And what natural numbers that obey that formula do not obey that formula? > > > > > > There is a subtle difference. With the diagonal proof we always > > > > > > remain in the finite. > > > > > > > > > > That's why it fails for numbers with infinitely many digits. > > > > > > > > No. > > > > > > Assertion, no argument. > > > > Yours was also an assertion without argument. > > My argument is: Why do you have to exclude the case 0.999... = > 1.000... if you remain always in the finite? For every finite index > both numbers are different. But when you use all finite indices they are the same. But the indices are still finite. And by the axiom of infinity we can use the set of all finite indices. > > > because for the whole set the result aleph_0. > > > > Pray give a proof of this statement. > > Pray look it up yourself. Hrbacek and Jech, p. 188, for instance. I will look it up (that sum{i = 1..oo} i = aleph_0). But I suspect that you are leaving out context. They specifically define limits on series of ordinal numbers, but that is (as far as I know) far from common. Without such a definition the above is *undefined*. > > > Do you think it would be > > > possible to sum an infinite set by (n + 1) * n / 2? > > > > No, as such has not been defined in mathematics. But you reject the > > formula above for any 'n' for which you did not ask. So each time you > > use the formula with a different 'n' you should prove it for that > > particular 'n'. > > It is obviously wrong for all n for which it holds, if such set of all > n exists. Again makes no sense. It is a formula not about a set of n, but about each individual n. So if the formula holds for all n, it does not need to hold for the set of all n, because the statement is not about a set. The statement is about an individual natural number n. > > > > Oh. A mathematical proof, please. > > > > > > The set of natural numbers is completed such that there is no natural > > > number outside of that set. Do you disagree? > > > > Ah, now you define completed, I think. > > That meaning should be obvious without definition. Well, with you I prefer to use definitions, otherwise you sometimes go one way sometimes another way. > > But it is not true that nothing > > remains. We have still the rational numbers, the real numbers, and > > whatever. > > Read again what I wrote, please: > The set of natural numbers is completed such that there is no natural > number outside of that set. I read again what you wrote: "completed" means nothing remains and that is why I ask for definitions. It is only with your *final* definition that you made sense. So your statement: "completed" means nothing remains was an error? > > > > > > sum{i = 1..n} i = (n + 1) * n / 2 > > > > > > for all n, but only for finitely many n. .... > > Again you misread what I state. I do not state "for finitely many i", I > > state "for finitely many n". The formula is a statement about a single n. > > Only on the right-hand side. The left-hand side formula is a statement > about a set. A set, yes, but a set defined by a single n. > And here we can see again, that "infinite set of finite natural > numbers" is a selfconradiction. Eh? The statement is *not* about infinite sets. The statement is about finite sets governed by a single finite number n. > > In analysis it is undefined. As it is in algebra. And also in many > > instances of set theory. But as I have stated already before, Hrbacek > > and Jech give their particular definition of limits in ordinal arithmetic, > > and with those definitions it makes sense. I think the statement is in > > the context of that definition. > > Here is a part of the introduction, but I have not bothered to > translate the Greek and other symbols: Sorry, that makes it unreadable to me. But from what I can decipher is that they define the infinite sum of cardinal numbers as the cardinality of the infinite union of disjoint sets. In common set theory the infinite union of sets is defined, and so this makes sense. But this has nothing to do with sum{i = 1..n} i = (n + 1) * n /2 where the sum is constrained to natural numbers n. Analysis and algebra are about real numbers, and in those fields infinite sums are not defined, except as a limit. Moreover, using their definition of limits of sequences of ordinal numbers we would get omega. It can make sense to make sum{i = 1..oo} i = aleph_0 but it does not follow in any way from the usual theorem, and requires some additional definitions. Remains the fact that the base formula is (I think) in your opinion only defined for those n that you can ask for (and those are finitely many), and so you have to reprove the formula each time you come up with new n. > > > > What is the relevance? But try to find the first occurrence where > > > > pi(x) and Li(x) switch place. Nevertheless, it *is* a natural > > > > number > > > > > > Are you sure? > > > > Well, it should be a natural number, otherwise we could not even talk > > about pi(x). > > We can talk about pink elephants. William enjoys it frequently. Yes. We can even talk about multi-coloured elephants. There is a good film about such a thing. But there is a difference. pi(x) is *defined* as a function on natural numbers. So if there is a switch between pi(x) and Li(x) it should be on a natural number, otherwise the function pi is not defined there. -- 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 15 May 2007 22:01 In article <1178805746.803872.67630(a)n59g2000hsh.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 10 Mai, 03:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1178716585.114658.276...(a)h2g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > ... > > > > > > There are subsets of a finitely definable set which are not > > > > > > finitely definitable. > > ... > > > There does not exist anything in mathematics, unless it is finitely > > > definable. In particular, any infinite definition is not a definition, > > > because there is a definition of "definition" which says: Definition: > > > A definition has an end, i.e., a last word and a point. > > > > Try to find a finite definition of the set of Ulam's lucky numbers. > > Here it is. Use the prime sieve of Eratosthenes but don't count the > numbers already eliminated. > What is the problem? I would expect a finite definition of the sieve. > The set is finitely defined. Not all lucky numbers can get finitely > defined. In general a recursive definition is not considered a finite definition. But if you want to use your own definition of terms, be sure to be misunderstood. -- 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 15 May 2007 22:13 In article <1178808468.554108.46820(a)y80g2000hsf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 10 Mai, 03:59, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > There is a slight difference. Finite definitions do not imply finitely > > definable. Consider the computable numbers as they are presented by halting > > Turing machines. The definition is indeed finite. > > The definition of the set is finite, not of the numbers. That is your interpretation. I would consider each of the numbers in the set to be finitely defined, but the set not finitely defined. > > Nevertheless they are > > considered not to be finitely defined because it is impossible to find out > > whether a particular Turing machine indeed *does* halt. > > Therefore only few of the numbers are finitely definable. Each of the numbers is finitely definable, because each is a halting Turing machine, and Turing machines are finitely definable. It is the set that is not finitely definable. > > > Yes, of the set. But it is only a definition of those real numbers for > > > which Cauchy sequences can be finitely defined. > > > > No. It is a definition of the set. It is not a definition of the real > > numbers at all. > > A well defined Cauchy sequence, for pi, for instance, makes pi well > defined - not as a number in the sense of MatheRealism, but as a > number in the sense of mathematics. Now you come up again with a new term. 'Well defined'. What is 'well defined'? > > > I gave a bijection between the set of nodes and the set of branching- > > > offs of paths bunches. > > > > You did not. What node is bijected with 0.1010101010...? > > Please read: > I gave a bijection between the set of nodes and the set of branching- > offs of paths bunches. Yes. I do not understand. 0.1010101010... is a path bunch. So there is a branching-off of this path bunch. And you claim a bijection with the nodes. What node is bijected with the branch-off of 0.101010101010...? > > > As there can be not more results of branching- > > > offs than branching-offs, the task is done. > > > > As in each node the number of path bunches decreases I wonder about this > > statement. > > In each node the number of path bunches increases. This is utter nonsense. *All* path bunches start at the root. Or are you now considering bunches that do *not* start at the root? > If we add one bunch > going into the first node, we an say: In all nodes one bunch goes in > and two come out. Makes no sense. As all bunches start at the root, there cannot be more bunches that come out of a node than go in. By your own definitions a bunch that comes out of a node, also comes in it (when it arrives there from the root). > It is remarkable how this simple fact could have been overlooked for > such a long time. It is remarkable how you could overlook the above simple facts for such a long time. -- 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 15 May 2007 22:17
In article <1178808886.504395.245480(a)w5g2000hsg.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 10 Mai, 04:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > I will answer there. > > > > I will not see it, so you clearly cut short the discussion. > > Please try, we could reduce the noise significantly and could kill any > polemics and insults. If you are not able to see it, I will repeat it > here. But I wonder why Google introduced this technique if it remains > inaccessible to most users and even to experts. It is not inaccessible to me. But I do not want to sign up with google just for the privilege of accessing it. And I do not want to use a web browser to access discussions. Why google introduced it beats me. I prefer my keyboard based access to newsgroups, no mouse for me. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |