Cantor Confusion
in [Math]
|
From: Dik T. Winter on 9 May 2007 09:50 In article <1178628554.931048.21720(a)y80g2000hsf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 7 Mai, 22:36, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > (1) Who is the author of that book? I'd like to look it up to see just > > what is written there. > > Karel Hrbacek and Thomas Jech: "Introduction to Set Theory" > Marcel Dekker Inc., New York, 1984, 2nd edition. They do not *use* the word "formula". Their informal statement is: "Function, as understood in mathematics, is a procedure, a rule, assigning to any object a from the domain of the function a unique object b, the value of the function at a." But that is not the definition, their formal definition is: "A binary relation F is called a function (or mapping, correspondence), if aFb_1 and aFb_2 imply b_1 = b_2 for any a, b_1, and b_2." -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Carsten Schultz on 9 May 2007 10:05 WM schrieb: > On 9 Mai, 13:05, Carsten Schultz <cars...(a)codimi.de> wrote: >> WM schrieb: >> >> >> >> >> >>> On 8 Mai, 20:32, Virgil <vir...(a)comcast.net> wrote: >>>> In article <1178636838.282834.114...(a)w5g2000hsg.googlegroups.com>, >>>> WM <mueck...(a)rz.fh-augsburg.de> wrote: >>>>> On 7 Mai, 23:50, Virgil <vir...(a)comcast.net> wrote: >>>>>>> http://books.google.com/books?id=Er1r0n7VoSEC&pg=PA23&ots=1afi1e6mts&... >>>>>>> k++Jech+set+theory+function&sig=Zx5hPqZZ2icNy3mkguHi9kyrFVA#PPA24,M1 >>>>>>> /Introduction to Set Theory/, by Karel Hrbacek, Thomas J. Jech >>>>>>> (1999) [pages 23-4] >>>>>>> : >>>>>>> : * 3. Functions * >>>>>>> : >>>>>>> : Function, as understood in mathematics, is a procedure, a >>>>>>> : rule, assigning to any object /a/ from the domain of the >>>>>>> : function a unique object /b/, the value of the function >>>>>>> : at /a/. A function, therefore, represents a special type >>>>>>> : of relation, a relation where every object /a/ from the >>>>>>> : domain is related to precisely one object in the range, >>>>>>> : namely, to the value of the function at /a/. >>>>>>> : >>>>>>> : * 3.1 Definition * A binary relation /F/ is called a >>>>>>> : /function/ (or /mapping/, /correspondence/) if /aFb_1/ >>>>>>> : and /aFb_2/ imply /b_1 = b_2/ for any /a/, /b_1/, and >>>>>>> : /b_2/. In other words, a binary relation /F/ is a function >>>>>>> : if and only if for every /a/ from dom /F/ there is exactly >>>>>>> : one /b/ such that /aFb/. This unique /b/ is called >>>>>>> : /the value of F at a/ and is denoted /F(a)/ or /F_a/. >>>>>>> : [F(a) is not defined if /a [not in] dom F/.] If /F/ is >>>>>>> : a function with /dom F = A/ and /ran /F/ [subset] B/, >>>>>>> : it is customary to use the notations /F: A -> B/, >>>>>>> : /<F(a)| a [in] A>/, /<F_a>_a[in]A/ for the function /F/. >>>>>>> : The range of the function /F/ can then be denoted >>>>>>> : /{F(a)| a [in] A}/ or /{F_a}_a[in]A/. >>>>>>> : >>>>>> As usual, WM includes only the irrelevant bits >>>>> I only wanted to avoid typing infinite definitions. >>>>>> and excludes the part >>>>>> that gives the formal definition, and, incidentally, proves him wrong >>>>> The second paragraph proves the first one wrong, in your opinion? >>>> A formal definition always REPLACES any informal ones, and governs the >>>> meaning and usage of the thing defined. >>> But the formal definition does not specify how a and b are related >>> other than by mentioning F. This F however is undefined unless you >>> know from the first paragraph that it is a procedure or rule. >> No, it says that F is a binary relation. Do you know how a binary >> relation is defined? > > By at least one set and a rule. Maybe you should look up the definition of a binary relation. >>> Further definition 3.1 contains: It is customary to use the notations / >>> F: A -> B/. Why do you think A and B were not two sets and F was not a >>> formula, as I said (and as anybody says who ever studied some >>> mathematics)? >> Why don't you go to a university, take a course in set theory, and come >> back once you have mastered the basics? > > Because it seems you have done this. And we see the deterrent result. I see, because you seem to derive no pleasure from understanding, but only from opposing things you do not understand, you should avoid to learn something. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: William Hughes on 9 May 2007 10:10 On May 9, 9:39 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 01:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > > > But why do you write down these paths: > > > > > > Let p'(1) be the path 0111... > > > > > Let p'(2) be the path 00111... > > > > > Let p'(3) be the path 000111... > > > > > > ? > > > > > Any set p(1) up to p(n) can be substitued by a single path. > > > > In fact? Why then did you write down so many useless paths? > > > > > Look! Over there! A pink elephant! > > > > > The set P' can be substitued by a single path. > > > > Oh, does the set P' consist of any paths which are not paths p(n) with > > > a finite argument n? > > > i: Any set of paths {p(1) ... p(n)) can be replaced by a single > > path. > > Why then did you write the unnecessary paths? > > > > > ii: The set P' is the union of sets {p(1) ... p(n)} > > and as such a set of nodes. > > In order to avoid the clumsy expression of path bunches, I call a path > bunch now a finite path. A finite path ends at some node. > > The infinite path {0.000...} is the union of all finite paths with > only 0's, namely {0.} U { 0.0} U {0.00} U ... > The union is one infinite path and, therefore, has not more paths than > were unified. > There are only a countable number of finite paths in an infinite path Look! Over there! A pink elephant! There are only a countable number of infinite paths. > All nodes of the tree are last nodes of finite paths. > The union over all finite paths is the set of all infinite paths. Not without a pink elephant. The union over all sets of finite paths contains the set of all infinite paths. Look! Over There! A pink elephant! The union over all finite paths is the set of all infinite paths. > This > set has not more elements than were unified. > > > > > Look! Over there! A pink elephant! > > > iii: The set P' can be replaced by a single path. > > You need a pink elephant to see that? Since it is trivial to see that no path p(n) can replace the set of paths P', Yes. - William Hughes
From: William Hughes on 9 May 2007 10:16 > On May 8, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > time does not play a role in set theory > > > and > > > To branch off is a temporal act. Look! Over there! A pink elephant! (we can now pretend I didn't make the first statement) - William Hughes
From: WM on 9 May 2007 10:22
On 9 Mai, 04:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178463202.150771.47...(a)y80g2000hsf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 6 Mai, 04:33, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > Your conclusion is invalid. Indeed, each node is passed by a > > > > > mutlitude of paths. But the path "0.10101010..." is nevertheless > > > > > separated from all other paths. > > > > > > > > No. Only from those few you can ask for. > > > > > > And those are (in your opinion) only finitely many. Or can I ask for > > > infinitely many paths? > > > > No. You may ask for a set of infinitely many paths, but not for > > infinitely many paths. > > So in your opinion there are only finitely many paths in the infinite tree. I did not say that. > Great. You reject the axiom of infinity, as I wrote already many times > before. (Note: reject, not refute.) I did not reject anything. I ask you whether you can put infinitely many questions. Please answer. If the answer is no, then you should recognize that you cannot ask whether all paths which accompany 0.101010... will get separated from it. > > > > > > So what? For each two different paths it is possible to state a > > > > > node where they do separate. For each path there is no node where > > > > > it separates from all other paths. There is no contradiction. > > > > > > > > No. Only from those few you can ask for. > > > > > > Opinion again. > > > > It is proven by Cantor that there are uncountably many real numbers. > > It is also clear that there are only countably many questions, no? > > I would think there are only finitely many questions. But all this is not > mathematics but philosphy. No. That is mathematics, because you imply to be able to ask infinitely many question. > > > > 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. > > > > With the paths we can also always remain in the finite because > > > any two different paths have a node where they differ *in the finite*. So > > > for each node there is a path p' that has it in common with p. But also > > > for each p' != p there is a node in p that is not in p'. > > > > No. Only for those p' you can ask for. There must be others, because > > there is no node which belongs only to p alone. > > I do not ask for each individual path. I show it for all. Or do you reject > the formula sum{i = 1..n} = (n + 1) * n / 2? You can not ask the validity > for all n, but only for finitely many n, according to your thinking. So for > each and every n you have to prove it again. This formula is obviously only correct for finite sets f nubers, becasue for the whole set the result aleph_0. Do you think it would be possible to sum an infinite set by (n + 1) * n / 2? > > > > "completed" means nothing remains. > > > > > > Perhaps. > > > > Sure. > > 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? > > > > > In the present state of affairs we know that at every node of path p > > > > which we look at, path p is not unique. And we know, that we can test > > > > and test for different paths p', but we cannot carry out more than > > > > countably many tests (our list is limted). > > > > > > You can do countably many tests? I thought you could only do finitely > > > many tests. > > > > Please read carefully. We cannot do more than countably many. And in > > finite time, we cannot do more than finitely many. > > Perhaps. You are getting more and more philosophical each turn. You refuse to talk and think about what mathematicians can do? > > > > But again, in mathematics it is *not* necessary to test > > > each and every individual. Otherwise you could not even prove that > > > sum{i = 1..n} i = (n + 1) * n / 2 > > > for all n, but only for finitely many n. > > > > Of course this is only true for finitely many n. For infinitely many > > natural numbers, we get aleph_0. > > You are wrong. We do not get that. If so, pray show a proof. See, for instance, Hrbacek and Jech, p. 188, if you do not believe me. (There are more books which treat that topic, but this in my possession: "It is not very difficult to evaluate infinite sums. For example, consider 1 + 2 + 3 + ... + n + ... + ... (n in N). It is easy to see that this sum is equal to aleph_0." What else should this sum be? It cannot be finite and it cannot be uncountable. > > 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? Look at What evidence is there that 2^65536 is a natural number? http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.ndjfl/1093634481&abstract= Regards, WM |