Cantor Confusion
in [Math]
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From: WM on 8 May 2007 08:27 On 7 Mai, 21:49, William Hughes <wpihug...(a)hotmail.com> wrote: > Since there is no last path which branches off, this list never ends. > So every node of p is shared by other paths. > However, every path branches of somewhere, so there > is no path p' which shares every node of p. You state: (1) Every node of p is shared by other paths. (2) Every path branches off somewhere. (1) ==> Not every path has banched off somewhere (because there must remain at least one "other path" and "is shared" is the opposite of has branched off). (2) ==> Every path has branched off somewhere. So your claim can be summarized: Not every path has branched off somewhere. Every path has branched off somewhere. Regards, WM
From: William Hughes on 8 May 2007 08:41 On May 8, 8:20 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 7 Mai, 14:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > Every node, n, of path p is shared with another path > > > > p'(n). Note that p'(n) can be different for every n. > > > > Note that p'(n) need not diffrent for every n. > > > However, it cannot be the same for all n. > > For which n must it change? > > Could you explain, please, how you define a set of at least two > different paths, p' and p'' (or even more) which are required to > accompany p for all n? > Here is a very simple, countable example. Let p be the path 000... Let p'(1) be the path 0111... Let p'(2) be the path 00111... Let p'(3) be the path 000111... .... let P' be the set of all the p'(n) In general p'(n) has a n 0 nodes, followed by only 1 nodes. (Note that there is no last p'(n)) A: for any node k, of p, there are an infinite number of paths in P' which contain node k B: No path in P' contains every node in p - William Hughes
From: Carsten Schultz on 8 May 2007 08:46 William Hughes schrieb: > On May 8, 8:20 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: [The usual stupid stuff that Mückenhein writes] > > Here is a very simple, countable example. > > Let p be the path 000... > > Let p'(1) be the path 0111... > Let p'(2) be the path 00111... > Let p'(3) be the path 000111... > ... > > let P' be the set of all the p'(n) > > In general p'(n) has a n 0 nodes, followed by only 1 nodes. > (Note that there is no last p'(n)) [...] You mean the set of all integers exists, and really and actually exists, and still does not have a largest element? Wow, that must be hard to swallow for Mückenheim. Best, Carsten -- 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: WM on 8 May 2007 08:49 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. > > (2) You can look at the most commonly used and respected textbooks in > set theory to see that they don't define a function as a formula, > domain, and range. Enderton; Suppes; Levy; Quine; Bernays; Kunen; > Stoll; Halmos; Moschovakis; Jech; Takeuti & Zaring; Shoenfield (in the > set theory chapter); Chang & Keisler (in the set theory chapter); > Mendelson (in the set/class theory chapter), Godel (in his small book) > do not define a function as a formula, domain, and range. So these people are simply wrong. No reason to change the definition of a function. > > (3) In such textbook proofs of Cantor's theorem (the theorem that no > set maps onto its power set), the definition of 'function; used is the > one I've mentioned and is not that of a formula, domain, and range. Perhaps that's the reason why these proofs are not worth the paper they are written on? > > (4) No matter what ANY author says, the formal Z set theory proof that > I point to (which agrees with those authors anyway) is using the > definition of 'function' that I mentioned and not that of a formula, > domain, and range. Perhaps that's the reason why you are wrong? > (5) Taking a function to be formula, domain, and range in a formal > proof of Cantor's theorem would be NONSENSE. We don't do that. We take > a function to be just what I've said it is and we prove that, for all > x, there is no function on x and onto Px. And if you think that that > proof uses the definition of a function as being a formula, domain and > range, then you have NO IDEA what the proof is. Perhaps that's the reason why your proofs are wrong? > I said that the proof in formal Z set theory of Cantor's theorem uses > nothing but first order logic applied to the axioms of set theory. You > have not refuted that, and you could see for yourself that it is true > if you only knew basic predicate calculus and even less than what one > learns by the end of a first semester course in set theory. Which university did you attend? > > Did I oppose to a function being a set? I don't know why I should have > > done that. Perhaps your understanding is not as sharp as you think? > > By saying that a function is a formula, domain, and range you > contradict the set theoretic definition of a function being a certain > kind of set of ordered pairs. That is wrong. > And, as to understanding, notice that I > did NOT say that you contradicted that a function is a set, but rather > I am informing you that in SET THEORY (such as ordinary textbook set > theory and as formalized as formal first order theory) a function is a > certain kind of set of ORDERED PAIRS, and you contradict that as you > take a function to be a formula, domain, and range, since whatever > that is, either formalized as a triple or not formalized at all, it is > not a set of ordered pairs and, a fortiori, not the kind of set of > ordered pairs that a function is in set theory. Of course, the formula connects the elements of domain and range to get a set of ordered pairs. > > > > No, a function is not ITSELF a triple of a formula, domain, and range, > > > but a formula may DEFINE a certain set of ordered pairs that is a > > > function. You need to understand such basics of set theory. > > > Do I need that, in fact? Best taught by you? Amusing. > > Apparently you do! You claim that in set theory a function is a > formula, domain, and range. So you DO need to be told that in set > theory that is NOT what a function is. You are in error. > > > > > Sum [n in N] (2-1-1) < 3. > > > > That is but another way of writing 2-1-1+2-1-1+2-1-1+-... < 3. > > > > That is not a DEFINITION of "2-1-1+2-1-1+2-1-1+-... " > > > Why do you dislike it? > > It's not a matter of dislike. It's just that it's not a definition! > > > There are definitions like Sum [n in N] (1) = aleph_0 in several books > > on set theory. > > You can mention anything in particular so that I can evaluate the > context of the notation. It is easy to see that this sum (SUM [n in N] n) is equal to aleph_0. (See quote above, p. 188). > > > Do you miss the sentence: let n be a finite ordinal, and let N be the > > set of all finite ordinals, then ...? > > That doesn't lend a definition to: > > "2-1-1+2-1-1+2-1-1+-... " What then is missing in your opinion? > > > > > If you have problems with infinite sums or products then you should > > > > look up set theory, for instance Koenigs theorem. > > > > I have no problem with infinte sums and products. I am just pointing > > > out that your notation is NOT justified as infinite summation until > > > you specify what F is in Sum(from n to oo) Fn. > > > Fn = (2-1-1) > > My notational slip actually. I meant to write: > > Sum(n=0 to oo) Fn I understood it that way and answered Fn = (2-1-1). > > or whatever range you wish instead of starting at 0. > > Anyway, your Fn = 2-1-1 is just a constant function! What infinite > summation do you think you get from a constant function?! Sheesh! I do not believe to get an infinite sum. That is your argument. > > > It is the number of separated path coming out of the n-th element of > > the tree minus the number of ingoing separated paths minus the number > > of nodes of this element. > > And you just said that for n it is 2-1-1. So it's a constant function. > So, for an infinite summation, just do the correct math given F is a > constant function! > > What do you think Sum(n=0 to oo) 2-1-1 = Sum(n=0 to oo) 0 is? I would say the result is less than 3, so there are not uncountably many more paths than nodes. Regards, WM
From: WM on 8 May 2007 08:53
On 7 Mai, 22:55, Virgil <vir...(a)comcast.net> wrote: > In article <1178565586.844949.42...(a)n59g2000hsh.googlegroups.com>, > > > > You are the one pushing "finitely definable", whatever that means. > > > That means: You cannot determine or define or construct or write down > > or tell me a bijection between the set of numbers which are defined > > by a definition identifying them uniquely and the set of natural > > numbers. > > In Zf, every set, including N, has a power set, and there is provably no > surjection from any set in ZF to its power set, but trivially an > injection, so that the cardinality of any set is less that that of its > power set. > > The paths in a complete infinite binary tree in ZF are easily seen to > biject with the power set of N, so that set of paths has cardinality > greater than N. > > And nothing that WM can say can change that. We know your opinion. But it is out of topic. What is remarkable is the following: You cannot determine or define or construct or write down or tell me a bijection between the set of numbers which are defined by a definition identifying them uniquely and the set of natural numbers. Regards, WM |