Cantor Confusion
in [Math]
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From: WM on 8 May 2007 11:09 On 8 Mai, 15:09, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 8, 8:27 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > 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. > > The statement > > "Every path branches off somewhere" > is true. > > However. this implies "Every path has branched > off somewhere" only after the last path has branched off. If "*there is* a set such that every path branches off somewhere", then every path *has branched off* somewhere. For static sets this must be true. Otherwise there would be a temporal development involved. Then we should be precise and say: Every path will branch off somewhere, after, say, first of June. But time does not play a role in set theory. Did you understand and can you explain your error now? Regards, WM
From: William Hughes on 8 May 2007 11:11 On May 8, 11:00 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 8 Mai, 14:41, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > 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 > > 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. Look! Over there! A pink elephant! The set P' can be substitued by a single path. - William Hughes
From: William Hughes on 8 May 2007 11:28 On May 8, 11:09 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 8 Mai, 15:09, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 8, 8:27 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > 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. > > > The statement > > > "Every path branches off somewhere" > > is true. > > > However. this implies "Every path has branched > > off somewhere" only after the last path has branched off. > > If "*there is* a set such that every path branches off somewhere", > then every path *has branched off* somewhere. No. 'every path branches off somewhere' needs only existence, no order (temporal or otherwise) is needed 'every path *has branched off* somewhere' needs some sort of order (temporal or otherwise) There are ordered sets that do not have a last element - William Hughes ..
From: Carsten Schultz on 8 May 2007 11:48 WM schrieb: > 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? The second paragraph proves your interpretation of the first paragraph wrong. Note that `procedure' and `rule' in the first paragraph are undefined, which is why axiomatization of set theory is a good thing. -- 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: MoeBlee on 8 May 2007 13:01
On May 8, 5:49 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > 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. In which book we find the standard set theoretic definition. Your mistake is in conflating their mention of the general non-rigorous notion with the actual definition they give. > > (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. Those people give the same definition as Hrbacek & Jech, you fool. And one of those listed above is Jech in his own book. And it's not a matter of "changing" the definition but rather of recognizing a non- rigorous general notion but then formulating a rigorous definition. You, however, are unable to cope with the rigorous definition. > > (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? They're the same proofs as in Hrbacek & Jech. And using a rigorous definition is what makes such proofs so very much worthy. > > (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? No, actually Hrbacke & Jech use the same definition I do. You're just either incapable or unwilling to understand the difference between stating a non-rigorous general notion and the rigorous definition that makes that general notion either formal or formalizable. > > (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? What proof of mine do you claim is wrong? Using the rigorous set theoretic definition of 'function' does not make proofs "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? I did not say that I have or attended or have not attended any university. It's irrelevent. The point is that you are spouting foolishly and idiotically about a subject of which you don't even understand at a first semester level. Whether you accomplish understanding the material thorugh a university course or by studying the textbooks on your own is of no matter to me, but you need to do one or the other if you are not to continue spouting misinformation and making a fool of yourself as you do it. And if you think the set theoretic definition of a function is that of a formula, domain, and range, then you need to go back to your Hrbacek and Jech to see that they present an informal notion that is then made rigorous by a definition. The informal notion is not itself the set theoretic definition. > > > 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. Look right in your Hrbacek and Jech. > > 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. 'formula' is not part of the definition of 'function'. Just learn the set theoretic definition of a function: f is a function <-> (f is a relation & Axyz((<x y> e f & <x z> e f) -> y=z)) Note: Hrbacek & Jech use the word 'binary relation' instead of 'relation' that is more commonly used, but that does not affect the important feature of this definition, especially that, indeed, AS I SAID, a function is a certain kind of set of ordered pairs and the set theoretic definition of a function is NOT that of a formula, domain, and range. > > > > 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. Look at the EXPLICIT definition in Hrbacek & Jech. Hint: The discussion of the general mathematical notion there is NOT the explicit definition. > > > > > 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). Now page 188 of Jech & Hrbacek. You FAILED at page 23. But if you want to discuss summations, and cardinal summation, or whatever, I'd be happy to do so, but not in any context in which you don't understand the definitions and theorems that all lead up to the subject of summations, starting at page 1, to page 23 (!) and the rest. > > > 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? An equality formula in which, on the left is your "2-1-1+2-1-1+2-1-1+-... " and on the right is a term in the extended language of set theory (extended by previous definitions) or in whatever language your theory is in. > > > > > 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). Fine. I will take "2-1-1+2-1-1+2-1-1+-... " to be a senselessly protracted manner of writing '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. MY argument? WHAT argument is that? > > > 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. As I said, if you ever care to state axioms and definitions for your notions about trees, time and patience (with your confusions such as you've shown in your abysmal misunderstanding and misREPRESENTATION of Hrbacek & Jech) enduring, I'll read your tree argument. MoeBlee |