Cantor Confusion
in [Math]
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From: William Hughes on 7 May 2007 06:56 On May 7, 6:23 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 6 Mai, 22:26, Virgil <vir...(a)comcast.net> wrote: > > > In article <1178461785.380972.277...(a)e65g2000hsc.googlegroups.com>, > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 6 Mai, 03:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1178190762.081101.159...(a)h2g2000hsg.googlegroups.com> WM > > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > > Wrong. If there is always a path p' with p, > > > > > What does this statement mean? > > > > Even in the *infinite* tree a path cannot be distinguished from all > > > other paths. That means, a real number which cannot be described by a > > > finite formula (and most of them cannot) does not exist. > > > What axiom requires that a number must be defined by a finite formula in > > order to exist? > > What kind of existence has an undefinable number? > > > > > > Cantor's diagonal proof fails, because the diagonal number is never > > > distinguished from all other real numbers (if uncountably many real > > > numbers exist). > > > Cantor's "diagonal" proof only involves binary strings, not numbers, and > > a very simple rule distinguishes the "diagonal" form every member of the > > list. > > Only finite strings have been tested so far. > > Regards, WM There are an infinite number of members to check. However, each of them only requires checking a finite string. Whoops Over there! A pink elephant! There are an infinite number of members to check. So to check them all you have to check an infinite string. - William Hughes
From: WM on 7 May 2007 06:56 On 6 Mai, 23:02, Virgil <vir...(a)comcast.net> wrote: > In article <1178463202.150771.47...(a)y80g2000hsf.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > 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 ma yask for a set of infinitely many paths, but not for > > infinitely many paths. > > So one can ask for, and get, a set with infinitely many elements but it > will not have infinitely many elements? You cannot ask for infinitely many elements. And yes, you never will have infinitely many elements, that is correct too, but does not matter here. > > That does not happen is a well regulated logical system. You cannot ask for infinitely many elements. > > It is proven by Cantor that there are uncountably many real numbers. > > It is also clear that there are only countably many questions, no? > > Define question! A question is an element of a countable set of words over a finite alpabet. That is enough to see that my statement is correct. > > > > > 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. > > It must be true for every pair of paths whatsoever or else one does not > actually have a complete infinite binary tree at all. That may well be possible. > > > But we can mention and test only finitely many paths. So you now say that > > > there are only finitely many paths? And that > > > sum{i = 1..n} i = (n + 1) * n / 2 > > > holds only for those n for which it has been tested? > > It holds for finitely many natural numbers. The natural numbers are a > > countable set. > > Then WM is saying that it does NOT hold for some members of N. It holds for all members of N, and it holds for every finite series 1 + 2 + 3 ... + n. That means every series out of N is finite. > > Thus in ZF and NBG, the kind of anomaly where > sum{i = 1..n} i = (n + 1) * n / 2 is only true for finitely many n in N > cannot happen, because one can show it true for n = 1 and also show that > if true for n = x it is also true for n = succ(c), thus it must be true > for all n in N Correct. Obviously this is a finite set. The inductive proof is valid for all series of natural numbers 1+2+3+...+n. It is not valid for an infinite series. That means "all series" of natual numbers do *not* include an infinite series. Regards, WM
From: William Hughes on 7 May 2007 07:02 On May 7, 6:16 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 6 Mai, 20:16, William Hughes <wpihug...(a)hotmail.com> wrote: > > > On May 6, 10:53 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > Fine. That means, the set is countable but there is no bijection with > > > N definable. > > > No it means that while a bijection exists it is not finitely > > definable. > > The bijection with this countable set is not finitely definable. > Why then do you require a finitely definable bijection between paths > and nodes in the tree? I don"t. I require that p' and p have a node not in common. I do not require that this node be finitely definable. A: simple true statement. Crank: <Stupid arguement that simple true statement implies dumb statement> Why do you make this dumb statement? - William Hughes
From: WM on 7 May 2007 07:19 > > It's not insane just to point out that a proof in formal Z set theory > is one that uses only first order logic applied to the axioms of > formal Z set theory. I only corrected your misprint which you deleted here. > > > > (3) In ordinary set theory, a function is a certain kind of set of > > > ordered pairs and, in ordinary set theory, a function is NOT a triple > > > of a formula, domain, and range. That you even think a set theoretic > > > function must have a FORMULA (!) as a component shows, again, > > > your > > > lack of understanding even the basics of set theory. > > > 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. > > That is a very common NON-formal definition and notion of function > found in a great amount of mathematics. However, it is NOT the set > theoretic defintion that is being used in formal Z set theory and is > NOT the definition that is used in formal Z set theory to prove > Cantor's theorem that there is no function from a set onto its power > set. It is a definition from a book on set theory, called "Introduction to Set Theory". So it gives basic set theory. It shows that your statement "> That you even think a set theoretic > function must have a FORMULA (!) as a component shows, again, your > lack of understanding even the basics of set theory." is as wrong as most of your statements. > > If you INSIST on using 'function' in a sense that is NOT the sense in > formal Z set theory, then you are not talking about formal Z set > theory. I never wanted to talk about astrology, why should I talk about formal set theory? > > > Please do not believe that I will repeat for another 1000 postings > > your controversy with other great mathematicians about the properties > > of a function. > > All you need to do is pick up a textbook on set theory to see that in > ordinary set theory 'function' is defined to be a certain kind of set > of ordered pairs and not as you define. 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? > > 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. > > > > > You are wrong to assume that I did not study set theory. > > > > I infer it by the ignorance you display. > > > Obviously you are not very good in concluding from given facts on > > hidden facts. > > I have no idea what "hidden facts" you're thinking of. I see. > > > If you want to talk about an infinite summation, then you need to be > > > able to eliminate the '...' and put it in the form: Sum(from n to oo) > > > F(n). > > > 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? There are definitions like Sum [n in N] (1) = aleph_0 in several books on set theory. Do you miss the sentence: let n be a finite ordinal, and let N be the set of all finite ordinals, then ...? > > > 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) 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. > > As I said, set theory is not answerable to your fingerpainting with > symbols. Nor am I. It seems to be a rather sad theory. Regards, WM
From: WM on 7 May 2007 07:24
On 7 Mai, 00:59, William Hughes <wpihug...(a)hotmail.com> wrote: > Look! Over there! A pink elephant! That seems necessary to make us believe that a paths that shares every node with another path does not share every node with another path. But it is not sufficient. Regards, WM |