Cantor Confusion
in [Math]
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From: Virgil on 7 May 2007 12:12 In article <1178533414.231814.211740(a)n59g2000hsh.googlegroups.com>, WM <mueckenh(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? What kind of existence does a defineable number have? Certainly not physical. So all numbers are equally mental, existing only in the imagination, and to those who can imagine undefineable numbers they exist on the same basis as any others. > > > > > 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. Physical testing of what transpires in the worlds of imagination is neither necessary nor possible.
From: Virgil on 7 May 2007 12:17 In article <1178534594.281830.256500(a)n59g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 6 Mai, 22:33, Virgil <vir...(a)comcast.net> wrote: > > In article <1178463181.745792.46...(a)y80g2000hsf.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 6 Mai, 04:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1178366406.361131.321...(a)y5g2000hsa.googlegroups.com> WM > > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > ... > > > > > Theorem 2: The set of finitely defined numbers cannot be put into a > > > > > list. > > > > > > > Proof: Assume such a list exists. This means, such a list has been > > > > > finitely defined. > > > > > > Wrong. Existence does *not* mean finitely definable. > > > > > Marthematical existence does mean definable. > > > > But not necesssarily "finitely" so. > > Who can give or understand an infinite definition? Some definitions of pi are infinite but quite well understood. > > > > > What else should it mean? > > > Definitions can only be finite. > > > > That may be one of WM's axioms, but does not appear in ZF or NBD. > > That is the basis of logic which ZF or NBG have to obey (if they are > logical systems). What logical rule in any predicate logic says that a thing must have a finite definition in order to exist? In ZF and NBG those things exist which are required to exist by the axioms themselves or by theorems derived from those axioms, and that requires that there exist a power set of the set of naturals, which all its members, in both ZF and NBG.
From: Virgil on 7 May 2007 12:27 In article <1178535408.151099.273790(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. ZF not only asks for what turns out to be infinitely many elements, it demands it. So does NBG. So do all standard set theories. > And yes, you never will have infinitely many elements, that is correct > too, but does not matter here. ZF not only asks for what turns out to be infinitely many elements, it demands it. So does NBG. So do all standard set theories. > > > > That does not happen is a well regulated logical system. > > You cannot ask for infinitely many elements. ZF not only asks for what turns out to be infinitely many elements, it demands it. So does NBG. So do all standard set theories. > > > > 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. According to WM's definition, it appears as if a question must be a word. > > > > > > > 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. Not in ZFor NBG , at least not in the Dedekind sense, as one can prove that it injects to a proper subset of itself. And not in ZF or NBG if one uses the von Neumann definition of N as one can prove that N is not a member of itself, but that every finite ordinal is a member of N. > > 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. No one said otherwise, but N is infinite and for each n in N, the sum of all naturals up to and including n is n*(n+1)/2, as provable by induction.
From: Virgil on 7 May 2007 12:50 In article <1178536779.509069.124620(a)l77g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > > > > 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 may very well be a very naive set theory for non-mathematicians. Who wrote it, and for what sort of students is it supposed to be an introduction? 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. In ZF, since everything is a set, functions, if they are to exist at all in ZF , MUST be sets. Thus, in ZF, the ONLY relevant definition of a function is as a set. > > > > > > 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? Because you set yourself up as a critic of formal set theory, and you DO talk about it all the time, and in unwarrantedly disparaging terms. > > > > > 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? Yes! > I don't know why I should have > done that. Neither do we. > Perhaps your understanding is not as sharp as you think? And perhaps our collective understanding is a good deal sharper than 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? If you set yourself up as a critic of set theory, it helps to be aware of what goes on in what you are criticizing. Since you choose not to be aware, you are bound to make a fool of yourself occasionally. If fact you do s frequently. Best taught by you? Amusing. Since you seem to be incapable of learning it without help, any help is better that you are doing on your own. > > > > > > > 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. I doubt it. > > > > > 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. There are all sorts of things in naive set theory texts that formal set theories do not allow. > > 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. In that case, it is irrelevant. Much more relevant wold be a count of the number of paths in a finite binary tree in which each path has n branchings (edges). I.e., the number of non-leaf levels is n, and the number of non-leaf nodes in any path is n, Then the number of paths equals the number of subsets of the set of levels. In the limit, the number of levels becomes N so the number of paths becomes the cardinal of P(N). > > > > As I said, set theory is not answerable to your fingerpainting with > > symbols. Nor am I. > > It seems to be a rather sad theory. Not anywhere near as sad as WM's futile attempts to build a theory.
From: Virgil on 7 May 2007 12:53
In article <1178537072.163162.176250(a)u30g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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 If WM believes that one path in any tree can share every node with another and different path, he has passed beyond reason into some sort of fugue state. |