Cantor Confusion
in [Math]
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From: mueckenh on 16 Feb 2007 06:42 On 16 Feb., 02:38, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171575875.054198.56...(a)q2g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 15 Feb., 14:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: I think we should first clarify the notation before continuing: P(n) is the set of all paths of the tree T(n) with n levels. P(0) is a set with one path, namely p(0) = {0.} P(1) is a set with two paths, namely p(1) = {0.0} and q(1) = {0.1} P(2) is a set with four paths, namely p(2) = {0.00}, and some others. The union of all sets of finite paths is P(0) U P(1) U ... is the set of all paths. In this union P(0) U P(1) U ... there is the path p(0) as an element and there is the path p(1) as an element and so on. Therefor the set of all the paths p(0), p(1), p(2), ... is a subset of P(0) U P(1) U ... The tree forms unions from sets of paths. The tree T(2) is the union of the sets of paths P(0) U P(1) U P(2). But the tree T(2) does not contain the paths of P(0) and P(1), namely p(0) and p(1) and q(1). Similar the tree T(oo) is the infinite union P(0) U P(1) U P(2) U ... Form the subset p(0), p(1), p(2), ... of this union the tree forms the union p(0) U p(1), U p(2) U ... = p(oo). So the tree contains the union of the paths p(0), p(1), p(2), ... which according to you is p(oo). This is the outmost left path of the union tree. > ... > > > > It need not be in the union. You say it is the union of {p(0)} U > > > > {p(1)} U {p(2)}. If so, then it is the union of a subset of P(0) U > > > > P(1) U P(2) U ... > > > > > > I never stated that p(oo) = {p(0)} U {p(1)} U ...; that can not be the > > > case, because on the left hand side we have a set of nodes, and on the > > > right hand side we have a set of paths. And a path is not a node. > > > Moreover, {p(0)} U {p(1)} U ... is not the union of a subset of > > > P(0) U P(1) U ...; it *is* a subset. > > > > {p(0)} U {p(1)} U ... is a subset of the union of P(0) U P(1) U .... > > Wrong. p(0) U p(1) U ... is a set of nodes. Every path is a set of nodes. You said p(0) U p(1) U ... = p(oo). Of course this is a set of nodes. The paths of this union, p(0), p(1), ... are elements of the union P(0) U P(1) U ... . Therefore these elements form a subset of P(0) U P(1) U ... . > P(0) U P(1) U ... is a set > of paths. Again, how can a set of nodes be a subset of a set of paths > (which is a set of sets of nodes)? One path can be a subset of a set of paths as a singleton or it can be an element of the set of paths. > > Do you not know the difference between a set of sets of nodes and a set > of nodes? This is the same as stating: > {{1}, {1, 2}, {1, 2, 3}} subset U {{1}, {1, 2}, {1, 2, 3}} = {1, 2, 3} > and at the same time: > {1, 2, 3} subset {{1}, {2, 3}}. > Please keep a clear head about what the *elements* of the unions are. Please see above. The union of paths cotains elements p(n). The tree forms unions from these elements. The tree T(oo) does not contain any finite paths, but the unions of finite paths. > > > This subset is unioned by the tree. It yields p(oo). > > P(oo) is composed of the sets p(oo), q(oo), r(oo). > > The tree is not interesting in this. Why do you think did I choose the tree? > If the finite sets of paths contain only finite paths, their union can *not* > contain an infinite path. Because when taking that union you are uniting > sets of paths, not paths. That it is a subset of the union of that union > does not matter at all. The union of that union is simply a set of nodes, > *not* a set of paths. See above. > > > > The tree contains a path which *is* the union of the p(i), namely > > p(oo). The tree, i.e., the union of finite trees is or establishes the > > paths q(oo), r(oo), ... too. Therefore it establishes all elements of > > the set P(oo). > > Perhaps, depending on what you mean with "establish". I thought we had > a common definition, but that is apparently not the case. It is. p(oo) = p(0) U p(1) U p(2) U ... The finite paths are elements of the union of sets of paths P(n). The tree unites some of these elements which fit together to form an infinite path. >But anyway, > yes, P(oo) contains infinite paths, and so can *not* be a subset of > U P(i), as that one contains (as you note) only finite paths. It need not be a subset of U P(i) in order to be in the tree. Regards, WM
From: mueckenh on 16 Feb 2007 06:51 On 16 Feb., 02:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171562525.973762.49...(a)v45g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > ... > > One set of curly brackets would be enough to be number 3. But it is > > correct that there are more obvious and less obvious numbers: > > ... > > An ellipsis. > > > ||| > > A sequence of bars. > > > {{{ > > A sequence of braces. > > > abc > > The first letters in some of the alphabets that use the standard order. > > In none of them I do immediately see the number three. What I see is that > there are three of something, not the number three. What is common to all? Regards, WM
From: mueckenh on 16 Feb 2007 06:58 On 16 Feb., 00:32, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > It is true for every form of existence E can assume. But in order to > > avoid any misunderstanding: > > > Every set of finite even numbers contains numbers which are larger > > than the cardinal > > number of the set, if the set has a cardinal number. Please insert "positive" or simply replace "finite" by "natural": Every set of natural even numbers ... Then it is correct. Every set of even natural numbers contains numbers which are larger than the cardinal number of the set, if the set has a cardinal number. (Not necessary to mention, 0 is not natural.) > > And since in W... all cardinal numbers are finite this > reduces to the trivial > This can be proven everywhere. > > The sparrow is fixed. It is a songbird and it sings: The number of > > elements of my set is not fixed. > > Your set is a member of the equivalence class, it is not > the equivalence class. > The sparrow of E is an equivalence class which contains > your set and many other sets. Any property your set may have > is totally irrelevent. The fact that the number of elements in > your set is not fixed does not mean that the number of sets > in the equivalence class is not fixed. No? Do you talk of sets like the naturals, the primes, the even numbers, and so on? Even the set of these sets is potentially infinite. The number of sets in the equivalence class oo is also oo. > > > > Because vectors and tensors are closely related to complex numbers, > > Quaternions and Cayley numbers, but are not called numbers. > > In my > > opinion, all numbers should be ordered by size. > > But we are not interested in what is true in Wolkenmuekenheim. > It is quite natural to call the algebraic completion of the real > a set of numbers. Is your aim being natural? It is quite natural to call 0 not a natural number.> > True and absolutely irrelevent. Which of the following do > you disagree with? > > All initial segments of E have a fixed maximum yes. > > E does not have a fixed maximum > yes. > If every initial segment of E has a fixed maximum > then E has a fixed maximum no. > Regards, WM
From: Dik T. Winter on 16 Feb 2007 08:24 In article <MPG.203ef87d4dd2aa7989cd0(a)news.rcn.com> David Marcus <DavidMarcus(a)alumdotmit.edu> writes: > Dik T. Winter wrote: > > In article <gn2at2565gmfg5kjlbpr2jn10l6mfuiht3(a)4ax.com> G. Frege <nomail(a)invalid> writes: > > > On Thu, 15 Feb 2007 20:29:22 -0500, David Marcus > > > <DavidMarcus(a)alumdotmit.edu> wrote: > > ... > > > >>> For one thing, WM is missing the knowledge that a name is not the object > > > >>> named. > > > >>> > > > >> Is this valid for names too? > > > >> > > > > I think so. My name is "David". ""David"" is a name for my name. Another > > > > name for my name would be "my name". > > > > > > Actually, again one of the many insights of Gottlob Frege [the real > > > one]. > > > > Actually it has already been used in Alice in Wonderland, published when > > Frege was 17 years old: > > Knight: The name of the song is called Haddocks' Eyes. > > Alice : Oh, that's the name of the song, is it? > > Knight: No, you don't understand, that is what the name is called, the > > name really is The Aged Aged Man. > > Alice : Then I ought to have said that's what the song is called? > > Knight: No, you oughtn't: that's quite another thing! The song is > > called Ways And Means, but that's only what it's called. > > Alice : Well, what is the song then? > > Knight: I was coming to that, the song really is A-sitting On A Gate. > > Lewis Carroll was not entirely stupid... > > As Martin Gardner pointed out, the Knight should have just started to > sing the song after saying "... the song really is". "A-sitting On A > Gate" is another name, not the song itself. It would also have been right if he had stopped at that point and not started singing after all... (And it is from Through the Looking Glass of course, 6 years after Wonderland.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: William Hughes on 16 Feb 2007 09:19
On Feb 16, 6:58 am, mueck...(a)rz.fh-augsburg.de wrote: > On 16 Feb., 00:32, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > True and absolutely irrelevent. Which of the following do > > you disagree with? > > > All initial segments of E have a fixed maximum > > yes. > > > E does not have a fixed maximum > > yes. > > > If every initial segment of E has a fixed maximum > > then E has a fixed maximum > > no. > Your claim is M: If all initial segments of E have [a] property and if M: no element of E is outside of every initial segment, then E has that M: property. So if i: All initial segments of E have a fixed maximum and ii: no element of E is outside of every initial segment then iii: E has a fixed maximum You have claimed that i is true, and iii is false. Given the ii is clearly true, we have i [true] plus ii [true] implies [Claim above] iii. How do you explain this contradiction? - William Hughes |