Cantor Confusion
in [Math]
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From: Dik T. Winter on 15 Feb 2007 20:06 In article <1171562525.973762.49300(a)v45g2000cwv.googlegroups.com> mueckenh(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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Marcus on 15 Feb 2007 20:29 mueckenh(a)rz.fh-augsburg.de wrote: > On 14 Feb., 20:20, Virgil <vir...(a)comcast.net> 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". -- David Marcus
From: Dik T. Winter on 15 Feb 2007 20:38 In article <1171575875.054198.56510(a)q2g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 15 Feb., 14:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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 paths. P(0) U P(1) U ... is a set of paths. U [P(0) U P(1) U ...] is a set of nodes. How can a set of paths (which is a set of sets of nodes) be a subset of a set of nodes? > > You are again confusing two things. If you replace in your paragraph > > {p(0)} U {p(1)} U ... > > by > > p(0) U p(1) U ... > > it becomes correct. But you stated that P(oo) was a subset of > > P(0)U P(1) U ..., and that is trivially false, because p(0) U p(1) U ... > > is an element of P(oo) but not of P(0) U P(1) U ... . > > p(0) U p(1) U ... is a subset of P(0) U P(1) U ... . Wrong. p(0) U p(1) U ... is a set of nodes. 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)? 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. > 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. > > > Take he union of this union, then you have > > > p(oo). This union is done in the tree. > > > > That does have no bearing on the correctness of your statement. > > It does have bearing on the presence of p(oo) and all other infinite > paths in the union of finite sets of paths, 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. > You cannot avoid this by formal retreat fights. Well, formally your statement was incorrect, so how is this a formal retreat fight? All over you are assuming that when you unite sets of paths you are also uniting paths. That is *wrong*. > > > > Yes. Why do you think I disagree with all this? I have stated again > > > > and again that I *did* agree with all this. What I disagree with is > > > > that P(oo) (the set of paths in T(oo)) is a subset of the union of the > > > > P(i) (the sets of paths in T(i)). > > > > > > Every element of P(oo) is the union of a subset of the union of the > > > sets P(i). > > > > Yes. But why do you think that P(oo) is a subset of that union? For > > that each element of P(oo) must be an *element* of that union. > > No. If p(oo) is established as the union of p(i), then it is > established by the P(i). You agreed with my definition of "establish": A path is established if it is the union of a collection of a sequence of paths with the property that each path is a subset of its successor. As you stated: "I accept your definition". But in P(0) U P(1) U ... we have not a sequence of paths with the required property, so it does not fit the definition, and so p(oo) is not established by that union. > This holds for every infinite path q(oo), > r(oo), ... which s element of P(oo). Hence, P(oo) is established by > the union of finite trees. A complete new meaning of the term "establish" here. > > So it has no bearing on whether P(oo) is a subset of the P(i) or not. > > UP(i) = {p(0), p(1), p(2), ..., and other finite paths} Indeed, U P(i) is the set of finite paths. There is *no* infinite path in that union. > 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. 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Feb 2007 20:40 In article <1171578226.939003.198460(a)l53g2000cwa.googlegroups.com> "MoeBlee" <jazzmobe(a)hotmail.com> writes: > On Feb 15, 1:42 pm, mueck...(a)rz.fh-augsburg.de wrote: .... > > There is only one type of convergence. Convergence means coming > > together, not staying apart. (Of course, in set theory convergence > > means divergence.) > > What theorem of set theory can be understand to say that "convergence > means divergence"? Stronger, where in set theory is convergence defined? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: G. Frege on 15 Feb 2007 20:47
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]. A related topic is the "use-mention distinction". "The /use-mention distinction/ is familiar to virtually everyone trained in professional philosophy, but to few others. It's a technical notion that is important particularly to philosophers of language and logic; it's the distinction between /using/ an expression or bit of language (e.g., a word or phrase or sentence) and /mentioning/ or talking about that expression. This pair of concepts proves to be rather difficult for some college students to grasp--in part, I suppose, because it requires distinguishing between, say, a name and the name of that name." Source: http://www.unconventional-wisdom.com/WAW/ROBERT.html F. -- E-mail: info<at>simple-line<dot>de |