Cantor Confusion
in [Math]
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From: mueckenh on 13 Feb 2007 06:11 On 12 Feb., 21:49, Virgil <vir...(a)comcast.net> wrote: > In article <1171288127.143257.142...(a)v45g2000cwv.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 12 Feb., 13:59, Franziska Neugebauer <Franziska- > > Neugeba...(a)neugeb.dnsalias.net> wrote: > > > mueck...(a)rz.fh-augsburg.de wrote: > > > > Set theory, however, says: If the union of all finite paths, > > > > 0., 0.0, 0.00, 0.000, ... , > > > > is in the tree, then the infinite path p(oo) = 0.000.... is in the > > > > tree too. > > > > No contemporary set theory or graph theory does posit such a nonsensical > > > claim. It is entirely *your* claim, better known as > > > It is easy to recognize from the union of all finite segments of N. > > Set theory says: If the union of all finite segments, > > {1,2,3,...,n} is present, then the infinite set N = {1,2,3,...} is > > present. (In fact set theory says: N is the union. But I don't know > > how this is accomplished.) Or do you see a difference betwee n the > > union of all finite segments and N? > > > Regards, WM > > If the union of all finite segments, including both 0's for left > branchings and 1's for right branchings leads to all infinite binary > sequences, as WM has esssentially claimed above, then WM has managed to > prove that there are uncountably many infinite paths. > > Good for you, WM Bad for set theory, because these uncountably many infinite paths are the limits of countably many sequences. Regards, WM
From: mueckenh on 13 Feb 2007 06:30 On 13 Feb., 02:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171283208.696664.25...(a)a75g2000cwd.googlegroups.com> mueck....(a)rz.fh-augsburg.de writes: > > > On 12 Feb., 04:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > > III > > > > > > ooo > > > > > > abc > > > > > > all that is understandable as 3 without other convention than that > > > > > > "all" is the start of this sentence and not another > > > > > > representation of 3. > > > > > > Well, I would concede that the above three things are representaions of > > > the number three, using some convention. Anyhow, they are *not* the > > > number three. > > > > What has "the number three" that is not expressed above? > The abstract > number three is not a representation. I asked what has the abstract number 3, that III has not? (I did not ask what the abstract number 3 has not.) > > > > > I missed one o. I corrected it above. Now it expresses all properties > > > > that the number three can express. Therefore it is the number three > > > > (and in addition it has the special form of o's). > > > > > > Numbers can express properties? You have lost me here. > > > > To have three elements is a property of a set. > > Oh. Your terminology is unfathomable. Indeed the number 3, when seen > as a set, can have three elements. In the von Neumann model. However, > I remember to also having seen another model, where the number three was > {{{}}} It was page 93 of my book. >(but I think now you can only model the natural numbers). Anyhow, > within this model 3 when seen as a set has only one element. When seen as a set of curly brackets it has 3 at the left sinde and 3 at the right. > > > > > > The third contains three letters from the Latin script, > > > > > but I do not see a suggestion of the number three at all. > > > > > > > > Each of them expresses everything that the number 3 can express. Each > > > > of them is number 3. > > > > > > Well, except that in a number of languages using the Latin script, "abc" > > > are *not* the first three letters. And I have *no* idea what you are > > > meaning with a statement that numbers can express something. > > > > If you think that numbers cannot expres anything, then you can use the > > empty set for all of them. > > I have *no* idea what you are meaning with a statement that numbers can > express something. I do not deny it, because I do not understand the > meaning. "sie [die Zahlen] dienen als ein Mittel, um die Verschiedenheit der Dinge leichter und schärfer aufzufassen." > > > A number can express how many wheels your car has, for instance. > > Perhaps, but I would not use this strange terminology. Why not? Regards, WM
From: mueckenh on 13 Feb 2007 06:42 On 13 Feb., 02:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171284657.431713.275...(a)l53g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 12 Feb., 04:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > You repeatadly misunderstand: > > > (1) The union of sets of finite elements does contain an infinite element > > > with > > > (2) The union of finite sets is infinite. > > > The first is false, the second is true. > > > > > > But you want to conclude (1) from (2). > > > > No. I do not conclude that p(oo) is in the union of P(0) U P(1) U P(2) > > U ... and I do not conclude that p(oo) is in the union of {p(0)} U > > {p(1)} U {p(2)} U .... > > > > What I conclude and what reasonably must be concluded is: > > {p(0)} U {p(1)} U {p(2)} U .... c P(0) U P(1) U P(2) U ... > > But, indeed, that is true. > > > If p(oo) is established by {p(0)} U {p(1)} U {p(2)} U .... > > Then p(oo) is also established by P(0) U P(1) U P(2) U ... > > But again, I ask you what you mean with "established" here. It is neither > an element of those unions, nor is it one of those unions. So what is > your meaning? As I have stated again and again: > p(oo) = p(0) U p(1) U p(2) U ... > but that one is quite different from > {p(0)} U {p(1)} U {p(2)} U ... This union contains (as a subset) all sets p(n). Their union is, as yo say, p(oo). They all are in the tree. Their union is in the tree. Hence p(oo) s in the tree. > > > > Great. I state that I do not know what you mean with the term "establish", > > > and you follow up with a paragraph having that word all over the place, > > > and ask me to explain it. With all those occurrences of the word > > > "establish" it makes no sense at all. And all that is followed by QED. > > > What do you *mean* with "establish"? > > > > In my opinion the path p(oo) does not exist. Set theory, however, > > says: If the union of all finite paths, > > 0., 0.0, 0.00, 0.000, ... , > > is in the tree, then the infinite path p(oo) = 0.000.... is in the > > tree too. > > Yes, because the union *is* the infinite path p(oo). > > > I do not know how this is accomplished. > > The axiom of infinity, and from that some simple theorems. May it be. > > > Therefore I leave it to you and > > I say only: "p(oo) is established" by the set of finite paths, i.e., > > p(oo) belongs to the tree which contains all the finite paths. > > And if this is correct, then it is also correct if we have the paths > > 0., 0.0, 0.00, 0.000, ... , and some others like 0.10 or 0.111. Then > > the binary tree contains also the path 0.000.... > > But I never denied that. But I now think I understand what you mean > with that word "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. > Correct me if I am wrong, and if so, please state what you actually *do* > mean with that word (not examples, a clear definition, please). I accept your definition: p(oo) is the union of all finite paths p(n). > Bit with > this definition above, p(oo) is *not* established by: > {p(0)} U {p(1)} U {p(2)} U ... > because the paths are not united, it is the sets containing paths as a > single element that are united. The tree T(oo) contains all pats p(n), hence the tree conains the union. The tree T(oo) contains all finite trees T(n). Do you think if we don't form the union, then T(oo) does not exist? It is established by the presence of all finite trees. Same is valid for the paths. But if you think so, then, please, form the union. Then you have T(oo) and all paths p(oo), q(oo), ... Regards, WM
From: Franziska Neugebauer on 13 Feb 2007 06:44 mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Feb., 17:38, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: >> mueck...(a)rz.fh-augsburg.de wrote: >> > On 12 Feb., 13:59, Franziska Neugebauer <Franziska- >> > Neugeba...(a)neugeb.dnsalias.net> wrote: >> >> mueck...(a)rz.fh-augsburg.de wrote: >> >> > Set theory, however, says: If the union of all finite paths, >> >> > 0., 0.0, 0.00, 0.000, ... , >> >> > is in the tree, then the infinite path p(oo) = 0.000.... is in >> >> > the tree too. >> >> >> No contemporary set theory or graph theory does posit such a >> >> nonsensical claim. It is entirely *your* claim, better known as >> >> > It is easy to recognize from the union of all finite segments of N. >> > Set theory says: If the union of all finite segments, >> > {1,2,3,...,n} is present, >> >> "Present"? Where? > > That question should be answered by those person who claim the > existence. Those who claim (within the framework of some calculus) that this or that exists mathematically do *not* claim that it is physically (i.e. spatio-temporally) located. They also do not use the verb "to be present". This is entirely your wording. > But to make it easy for you: It is claimed that N exists as > the union of its initial segments. Noone who is practicing contemporary set theory de lege artis is using a phrase like "exists as" but simply "exists". >> > then the infinite set N = {1,2,3,...} is present. >> >> Where? I don't know of any contemporary set theory which states that >> entities are "present". > > It is stated that sth "exists". Can something exist without being > present somewhere? If you mean a spatio-temporally locus -> sci.physics If you mean mental representation -> sci.neurosciences >> > Or do you see a difference betwee n the union of all finite >> > segments and N? >> >> First of all I don't see that contemporary set or graph theories >> posit >> >> | Set theory, however, says: If the union of all finite paths, >> | 0., 0.0, 0.00, 0.000, ... , >> | is in the tree, then the infinite path p(oo) = 0.000.... is in the >> | tree too. [(*)] > > Then you deny that N is the union of all of its initial segments? > Or do you deny that the nodes corresponding to the paths 0., 0.0, > 0.00, 0.000, ... , can be mapped on the initial segments of N? > > Perhaps certain mappings are forbidden? Everyone who is practicing contemporary set theory de lege artis would complain about your wording in (*). And justifiably so. F. N. -- xyz
From: Franziska Neugebauer on 13 Feb 2007 07:11
mueckenh(a)rz.fh-augsburg.de wrote: >> >> > Given my fixed coordinate system, there remains no ambiguity at >> >> > all. >> >> >> Then we are no longer talking about binary trees in general. >> >> > I never did so. >> >> I couldn't agree more. > > Why then did you claim the opposite? Did you lie? Cause you insist using mathematically predefined words (trees, paths, union) in a mathematical newsgroup. It is obviously true that you never really write about these mathematical issues. Hence I agreed to your statement "I never did so." F. N. -- xyz |