Cantor Confusion
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From: Dik T. Winter on 6 Jan 2007 20:44 In article <1168107244.731296.9350(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > Name a level or a node or and edge which are in one of the trees only. > > > > Why should I do that? > > In order to prove that there is evidence for at least one path being > not in both trees. But for that I do not need to mention a node or edge, I only need to mention a path. > > I can name a *path* that is in the complete infinite > > tree but not in the union of finite trees. > > Without different nodes that is an empty assertion. Well, I have shown it by proof. So it is not an empty assertion. > > How many times do I have to explain that to you? Use the definition of > > the union of an infinite collection of sets. If an element is in one > > of the sets from the infinite collection it is in the union, if it is > > in none of the individual sets it is also not in the union. > > Do you claim that the union of all finite trees is not an infinite > tree? It is. > Why do you deny the same for the paths? The union of the sets of paths is an infinite set of paths. But there is no infinite element in there. > The infinite paths in the union of trees are the unions of the > respective finite paths. So much should be clear to every reader. What would be clear to every reader is that you are confusing issues. Unions are about sets. A tree has a set of nodes, a set of edges and a set of paths. When we do a union of trees we can do a union of those three sets for each tree. Unless you define union completely different (and if so, pray give a definition) we can not use unions of elements of those sets. So you can not take the union of paths but only the union of sets of paths. On the other hand, once you define your union such that we also must consider unions of individual paths, I also want to consider unions of individual edges. Can I? If not, why not? > What > not has been clear, until most recently, is that these unions cannot be > more abundant than the finite paths. The unions can at most be as > abundant as the finite paths. That is why you try to come up with your > fairy tales of different path-systems in identical trees. The unions are about sets of paths, not about paths. > > You were talking about the union of sets of finite paths. Asking for > > an infinite path in that union is the same as asking for an infinite > > number in the (particular) union of sets of finite numbers. > > I do not ask for an infinite path in that union. I claim that the union > constitutes infinite paths. (That is obvious.) And I claim that these > infinite paths cannot be more than the elements of the union. (That is > obvious too.) In that case you have not properly defined what you *mean* with the union of finite trees. Above I gave a proper definition for which my conclusion holds. Now give your definition. -- 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 6 Jan 2007 21:31 In article <1168107683.211353.227740(a)q40g2000cwq.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > Infinite node numbers are not in any of the finite trees, if you cut > > > them at a certain level (contrary to my initial definition with > > > sequences of 111... and 000... ). Does the union of such "cut-off" > > > trees not contain infinitely many nodes? > > > > Yes, but not infinite node numbers. > > What is the difference, please? The set N contains infinitely many naturals but no infinite natural. > Is there a tree with more nodes than the union of all finite trees has? Yes. See J. H. Conway: "On Numbers and Games". His number system is based *only* on trees. > > > You said that in the union of all finite trees there is no infinite > > > path. > > > Would you say that in the union of all finite trees there is a path > > > which contains infinitely many nodes? > > > > No. > > After which node does the first non-infinite = finite path of the union > end? All paths in the union terminate at some finite node. Or (with another definition) continue indefinitely by either all 0's or all 1's. So for all paths in the union there is a finite point after which follows a continuous stream of either 0 or 1. As this is not true for the path of 1/3, it is not in the union. > > Because in none of the finite trees there is a path that contains > > infinitely many nodes. So the two questions following are irrelevant. > > None of the sequences {1,2,3,...,n} contains infinitely many numbers. > Nevertheless, you assert that the union of all of them contains > infinitely many numbers. Right. But that is different. In none of the sets of paths from finite trees there is an infinite path. The union of those sets gives an infinite set of paths, but not an infinite path. > > > How can an infinite set N get into the union of all finite initial > > > segments? > > > > It *is* the union. It does not contain any infinite element because none > > of the constituents contains an infinite element. And if you think that > > N elementof N, you are wrong. > > The same is valid for the infinite paths in the complete tree. The > complete tree is the union of all finite trees. The infinite path is > the union of all respective finite paths. But in that case you have to properly define what you are doing. So pray tell, how do you *define* the union of trees. Because you are talking about the cardinality of the sets of paths, I was thinking that you were talking about sets of paths. > > You are dense. The well-ordering is given under the conditions of the > > axiom V=L. > > THEN GIVE IT PLEASE, HERE AND NOW ! Simply reproduce it. I have not looked into it, but the axiom V=L immediately gives a well-ordering, not only of R. > > If the axiom is false the well-ordering is not given. > > How could it disappear after having existed? Who would force us to > forget it, if it had been given? Did you look at that axiom? No, I think not. > > > (If you really have two parallels, then you do not need the axiom that > > > parallels exist.) > > > > How do you prove they are parallel? > > By measuring their distance in a plane. How large is that plane? Stated in Euclidean coordinates? > > Note moreover that Euclid postulated > > that there was only a single parallel. So in a finitistic world Euclid's > > postulate is false. There is more than a single parallel. > > False. Oh. Strange enough one the models of hyperbolic geometry is a finitistic world (a world bounded by an ellipse). > Consider your screen. > > _________________ > > x > > Draw more than one parallel to the line through the centre of the x. Terminate the line at the right hand slightly above the horizontal line through x. > > > I have not to allow for anything. Every path which is possible does > > > exist as soon as all its edges exist. > > > > No, you are talking about the union of sets of paths. That union is > > different from the complete set of paths, as is easily shown. > > What counts is only: There cannot be more unions than elements. Eh? As I understand it, there is only one union. But, what is the relevance? > > In that case do *not* talk about unions of sets of paths. > > I have a union of all finite trees. The result is an infinite tree > (which need not be an element of the union). But you are talking about the union of sets of paths. What is the relevance? > > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, > > home: bovenover 215, 1025 jn amsterdam, nederland > > Which is the postal address by which you would like to receive my book? > Please write capitals where appropriate. You could use either, but the postal authorities here like it when you use capitals in the postal code, but that is not really necessary. As this is all slightly work related, I would prefer the first: > > dik t. winter, cwi, kruislaan 413, 1098 SJ amsterdam, nederland that would find its way to my desk pretty well (insert new-lines were I use a comma). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Andy Smith on 7 Jan 2007 18:48 A slightly different question. A real point, such as pi or e has a genuine existance independent of its e.g. binary representation. Also integers e.g. 0 = 0.00... does not need to be defined as the limit point between 1/2^n and -1/2^n. What about .11111.... ? That has a definite existance on th line - can one define that as the point that has no zeroes in its binary representation? If so then one can argue that .1111... is different from 1.0000... and that there are no rationals between these two pints (because any rational must have a terminating string of 0000... or a repeating string which is other than all ones?
From: Dave Seaman on 8 Jan 2007 08:29 On Mon, 08 Jan 2007 04:48:49 EST, Andy Smith wrote: > A slightly different question. > A real point, such as pi or e has a genuine existance > independent of its e.g. binary representation. Also integers > e.g. 0 = 0.00... does not need to be defined as the limit point > between 1/2^n and -1/2^n. You might look at <http://www.cs.uwaterloo.ca/~alopez-o/math-faq/node11.html#SECTION00320000000000000000>, concerning the construction of the number systems. > What about .11111.... ? That has a definite existance on > th line - can one define that as the point that has no > zeroes in its binary representation? Oh? What if I write it as 0.111...? > If so then one can argue that .1111... is different from > 1.0000... and that there are no rationals between these > two pints (because any rational must have a terminating > string of 0000... or a repeating string which is other > than all ones? A number is not a digit string. The fact that there are no rationals between two reals constitutes proof that those reals are identical. See the reference above for an explanation of Dedekind cuts. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Andy Smith on 7 Jan 2007 23:44
Ta. |