Cantor Confusion
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From: Newberry on 18 Dec 2006 21:27 Virgil wrote: > In article <1166474763.304897.177520(a)80g2000cwy.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > Virgil wrote: > > > In article <1166451538.861033.303300(a)48g2000cwx.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Newberry schrieb: > > > > > > > > > Sorry that I joined a bit late. > > > > Doesn't matter. > > > > > > > > >Are you saying that (in an infinite > > > > > binary tree) the set of paths is uncountable but the set of edges is > > > > > countable? > > > > > > > > The set of edges is obviously countable by, e.g., > > > > > > > > 1, 2, > > > > 3,4,5,6, > > > > 7,... > > > > > > > > As no path can separate from another one without the existence of two > > > > more edges, the number of edges is an upper bound for the number of > > > > paths. > > > > > > > > > That is only the case in finite trees. > > > It fails miserably in infinite binary trees in which no path has a last > > > node or last edge. > > > > Do you agree that the number of edges is the upper bound of the number > > of paths but disagree that the edges are countable? > > I do not agree that in an infinite binary tree, such as may be > constructed in ZFC, the number of edges is an upper bound for then > number of paths. > It is my contention that the number of paths through every edge is in > that case uncountable. > > > > I guess if the number of edges > number of paths and you accept the > > diagonal argument then it follows that the edges are uncountable. I do > > not know if there is any way to prove directly that the edges are > > countable. > > Actually, it is not too hard to enumerate the edges. > > At the first level, issuing from the root node there are two edges, > which can be numbered from "left" to "right" as 1 and 2. > > At the next level, issuing from child nodes of the root node, there are > 4 edges numberable 3,4,5 and 6. > > At the next level 8, numberable from 7 through 14. > > And so on. > > General rule: at the nth level there are 2^n edges, which can be > numbered from 2^n - 1 to 2*(2^n-1). OK, let's recapitulate. Edges are countable, paths are not countable. The number of edges is not an upper bound for the number of paths. But at any given finite level there are more edges than paths.
From: Newberry on 18 Dec 2006 21:32 Dik T. Winter wrote: > In article <1166474763.304897.177520(a)80g2000cwy.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > ... > > > That is only the case in finite trees. > > > It fails miserably in infinite binary trees in which no path has a last > > > node or last edge. > > > > Do you agree that the number of edges is the upper bound of the number > > of paths but disagree that the edges are countable? > > It is the case in finite trees, but that makes it not true for infinite > trees. So for finte trees there are more edges than paths but for infinite trees there are more paths than edges? > > > I guess if the number of edges > number of paths and you accept the > > diagonal argument then it follows that the edges are uncountable. > > No. You can not apply the diagonal argument to the edges. Nobody said anything about applying the diagonal argument to the edges. > > > I do > > not know if there is any way to prove directly that the edges are > > countable. > > There is. It is easy to assign natural numbers to each edge so that > all edges get assigned different natural numbers, and so there is > an injection from the set of edges to the natural numbers, and so > the set is countable. So for finte trees there are more edges than paths but for infinite trees there are more paths than edges? > -- > 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 18 Dec 2006 22:55 In article <1166495541.239288.43640(a)48g2000cwx.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > Dik T. Winter wrote: > > In article <1166474763.304897.177520(a)80g2000cwy.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > > ... > > > > That is only the case in finite trees. > > > > It fails miserably in infinite binary trees in which no path has a last > > > > node or last edge. > > > > > > Do you agree that the number of edges is the upper bound of the number > > > of paths but disagree that the edges are countable? > > > > It is the case in finite trees, but that makes it not true for infinite > > trees. > > So for finte trees there are more edges than paths but for infinite > trees there are more paths than edges? Indeed. > > > I guess if the number of edges > number of paths and you accept the > > > diagonal argument then it follows that the edges are uncountable. > > > > No. You can not apply the diagonal argument to the edges. > > Nobody said anything about applying the diagonal argument to the edges. Well, you brought the diagonal argument in it, so I thought you wanted to apply it to the edges. Apparently not. So I have no idea where your "it follows" comes from. > > > I do > > > not know if there is any way to prove directly that the edges are > > > countable. > > > > There is. It is easy to assign natural numbers to each edge so that > > all edges get assigned different natural numbers, and so there is > > an injection from the set of edges to the natural numbers, and so > > the set is countable. > > So for finte trees there are more edges than paths but for infinite > trees there are more paths than edges? Right. It is taking conclusions about the infinite from the finite cases when you think it is otherwise. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Han de Bruijn on 19 Dec 2006 03:10 Virgil wrote: > In article <1a591$458667c1$82a1e228$22650(a)news1.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>Virgil wrote: >> >>>In article <1166090594.020341.42340(a)80g2000cwy.googlegroups.com>, >>> mueckenh(a)rz.fh-augsburg.de wrote: >>> >>>>I do not understand why you argue that in >>>>lim{x -> oo} lim{y -> oo} (2x + 3y)/xy = 0 = lim{y -> oo} lim{x -> oo} >>>>(2x + 3y)/xy >>>>interchanging limits is not possible. >>> >>>Nor do I. >>> >>>But for lim{x -> oo} lim{y -> oo} (2x + 3y)/(x + y) = 3 >>>and lim{y -> oo} lim{x -> oo} (2x + 3y)/(x + y) = 2, >>>one cannot exchange the order of the limits without changing the value >>>of the result. >> >>This is highly misleading. > > What is misleading about a true and relevant statement? > Both double limits exist but they have different values. > > The issue is whether such double limits are always reversible, > and the answer, as demonstrated by the example above, is "NO". The issue is that your limits are actually an ill-posed problem. Han de Bruijn
From: mueckenh on 19 Dec 2006 06:10
Dik T. Winter schrieb: > > I do not think that link makes much sense for somebody who does not > > understand the German sentence above. However, I do understand why > > you post that link. How much did publication cost you? My estimate > > is about EUR 1000. > > For those who wonder. Shaker Verlag is nothing more than a printing > company. It was started to aid universities to publish their books in > a nice format and to publish thesises. In addition to the printing it > does something about ISBN's, some publishing and so on. But they appear > to be going into the field of vanity press. Whatever you wish to publish, > they will do it, if you pay what they ask you. If you have a manuscript > of about 150 pages, they will publish with a print run of 200, for about > EUR 1000. Do you report own painful experience, Dik? No. I can recommend Shaker warmly. There is *no* fee. Of course, they would be mad if they published on their own risk a book in German on a topic which is not taught at any German school or university except my own. But they seem to work very economically. All they require from the author is to buy 30 copies of the book on half the selling price. You may calculate that your estimation was far too high. (My costs were somewhat higher because I required hard cover and additional dust cover which do not belong to the regular equipment. But I thought that necessary in order to prevent angry set theorists from tearing it into pieces too easily. A full metal jacket to prevent burning was too expensive, however.) > No review, no editing and no print allowance by the celebrities and high priests of the mathelogical church is required. That is correct and very important. So I am in a better position than, e.g., Galilei was. > Oh, and as a nicety, the author will receive parts of the > sale cost for each book outside the initial print run. It is 10%, but > I do not know whether that is before taxes or after. I would be glad to give these 10 % to any customer who really understands my ideas. Regards, WM |