Cantor Confusion
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From: Gc on 18 Dec 2006 16:55 mueckenh(a)rz.fh-augsburg.de kirjoitti: > 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. The edges and paths are there completely from the beginning. Every two paths separete on some finite level edge, but only when you got the whole countably infinite path (the union of it`s all finite subpaths starting from the beginning) all infinite long pathes separate from each other. (In what you are doing you don`t have to care about finite pathes) > Regards, WM
From: Virgil on 18 Dec 2006 17:06 In article <1166478692.220101.200880(a)f1g2000cwa.googlegroups.com>, "Jonathan Hoyle" <jonhoyle(a)mac.com> wrote: > > WM wants to divide something into a countably infinite number of equal > > shares. > > That would be a neat, considering that it has been proven in Measure > Theory than an interval (of finite but positive length) cannot be > partitioned in countably infinitely many intervals each of the same > size. > > > If s is the size of one share then WM wants to have s + s + s ...= 1. > > > > What is the numerical value of s, WM? > > > > Even in ZFC and NBG, we know better than that. > > Yup, what you have written is essentially a sketch of the proof. > > Good luck trying to wrangle out of this one, WM. WM usually ignores what he cannot find a way to wriggle out of.
From: Lester Zick on 18 Dec 2006 17:39 On 18 Dec 2006 13:51:32 -0800, "Jonathan Hoyle" <jonhoyle(a)mac.com> wrote: >> WM wants to divide something into a countably infinite number of equal >> shares. > >That would be a neat, considering that it has been proven in Measure >Theory than an interval (of finite but positive length) cannot be >partitioned in countably infinitely many intervals each of the same >size. > >> If s is the size of one share then WM wants to have s + s + s ...= 1. >> >> What is the numerical value of s, WM? >> >> Even in ZFC and NBG, we know better than that. > >Yup, what you have written is essentially a sketch of the proof. > >Good luck trying to wrangle out of this one, WM. Or the rest of use could just wait for you to wrangle out of what you claim to be true. ~v~~
From: Dik T. Winter on 18 Dec 2006 18:30 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. > 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. > 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. -- 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 R Tribble on 18 Dec 2006 19:32
William Hughes schrieb: >> Do you now claim the natural numbers do not exist? > muecken h wrote: >> More than enoug do exist. (More than we will ever need could be brought >> to existence.) > David R Tribble schrieb: >> How is a number "brought into existence"? >> >> I'm thinking of a natural number, call it q. It has the distinct >> property that it is not equal to any number you (MH) can think of. >> In fact, it is larger than any natural number that has been "brought >> into existence yet". So what is q? > muecken h wrote: > Not yet a number, unless you can specify it such that one (at least > you) can decide whether q < n or q = n or q > n for any natural number > n given to you. Only saying that it is not equal to any natural number > given is not enough. I did not say that q is not equal to any given natural number. I said that q is not equal to any natural number that has been "brought into existence". Trichotomy is well defined for q. For any natural n given, if n has been "brought into existence", then q > n. Since you believe that only those numbers that have been "brought into existence" exist, you should readily grasp the definition of q. Assuming that the largest natural that has been "brought into existence" is m, then we could easily define q as m+1. But then we have a contradiction: since m "exists", so does q, but this contradicts your statement that q is "not yet a number", i.e., q does not exist yet. m exists, but m+1 does not. So one of your assumptions about the "existing" naturals must be wrong. |