Cantor Confusion
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From: William Hughes on 24 Jan 2007 08:41 On Jan 24, 6:58 am, mueck...(a)rz.fh-augsburg.de wrote: > On 23 Jan., 13:42, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > There is no path which ends at a node which you can specify. Every path > > > > > of the union tree has maximum length. 0.11 is not a path in the tree of > > > > > three or more levels. > > > > > Since T1 is not a tree, you cannot use a statement about the > > > > properties of of a tree to say something about the properties > > > > of T1. 0.11 is a path in T1. > > > > Whatever T1 may be called. 0.11 is, by definition, not a path in T1.It does not matter what we call T1. It does matter what properties T1 > > has. > > Since T1, the union of all finite trees, is the union of all finite > > paths, and 0.11 is a finite path, > > T1 has the property that 0.11 is in T1. >No. The union of {1} and {1,2} does not contain the maximum sequence > {1}. Paths in trees without leaves are maximum sequences. T1 is a tree > without leaf-nodes. > > > > > I guess you have a different definition for the union of all finite > > trees. > > > Let R be a set of finite trees with the property that: > > there is a (fixed) tree t_D in R, such that: > > if s is in R, then > > s is a subtree of t_D > > > Definition i: Here you snipped both definitions that I gave without noting this fact. Naughty! I will restore this section. Definition i: The union of all finite trees in R is the tree t_D. (Note that using this definition, the union of all finite trees in R is a finite tree.) Defintion i': The union of all finite trees in R is a set of paths, S, where S contains any path that is in a tree in R. (Note that i and i' are almost equivalent.) Now let W be the set of all finite trees. We know that there does not exist a (fixed) t_D in W such that every tree in W (that is every finite tree) is a subtree of t_D. So we cannot use definition i. Instead I use definition i'. By definition i', T1 contains every finite path. If you have a different definition for T1 would you care to share it? > > > The union of all finite treesis the tree which has all nodes and edges which are in at least one > finite tee. > As I noted, you have a different definition for the union of all finite trees. As before, let the set of all finite trees be W. Let P1 be the set of all finite paths, that is if p is an element of P1, then there is a tree t in W, such that p is a path in t. We will adopt your definiton of the union of all finite trees. Under this definition the union of all finite trees, T1, is the infinite tree T2. However, the set of paths in T1 is not the union of paths in P1. Each path in T1 is the limit of a sequence of paths from P1. Union and limit are two different things. Let us call the union of all finite trees (your definition) T1. P1 contains every finite path, so P1 contains 0.11, T1 does not contain 0.11, so P1 is not the same as T1. <snip> > Summary > > 1) Every complete infinite binary tree T (containing all nodes and > edges) contains all paths. This is true using your definition of a tree. We will use this definition. However, you should be aware that there are other defintions. > 2) The union tree T(oo) of all finite trees is well defined (as I have > shown elsewhere) and yields the complete infinite binary tree > containing all nodes and edges: T = T(oo). Again this is true using your defintion of union. We will use this definition. However, you should be aware that there are other defintions. > 3) The union of all finite trees includes the union of all nodes and, > with it, the union of all such subsets which are paths (because every > path is a well defined subset of the set of nodes if the structure of > the tree is well defined). With the proviso that every path in T1 cannot end at a node if this path can be continued (i.e. every path in T1 is infinite) > 4) The set of paths in T(oo) is a subset of the countable set of finite > sets of all paths in the finite trees. No. Each path in T(oo) is the limit of a (potentially) infintie sequence. So each path in T(oo) corresponds to a (potentially) infinite set of paths in the finite trees. > 5) A countable union of countable sets is a countable set (according to > ZF with AC). > ==> The set of all path is countable. (==> The real numbers are > countable.) No, the set of (potentially) infinite sets of paths is not countable. > > Going on, we can say: > > 6) T(oo) = T contains only finite paths. No. T does not contain a finite path. > 7) T(oo) = T contains all paths including all infinite paths. No. T contains only infinite paths. - William Hughes
From: imaginatorium on 24 Jan 2007 08:47 Andy Smith wrote: > In message <ep7kn4$79j$2(a)mailhub227.itcs.purdue.edu>, Dave Seaman > <dseaman(a)no.such.host> writes > >On Wed, 24 Jan 2007 09:48:01 GMT, Andy Smith wrote: > >> In message <9b2er2p5taadlea28n6fb3g4nuvmqeijhs(a)4ax.com>, G. Frege > >><nomail(a)invalid.?.invalid> writes > >>>On Tue, 23 Jan 2007 23:34:03 GMT, Andy Smith > >>><Andy(a)phoenixsystems.co.uk> wrote: <snip> > If you have a transcendental, you need to specify an infinite number of > bits to distinguish it from the set of all alternative transcendentals. > You specify reals as a Cauchy sequence, which unambiguously points > towards the point, but the point itself needs an infinite number of bit > positions. But you can't label an infinite number of bit positions - you > need to have all of the bits as a completed set to define the real - and > that is not a finite number. What do you mean by a "completed set"? Is an unending sequence such as the following a "completed sequence"? 1, 11, 111, 1111, 11111, 111111, 1111111, ... If "completed" means "having an end", then this is _not_ "completed", because it continues without end. But if "completed" means "complete" - that is, that there is nothing "missing" - then the sequence above includes every two-ended string of 1s, even though there are an unending number of them. Do you disagree? You do need to S L O W D O W N; too much of your recent posting is over-excitable, verging on babble, and it is babble that surely starts the slippery slope to crankhood. You seem to be claiming that it matters that "all integers are finite" when considering the representation of sqrt(2). Well, it would make no difference if all integers were less than 57, since you only need to use 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 to represent it, just so long as you don't might having an unending sequence of them. Brian Chandler http://imaginatorium.org > > > -- > Andy Smith
From: G. Frege on 24 Jan 2007 08:52 On 24 Jan 2007 05:47:10 -0800, imaginatorium(a)despammed.com wrote: Ah, I've just spotted another one of Andy Smith's misconceptions: >> >> But you can't label an infinite number of bit positions [...] >> But exactly that is WRONG. We can label them with the natural numbers 1, 2, 3, ... F. -- E-mail: info<at>simple-line<dot>de
From: Andy Smith on 24 Jan 2007 09:03 In message <1169646430.885558.7060(a)m58g2000cwm.googlegroups.com>, imaginatorium(a)despammed.com writes > >Andy Smith wrote: >> In message <ep7kn4$79j$2(a)mailhub227.itcs.purdue.edu>, Dave Seaman >> <dseaman(a)no.such.host> writes >> >On Wed, 24 Jan 2007 09:48:01 GMT, Andy Smith wrote: >> >> In message <9b2er2p5taadlea28n6fb3g4nuvmqeijhs(a)4ax.com>, G. Frege >> >><nomail(a)invalid.?.invalid> writes >> >>>On Tue, 23 Jan 2007 23:34:03 GMT, Andy Smith >> >>><Andy(a)phoenixsystems.co.uk> wrote: > ><snip> > >> If you have a transcendental, you need to specify an infinite number of >> bits to distinguish it from the set of all alternative transcendentals. >> You specify reals as a Cauchy sequence, which unambiguously points >> towards the point, but the point itself needs an infinite number of bit >> positions. But you can't label an infinite number of bit positions - you >> need to have all of the bits as a completed set to define the real - and >> that is not a finite number. > >What do you mean by a "completed set"? Is an unending sequence such as >the following a "completed sequence"? > >1, 11, 111, 1111, 11111, 111111, 1111111, ... > >If "completed" means "having an end", then this is _not_ "completed", >because it continues without end. But if "completed" means "complete" - >that is, that there is nothing "missing" - then the sequence above >includes every two-ended string of 1s, even though there are an >unending number of them. Do you disagree? Um, well I thought there was a distinction between the sequence 1,2,3,4,n,.. and {1,2,3,4,n,..}. All the terms in the sequence are finite, but the set of all of them is infinite. And that was what I was trying to say, was that a real point is not the sequence b0,b0b1,b0b1b2,.. but determined by {b0,b1,b2,...} 9and that actually was the whole point about indexing the reals, which frankly I wish I hadn't mentioned by now). > >You do need to S L O W D O W N; too much of your recent posting is >over-excitable, verging on babble, and it is babble that surely starts >the slippery slope to crankhood. OK, fair enough. Not knowledgeable enough to be a crank ... people keep posting things back.... -- Andy Smith
From: Dik T. Winter on 24 Jan 2007 09:51
In article <1169642941.554024.145640(a)k78g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 23 Jan., 19:27, Dave Seaman <dsea...(a)no.such.host> wrote: > > On Tue, 23 Jan 2007 17:15:52 GMT, Andy Smith wrote: > > > By definition, a set is "finite" if it has the size of some natural > > number. If a set isn't finite, then it's called "infinite". > > If a colour is not red, then it is called green by set theorists. Wrong. not-red. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |