Cantor Confusion
in [Math]
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From: Virgil on 19 Mar 2007 16:15 In article <1174328592.469787.62400(a)y66g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 19 Mrz., 16:18, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174253026.516861.271...(a)d57g2000hsg.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 18 Mrz., 02:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > The function Card(paths) over n cannot jump from finite to > > > > > uncountable. > > > > > > > > It does not when you consider finite paths. There is another jump, > > > > from > > > > finite paths to infinite paths. > > > > > > Where does that happen? > > > > When you consider the union of all finite trees. > > The union of all finite numbers gives an infinite number? If one is dealing with ordinal numbers, yes. If one is dealing with cardinal numbers yes. If one is dealing only with natural numbers, no. If one is to have a union of all finite cardinals or of all finite ordinals, as one must in any system with a standard axiom of union, then the result cannot be finite whatever one wants to call it otherwise. > > Yes. But the path-lengths are natural numbers, and there is no > infinite natural number. The path lengths are also cardinal number and ordinal numbers, and there are infinite cardinals and infinite ordinals. > The axiom of infinity grants the existence of all natural numbers, but > not the existence of an an infinite natural number. But it REQUIRES existence of infinite cardinals and ordinals, which is all that is needed. > > You try to get an infinite element out of a union of finite elements. > That does not work. It does not work for WM because he asssumes that it does not work, but there is no reason in ZF or NBG why not and reasons in them why it has to work. > > Further it is not sufficient, because the countable union of countable > sets is countable. But the set of all subset of a countable set is not, and that is where the uncountability of the set of paths in the CIBT arises. > The set of paths of every finite tree is countable. it is even finite. > The set of finite trees is countable. Quite so. > > Can we agree upon the following facts? > (1) The infinite paths 0.000... and 0.111... are completely contained > in the complete tree T(oo), i.e., every node belonging to one of > these > paths is a node belonging to the tree T(oo). > (2) The set of nodes of the tree T(oo) is countable. In particular, > the number of nodes belonging to one level L(n) is countable for > every > n. > (3) There is no node in the tree (including every node of the paths > 0.000... and 0.111...) which belongs to a level where uncountably > many paths arrive at or start off from or cross through. Since (3) is false, I, for one, will not agree to it. In any CIBT there are uncountably many paths through any node whatsoever. One can always take any given node and all its progeny as a new CIBT isomorphic to the original so there are as many paths through the given node as through the root of the original tree.
From: mueckenh on 20 Mar 2007 02:40 On 19 Mrz., 20:31, Virgil <vir...(a)comcast.net> wrote: > In article <1174292997.710730.80...(a)n76g2000hsh.googlegroups.com>, Can we agree upon the following facts? > (1) The infinite paths 0.000... and 0.111... are completely contained in the complete tree T(oo), i.e., every node belonging to one of these paths is a node belonging to the tree T(oo). (2) The set of nodes of the tree T(oo) is countable. In particular, the number of nodes belonging to one level L(n) is countable for every n. (3) There is no node in the tree (including every node of the paths 0.000... and 0.111...) which belongs to a level where uncountably many paths have been separated or will separate in the next level. If you agree to these three points, then you may continue your dreams of whatever you like. I will not further disturb you. And I am sure every sensible person will share my attitude. If you don't agree, please state which of the three points is/are wrong in your opinion. Regards, WM
From: Virgil on 20 Mar 2007 04:30 In article <1174372833.140688.156790(a)p15g2000hsd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 19 Mrz., 20:31, Virgil <vir...(a)comcast.net> wrote: > > In article <1174292997.710730.80...(a)n76g2000hsh.googlegroups.com>, > > Can we agree upon the following facts? > > > (1) The infinite paths 0.000... and 0.111... are completely contained > in the complete tree T(oo), i.e., every node belonging to one of these > paths is a node belonging to the tree T(oo). > > (2) The set of nodes of the tree T(oo) is countable. In particular, > the number of nodes belonging to one level L(n) is countable for every > n. > > (3) There is no node in the tree (including every node of the paths > 0.000... and 0.111...) which belongs to a level where uncountably many > paths have been separated or will separate in the next level. > > If you agree to these three points, then you may continue your dreams > of whatever you like. I will not further disturb you. And I am sure > every sensible person will share my attitude. If you don't agree, > please state which of the three points is/are wrong in your opinion. > > Regards, WM I fail to see that any of the points WM presents has any relevance whatsoever to the issue of whether the number of paths in a CIBT is countable or uncountable. Does WM agree that EVERY endless sequence of 0's and 1's following that header of 0 is an infinite path belonging to the tree T(oo)? If not, which uncountable subset of of those uncountably many endless sequences does WM argue have been left out?
From: mueckenh on 20 Mar 2007 09:12 On 20 Mrz., 09:30, Virgil <vir...(a)comcast.net> wrote: > In article <1174372833.140688.156...(a)p15g2000hsd.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 19 Mrz., 20:31, Virgil <vir...(a)comcast.net> wrote: > > > In article <1174292997.710730.80...(a)n76g2000hsh.googlegroups.com>, > > > Can we agree upon the following facts? > > > (1) The infinite paths 0.000... and 0.111... are completely contained > > in the complete tree T(oo), i.e., every node belonging to one of these > > paths is a node belonging to the tree T(oo). > > > (2) The set of nodes of the tree T(oo) is countable. In particular, > > the number of nodes belonging to one level L(n) is countable for every > > n. > > > (3) There is no node in the tree (including every node of the paths > > 0.000... and 0.111...) which belongs to a level where uncountably many > > paths have been separated or will separate in the next level. > I fail to see that any of the points WM presents has any relevance > whatsoever to the issue of whether the number of paths in a CIBT is > countable or uncountable. I am not so much interested in what you may feel relevant to prove your opinion concerning the multitude of paths. I am only interested in your boldness to refute the points 1, 2, or 3. > > Does WM agree that EVERY endless sequence of 0's and 1's following that > header of 0 is an infinite path belonging to the tree T(oo)? Every existing path belongs to the complete tree. And at no point of the tree more than countably many infinite paths have separated. Regards, WM
From: Dik T. Winter on 20 Mar 2007 12:07
In article <1174328592.469787.62400(a)y66g2000hsf.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 19 Mrz., 16:18, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174253026.516861.271...(a)d57g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > On 18 Mrz., 02:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > The function Card(paths) over n cannot jump from finite to > > > > > uncountable. > > > > > > > > It does not when you consider finite paths. There is another jump, > > > > from finite paths to infinite paths. > > > > > > Where does that happen? > > > > When you consider the union of all finite trees. > > The union of all finite numbers gives an infinite number? An infinite set. If you want to call that set a number, go ahead. But you are confusing yourself using the von Neumann model. In that model, indeed, the union of all finite integers >= 0 is omega, an infinite number. > > > How can they enter a union of finite trees U(T(n)) or a union of all > > > paths of all finite trees? > > > > By the definition of that union. Let us consider a node indexed by a pair > > of numbers (i, j), where i indicates the level and j indicates the node > > at that level. Let's have i go from 1 and j from 1 to 2^(i-1). The tree > > T(n) is the set of all nodes with i <= n. A terminating path in the tree > > T(n) is the set of nodes {(k, l)} where k ranges from 1 to n and each k > > occurs only once, and l is within the bounds with some additional > > requirements: if (k1, l1) and (k2, l2) are in a path, and k2 = k1 + 1 > > then l2 is either l1 * 2 - 1 or l1 * 2. This is a complete definition > > of finite paths and finite trees. U(T(n)) is by definition the set > > of pairs that contains all pairs contained in all of the T(n). So it > > is the infinite set {(i, j) | i in N and 1 < j < 2^(i - 1)}. The > > existence of this set is granted by the axiom of infinity. > > Yes. But the path-lengths are natural numbers, and there is no > infinite natural number. The path-length of a non-terminating path is *not* a natural number. > The axiom of infinity grants the existence of all natural numbers, but > not the existence of an an infinite natural number. Indeed. > > Now an > > infinite paths is, for instance, the set {(i, 1) | i in N}, which is > > a subset of U(T(n)). > > You try to get an infinite element out of a union of finite elements. > That does not work. Where is the infinite element? > Further it is not sufficient, because the countable union of countable > sets is countable. The set of paths of every finite tree is countable. > The set of finite trees is countable. Yes. But the *set* of infinite paths is *not* in a countable union of countable sets. > Can we agree upon the following facts? > (1) The infinite paths 0.000... and 0.111... are completely contained > in the complete tree T(oo), i.e., every node belonging to one of > these paths is a node belonging to the tree T(oo). Yes. > (2) The set of nodes of the tree T(oo) is countable. In particular, > the number of nodes belonging to one level L(n) is countable for > every n. Yes. > (3) There is no node in the tree (including every node of the paths > 0.000... and 0.111...) which belongs to a level where uncountably > many paths arrive at or start off from or cross through. No. From the root node of the tree there start uncountably many paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |