Cantor Confusion
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From: Andy Smith on 18 Jan 2007 02:42 I > > > > The union of all finite binary trees contains all levels which can be > > > > enumerated by natural numbers: > > > > > > > > 0 0. > > > > / \ > > > > 1 0 1 > > > > / \ / \ > > > > 2 0 1 0 1 > > > > ............................... > > > > Out of interest, aren't the set of all numbers defined by the union of all paths through a finite binary tree with N levels just all the numbers addressed by the first N bits? If so, why do you bother with the tree construction - does it have some special significance? -- Andy Smith
From: Virgil on 18 Jan 2007 02:57 In article <hyIrji07TyrFFwxz(a)phoenixsystems.demon.co.uk>, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > I > > > > > The union of all finite binary trees contains all levels which can be > > > > > enumerated by natural numbers: > > > > > > > > > > 0 0. > > > > > / \ > > > > > 1 0 1 > > > > > / \ / \ > > > > > 2 0 1 0 1 > > > > > ............................... > > > > > > > Out of interest, aren't the set of all numbers defined by the union of > all paths through a finite binary tree with N levels just all the > numbers addressed by the first N bits? If so, why do you bother with > the tree construction - does it have some special significance? WM tried, and failed, to use an analogy with binary trees to argue that there were only countably many real numbers (only countably many infinite paths in an infinite binary tree or only countably many infinite binary strings). But once WM allows an infinite binary tree at all, as he has done,he automatically allows the inevitability of uncountability.
From: Han de Bruijn on 18 Jan 2007 03:01 David Marcus wrote: > Han de Bruijn wrote: > >>Set theory is quite useful as a limited, relatively unimportant, part of >>mathematics. > > You sound like Charlie-Boo. Are you related? Define "related". Family? No. Han de Bruijn
From: Han de Bruijn on 18 Jan 2007 03:09 MoeBlee wrote: > [ ... ] And meanwhile, I always have my ears open for ideas not in > the mainstream, including intuitionism, finitism, ultra-finitism, > non-standard logics, paraconsistent logic, as well as the many > philosophical approaches such as constructivism, realism, > structuralism, fictionalism, and even as farflung as certain mystical > views of logic and mathematics. (But that doesn't entail that I don't > also exercise my prerogative to skewer postings by cranks.) How about materialism, engineering, applications, numerical analysis, computer graphics? You have't seen anything of the latter kind, huh? > I haven't found your questions to be stupid. But I do think your > questions would benefit by being put in the light of mathematical > definitions and some familiarity with certain basics of the subject. Especially the "light" is what bothers some of us .. Han de Bruijn
From: Franziska Neugebauer on 18 Jan 2007 03:24
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > Franziska Neugebauer schrieb: >> >> > What I defined as the union of finite trees is the union in the >> >> > sense of set theory. >> >> >> >> I remember that you refused to use usual graph theoretical >> >> approach where a tree is a directed graph. A graph is an ordered >> >> pair of a set of nodes _and_ a set of edges. Please show that in >> >> general a union of trees (ordered pairs of sets of nodes and sets >> >> of edges) is a tree. >> > >> > I do not assert that this is so in general. Therefore I will not >> > prove it. >> >> Then please don't call it "union". >> >> > What I said is that the union of the trees with n levels and the >> > tree with m >= n levels is the tree with m levels. >> >> So your "union" is merely a kind of "maximum". For there is no max. >> element n e N the "union of all finite trees" is undefined in the >> first place. > > The union of {1} and {1,2} is {1,2}. This holds for initial segments > of N as well as for finite trees, cut or as Trauerweide. { 1 } U { 1, 2 } = { 1, 2 } is simply true, but does not "hold". [...] >> >> > The Equilateral Infinite Triangle (EIT) contains all lines which >> >> > can be enumerated by natural numbers: >> >> > >> >> > 1 0.1 >> >> > 2 0.11 >> >> > 3 0.111 >> >> > ........... >> >> >> >> This is simply a(n infinite) list (sequence) of unary >> >> representations. >> > >> > The diagonal has an infinite number of digits. >> >> > (Otherwise, Cantor's diagonal proof would fail.) >> > >> >> 2. Cantor's proof does not "fail" in that case cause 0.1... is not >> >> _in_ the "EIT"-list (i.e. there is no pair (n, 0.111...) e EIT, >> >> n e N). >> > >> > I said *otherwise* Cantor's proof would fail. "Otherwise" means: If >> > there is no infinite diagonal. >> >> Your claimed (please take a look into the restored context): >> [WM:] Not necessary. Take care of where you paste. ,----[ context restored ] | > In addition it contains the complete diagonal 0.111... [(1)] | | Since Sigma^+ does not contain an element "1..." your "EIT" sequence | does not contain it as element either. So your claim is _wrong. | | > (Otherwise, Cantor's diagonal proof would fail.) [(2)] | | 1. Non sequitur. | 2. Cantor's proof does not "fail" in that case cause 0.1... is not | in the "EIT"-list (i.e. there is no pair (n, 0.111...) e EIT, | n e N). `---- >> i. (1) is true. There is an infinite diagonal (by definition). ^should read: false (your claim is wrong) > But there is no infinite path by definition? In a tree of finite hight: no. >> > If there is an infinite diagonal, however, then there is no reason >> > for the path 0.111... of the union of finite trees to be finite. >> >> The question is whether 0.[01] is in the "union of all finite paths". > > If it exists, It exists and it is 0.[01]. > then it must be in the union, 1. Non sequitur. 2. Wrong. > because there is nothing remaining outside of the union of all finite > trees Define "union of all finite trees". We are currently talking about finite paths. > and all finite paths which could extend any of its paths. > > The union of all finite paths is an infinite path, ,----[ http://en.wikipedia.org/wiki/Path_%28graph_theory%29 ] | In graph theory, a path in a graph is a sequence of vertices such that | from each of its vertices there is an edge to the next vertex in the | sequence. `---- Don't see that even a union of two paths is a path. > because for every end node n of a path, there is a path crossing n. Please elaborate. F. N. -- xyz |