Cantor Confusion
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From: Virgil on 14 Jan 2007 14:58 In article <1168779496.105249.289900(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > >> >> > This and your further questions are answerd by the following > > >> >> > texts. > > >> >> > > >> >> I did not pose any questions. I have informed you about the fact > > >> >> that there is no time and hence no temporal process _in_ math. > > >> > > > >> > Be informed then that mathematics is in time. > > >> > > >> From that it does not follow that there is a notion of time in > > >> mathematics. > > > > > > A failure > > > > Thanks again for confirming that there is no time im mathematics _now_. > > There is no time in what you may erronesously understand by > mathematics. So a non-mathematician , actually an anti-mathematician, now claims authority to decide what is and is not correct in mathematics. Such chutzpah! > > I do not discuss _with_ or _instead_ _of_ or _against_ Cantor or Brouwer > > both of which are dead. If _you_ want to argue _you_ must rephrase them > > in your own words. > > I should create new words? Why that, if there are good old well-known > words? Parroting the words of others does not give evidence of understanding. And inability to express one's ideas in one's own words is evidence of lack of understanding.
From: Virgil on 14 Jan 2007 15:08 In article <1168779834.276761.153130(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Recall Cantor's diagonal. It consists only of finite initial segments. Hardly. If the diagonal construction did not produce an endless string of digits, it would not produce a number known not to be in the list, so finite initial segments don't cut it. > Either it does not exist - or there are infinite paths in the union of > finite trees. False dichotomy. > > >> 2. To "unary represent" every real in [0, 1] > > > > > > You cannot "unary represent" every real in [0, 1]. > > > > It's rather unimportant whether I "_can_" do so. > > It is important that it is in principle impossible. Not if one can do it in binary or other bases. Unary representations are obviously limited to countable sets, which binary, decimal and other bases are not. > > "mutable sets" is a notifiable disease. > > "actually infinite sets" is a disease from which maths will recover. Only by dying entirely. WM would prefer to kill the "patient", rather that allow him to survive with his present "actually infinite sets".
From: Virgil on 14 Jan 2007 15:14 In article <1168780521.340713.316740(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 0.111... and its initial segments are *representations* (according to > you, they cannot be numbers). They can represent natural numbers as > well as certain real numbers. Obviously both sets of numbers are > isomorphic. What does "isomorphic" mean here? The sets are bijectable with each other in an obvious way, but as they have virtually no arithmetical properties in common, the word "isomorphic" is highly misleading.
From: Franziska Neugebauer on 14 Jan 2007 15:15 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > Franziska Neugebauer schrieb: >> [...] >> >> > The set of paths is countable. >> >> >> >> 1. The set of all rooted _finite_ paths in an infinite tree is >> >> countable. The union of all finite paths is exactly this set of >> >> all rooted finite paths. >> > >> > Yes, of infinitely many paths, each of which has infinitely many >> > nodes. >> >> _finitely_ many nodes. Your trees to be united do only have finitely >> many nodes. There are only finitely deep trees. Hence the union does >> not contain any infinite tree and hence not any infinite path. >> >> Recall: Do get it out you must have put it in. > > Recall Cantor's diagonal. It consists only of finite initial segments. > Either it does not exist - or there are infinite paths in the union of > finite trees. False dilemma. It "consists" of only finite initial segments (not finitely many therof, of course) _and_ it does exist _and_ there are only _finite_ paths _as_ _members_ in the union of all finite trees. Again: To get it out you must have put it in. Since noone put infinitely deep trees into the union of trees it does not have such paths as members. >> [...] >> >> 2. To "unary represent" every real in [0, 1] >> > >> > You cannot "unary represent" every real in [0, 1]. >> >> It's rather unimportant whether I "_can_" do so. > > It is important that it is in principle impossible. A mathematical argument is not that "I 'cannot' do this or that" but "that no unary representation of every real in [0, 1]" exists. ,----[ <45a641d2$0$97241$892e7fe2(a)authen.yellow.readfreenews.net> ] | 2. To "unary represent" every real in [0, 1] you also need infinite | paths. (i.e. 1/3 is not represented as finite path and hence not by | any path in your union). `---- >> But for the sake of argument let us assume, you "could" *not* >> represent every real in [0, 1] in terms of representations. Then your >> proof is meaningless already before writing it down. > > Obviuosly yoou intermingle "unary" and "binary". Indeed I mean "binary represenation". To resume: If your set of binary representations does not contain the representation of 1/3 your proof is meaningless already before starting to write it down. Without notice you paste from a different post. This is considered bad practice. It is originally from <45a6c9da$0$97224$892e7fe2(a)authen.yellow.readfreenews.net>. >> "mutable sets" is a notifiable disease. > > "actually infinite sets" is a disease from which maths will recover. Imaginary disease. F. N. -- xyz
From: Franziska Neugebauer on 14 Jan 2007 15:21
mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > >> Which ones of those specifically "rooted finite paths" does WM claim >> contain infinitely many nodes? > > Which segment {1,2,3,...,n} of N contains infinitely many elements? None of the segments {0, ..., n} n e N has infinite cardinality. But noone claims that there is one. Or do you? > Is N the union of all [finite] initial segments? It is. Obviously this union is not a member of itself. >> > Why do you call the set {{1}, {1,2}, {1,2,3}, ...} which is >> > equivalent and even isomorphic to the set{1, 2, 3, ...} = N finite >> > (finite path)? >> >> Each of the sets {1}, {1,2}, {1,2,3}, etc., is both a member and a >> subset of N, but N is not. > > And, in addition, it is not green. Yes. 1. Definition of "green"? 2. Proof? > Relevance? Good question. It is you who coined "green". F. N. -- xyz |