Cantor Confusion
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From: Virgil on 28 Jan 2007 15:09 In article <1169987266.609036.283220(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 28 Jan., 03:31, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1169905223.354212.199...(a)a75g2000cwd.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > On 26 Jan., 02:21, "Dik T. Winter" > > <Dik.Win...(a)cwi.nl> wrote: > > > > > A path-length which has no upper bound may be sufficent to be called > > > > > infinite. > > > > > > > > Sorry, a path-length is something fixed, so the wording is inadequate. > > > > But if you call a path-length infinite you do not satisty (1), because > > > > there you associate path-lengths with natural numbers. > > > > > > We can boil down the discussion about trees to the following simple > > > question, considering only one path, for instance the path p on the > > > outmost left hand side of the tree. This path p (in terms of nodes) is > > > the union of all paths of finite trees with length n, n in N. > > > > If you consider the union of paths as the union of sets of nodes, you are > > right. > > If I consider the path length as the union of path length: = > {0,1,2,...,n-1}, I am right too. > > > > > Therefore all the path-*lengths* in the union are natural numbers. > > > > Yes, the cardinality of all those paths (as sets of nodes) is a natural > > number. > > > > > Notwithstanding the question whether there are infinitely many paths > > > in the union or not: If the union path p is infinite, > > > > Note that the union path is the union of sets of nodes. > > > > > then at least > > > one of the paths in the union must be infinite. > > > > Wrong. The union of a collection of finite sets can be infinite, and in > > this case is. > > Here we are not concerned with the cardinal number of the union of > infinitely many elements but with the length of the union of all > finite path lengths, where each path length is measured from the > common origin, namely 0 at the root of the tree. If, as I suspect, Dik's definition of a path requires it to have a terminal as well as an initial node, then no endless chain of nodes can be a path. Alternately, Dik may regard the union of two ormore different paths, whatever his definition of path may be, not to itself be a path. In either case Dik is right according to his own terms. WM is, as usual, being deliberately ambiguous about definitions.
From: G. Frege on 28 Jan 2007 15:16 On Sun, 28 Jan 2007 19:53:56 GMT, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: >> >> Of course the difference between a beginner and a crank is that the >> beginner is able to (listen and) learn. >> > Well, please keep me in the beginner bin then, even if I satisfy your > criteria 1-3. > Fine. Glad to hear that. > > I am aware of my ignorance, but if some line of argument > seems solid to me, I don't see it's unreasonable to advance > it (even in the expectation of being told its nonsense)? > Well, my advice would be to try hard to understand why some guys much more knowledgeable than you (concerning theses things) tell you that something is nonsense. (Since it's highly probable that they are right and you are wrong.) This way you might learn something. (Cranks are not interested in learning anything.) F. -- E-mail: info<at>simple-line<dot>de
From: Virgil on 28 Jan 2007 15:20 In article <1169987531.365755.18830(a)p10g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > Your notation is > > ambiguous, to say the least. > > > > > > New ideas often require new notations. > > > > In that case you have to *explain* your notation. > > I did. I defined: > L is a set of limits L_k > L = { L_k | k in N } But what are the L_k ? > and S is the set of corresponding sequences > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > This well defined. > In general one does not use further symbols to denote outwritten > sequences. I could have used > <a_1k, a_2k, a_3k, ...> > but need not. How is WM's S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } different from S' = { a_nk : a in {0,1}, n in N, k in N} or different from S" = { a_{n,k} : a in {0,1}, n in N, k in N}
From: Virgil on 28 Jan 2007 15:22 In article <1169988521.201568.219710(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 28 Jan., 11:05, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > > G. Frege <nomail(a)invalid.?.invalid> writes>On 27 Jan 2007 21:10:06 -0800, > > davidmar...(a)alum.mit.edu wrote: > > > > >>>> I think that Cantor's diagonalisation argument has a terminal flaw. > > > > >>> You have earned crank status now. Congratulations! > > > > >> It would seem so. Andy keeps posting the same argument without taking > > >> into account the comments made on his previous posts. > > > > >Right. I've also noticed that. Btw, a rather interesting experience: > > >to meet a crank in statu nascendi! > > > Fair enough. But most cranks are probably convinced that they are right > > In particular this one. One crank denigrating another? Shameful!
From: davidmarcus on 28 Jan 2007 16:01
On Jan 28, 2:59 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > davidmar...(a)alum.mit.edu writes > > >Of course, you can attempt to enumerate the reals any way you wish. > >However, we don't see why the fact that some specific enumeration > >fails is evidence that the reals are uncountable. > I had understood that countability or otherwise was independent of a > specific arrangement . So if a specific arrangement clearly doesn't > allow all the reals to be countable (if they were), then that seems > like good evidence for the initial hypothesis that the reals were > countable to be false. If by "arrangement" you mean enumeration (or listing), then this is trivially false. Cranks use non-standard terminology without defining it. Non-cranks use standard terminology if such exists and/or define the words that they use. |