Cantor Confusion
in [Math]
|
From: Franziska Neugebauer on 12 Feb 2007 11:38 mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Feb., 13:59, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: >> mueck...(a)rz.fh-augsburg.de wrote: >> > Set theory, however, says: If the union of all finite paths, >> > 0., 0.0, 0.00, 0.000, ... , >> > is in the tree, then the infinite path p(oo) = 0.000.... is in the >> > tree too. >> >> No contemporary set theory or graph theory does posit such a >> nonsensical claim. It is entirely *your* claim, better known as > > It is easy to recognize from the union of all finite segments of N. > Set theory says: If the union of all finite segments, > {1,2,3,...,n} is present, "Present"? Where? > then the infinite set N = {1,2,3,...} is present. Where? I don't know of any contemporary set theory which states that entities are "present". Perhaps you borrowed that work from M�ckenmatics. > (In fact set theory says: N is the union. But I don't know > how this is accomplished.) "Accomplish" is certainly a word from M�ckenmatics, too. > Or do you see a difference betwee n the union of all finite segments > and N? First of all I don't see that contemporary set or graph theories posit | Set theory, however, says: If the union of all finite paths, | 0., 0.0, 0.00, 0.000, ... , | is in the tree, then the infinite path p(oo) = 0.000.... is in the | tree too. F. N. -- xyz
From: Franziska Neugebauer on 12 Feb 2007 12:08 mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Feb., 14:29, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> > The connectedness is given by the stucture of the tree. >> >> Your claim is: >> >> | >One cannot determine merely from a set of nodes for a given >> | >tree whether that set of nodes is or is not a path. [<-- FN] >> | One can. [<-- WM] >> >> As I have shown in my example one cannot. How long do you want to >> argue against facts? > > All you have shown is that some clumsy notation is always capable of > destroying meaning. "For they sow the wind, and they will reap the whirlwind. He has no standing grain. The stalk will yield no head. If it does yield, strangers will swallow it up." >> > The connectedness of elements of a matrix is given by the structure >> > of the matrix. >> >> 1. I don't see any relevance to your claim. > > Sorry, to make it easier seems impossible to me. Surely. >> > Given my fixed coordinate system, there remains no ambiguity at >> > all. >> >> Then we are no longer talking about binary trees in general. > > I never did so. I couldn't agree more. > I use one very special tree and its finite subtrees as > a tool for a general proof. Strike off "general" and "proof". > (I like trees in nature, like the trees in the Harz mountains, but I > have no special connection to trees of graph theory.) That confirms my assessment. F. N. -- xyz
From: Franziska Neugebauer on 12 Feb 2007 12:10 mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Feb., 13:59, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: >> mueck...(a)rz.fh-augsburg.de wrote: >> > Set theory, however, says: If the union of all finite paths, >> > 0., 0.0, 0.00, 0.000, ... , >> > is in the tree, then the infinite path p(oo) = 0.000.... is in the >> > tree too. >> >> No contemporary set theory or graph theory does posit such a >> nonsensical claim. It is entirely *your* claim, better known as > > It is easy to recognize from the union of all finite segments of N. > Set theory says: If the union of all finite segments, > {1,2,3,...,n} is present, "Present"? Where? > then the infinite set N = {1,2,3,...} is present. Where? I don't know of any contemporary set theory which states that entities are "present". Perhaps you borrowed that word from M�ckenmatics. > (In fact set theory says: N is the union. But I don't know > how this is accomplished.) "Accomplish" is certainly a word from M�ckenmatics, too. > Or do you see a difference betwee n the union of all finite segments > and N? First of all I don't see that contemporary set or graph theories posit | Set theory, however, says: If the union of all finite paths, | 0., 0.0, 0.00, 0.000, ... , | is in the tree, then the infinite path p(oo) = 0.000.... is in the | tree too. F. N. -- xyz
From: MoeBlee on 12 Feb 2007 14:21 On Feb 10, 5:38 am, mueck...(a)rz.fh-augsburg.de wrote: > On 9 Feb., 21:26, "MoeBlee" <jazzm...(a)hotmail.com> wrote: > > > > Reality. > > > What reality? Empirically testable physical reality? A mathematical > > statement is not of that kind. > > Every mathematical statement is of that kind. You snipped the rest of my argument regarding this. > You would recognize this > (if you could, but you couldn't) if all reality ceased to exist. No > mathematics would remain. Yes, to grasp these facts, without such a > bad experience, i.e., during the continued existence of reality, one > needs some deeper thinking than is required to understand the meaning > of some crazy axioms. Yes, if nothing existed, then mathematics would not exist, whatever it might be for there to be nothing that exists. Fortunately, we can do mathematics without consternation over such mind-boggling hypotheticals. MoeBlee
From: Virgil on 12 Feb 2007 15:02
In article <1171281091.730079.207810(a)a34g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 11 Feb., 18:32, "Mike Kelly" <mk4...(a)bris.ac.uk> wrote: > > On 11 Feb, 14:58, mueck...(a)rz.fh-augsburg.de wrote: > > > > I say 111 is number 7. Now what? > > The binary meaning is a convention and does not refute or change the > unary meaning of 111. But it does show that /the/ meaning of "111" lies purely in the eye of the beholder, and not in the expression itself. |