Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Virgil on 27 Jan 2007 22:05 In article <lrK6FmtmG9uFFwIl(a)phoenixsystems.demon.co.uk>, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > Randy Poe <poespam-trap(a)yahoo.com> writes > > > > > > > (snip) > >That's not an argument. It's a statement about why several of > >your intuitions combine in a certain way to lead you to an > >intuitive conclusion. I don't know what "uncountable" means > >to you, or why "necessarily infinite binary expansion" implies > >whatever that means to you. > > > >I do know that the set of infinite binary strings is uncountable, > >but I know that because it is easily proven. > > > Well, at the high risk of ridicule and a rather lower probability of > assassination by irate set theorists, I think that Cantor's > diagonalisation argument has a terminal flaw. > > The flaw is that Cantor's argument, disregarding any qualms about the > infinite sets involved, just demonstrates that one cannot construct an > infinite list of all permutations of binary strings. this is not the > same as showing that you cannot construct an infinite list of all reals, > because, as has been pointed out, reals may have more than one > representation in any base. To prove the countability of the reals requires, by definition, that one be able to show that there is a list of reals that contains all reals. Any set which can be listed in its entirety is, by definition cuntable and any set which is not so listable is, by definition, uncountable. Anything else is irrelevant. Cantor showed with two entirely different proofs that the reals cannot be listed. So Andy is talking nonsense. > > Please consider the following, and then abuse me; I hope that this will > cause intense discomfort, but I doubt that it will. > > Consider first Cantor's diagonalisation argument applied to the set of > reals in [0,1), i.e. excluding the point 1.00.... > > Assume that we can form a hypothetically infinite list of the reals. The > countability or otherwise is not a finction of arrangement, so consider > the list set out in the following systematic order, defined by the > mirror inverse of the indices to the hypothetical list: > > Index no. Index in binary Corresponding real number > 0 0 .0000... > 1 1 .1000.. > 2 10 .0100.. > 3 11 .1100.. > 4 100 .0010.. > > etc. > > This arrangement systematically covers all the reals (subject to our > assumption) - it is just the mirror inverse of counting from 0 upwards > in binary. Already False! There is no (finite) binary index for any of those uncountably many reals whose binary expansions require infinitely many 1's. For example 1/3. So that we already know of uncountably many real numbers not listed.
From: davidmarcus on 28 Jan 2007 00:10 On Jan 27, 6:47 pm, G. Frege <nomail(a)invalid> wrote: > On Sat, 27 Jan 2007 22:26:52 GMT, Andy Smith > <A...(a)phoenixsystems.co.uk> 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.
From: G. Frege on 28 Jan 2007 02:52 On 27 Jan 2007 21:10:06 -0800, davidmarcus(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! F. -- E-mail: info<at>simple-line<dot>de
From: mueckenh on 28 Jan 2007 04:23 On 27 Jan., 18:38, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Jan 27, 11:58 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > > On 27 Jan., 17:38, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > On Jan 27, 8:33 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > > > > > > Induction can possibly > > > > > prove that all the members of V* have some property, but can prove > > > > > nothing about V* itself. > > > > 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. > > > > Therefore all the path-*lengths* in the union are natural numbers. > > > > Notwithstanding the question whether there are infinitely many paths > > > > in the union or not: If the union path p is infinite, then at least > > > > one of the paths in the union must be infinite. > > > > > Is this so? > > > No. > > > > Think of the EIT. The diagonal is the union of the lines. > > > None of the lines is an (potentially) infinite set. > > > The diagonal is an (potentially) infinite set. > > > > A union of finite sets can be (potentially) infinite. > >Fine. (Please use "numbers". The paths-lengths are numbers (of a > > unit). ) But set theorists deny this. They say: A union of finite > > numbers cannot yield an infinite number. > No. They say exactly the opposite. They say the union > of any unbounded set of natural numbers (e.g. all natural numbers) is > an > infinite number. Please do not mix up numbers of elements and natural numbers concerning path lengths. Translated to our problem they say: There is an infinite number of finite path length. Now I am in doubt whether these path length when put together (such that every path starts at 0 and ends at n) yield a finite length or not. > > > So a union of different > > natural numbers must contain an infinite number should it exíst? > > > > The fact that p is a union of finite paths does not tell > > > us whether p is finite or (potentially) infinite. > > Fact is that p has no upper bound. Ok?So p does not have length n for any finite natural > number n. So p does not have a finite path length. But the length of p *is* a natural number. > > Either we say that p does not have a path length, > or we say that the path length of p is an infinite number. > That's it! Either there are not all natural numbers ( = the union p does not ave a length) or there is an infinite number ( = there is an ininiutr finite number). Yes, that is the implication of finite infinity: infinite finity. Regards, WM
From: mueckenh on 28 Jan 2007 04:27
On 27 Jan., 18:29, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > > > Induction covers (is valid for) all natural numbers, but not "the set > > of all natural numbers". > We know this for quite some time. The point is that you claim induction > allows any assertion of U { T(i) | i e N }. So you eventually agree > that is does not. Cant't you read? Please look closer. There is clearly spelled out that every number *in* N is concerned, i.e., every finite numbver, not the infinite number N. > > >> > You only try to huddle around, avoiding any concrete discussion. > > >>I have a different understanding of what concrete means. Do you mean > >>the > >> concrete in your head? > > > You seem to know only one aspect of many things which have more. > > There are three meanings in English: noun, verb and adjective. The > > adjective has the same meaning as in German. > The noun precisely represents the modus operandi you present in this > newsgroup. The verb is very well representing your efforts. > > > > Is this so?An in-depth discussion of your question you will find > in <45bb7541$0$97231$892e7...(a)authen.yellow.readfreenews.net> Already refuted. But especially for you I devised the example with the path length of the path at the outmost *right* hand side of the tree. There are only 1s instead of 0s. But without "trailing zeros". Regards, WM |