Cantor Confusion
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From: davidmarcus on 28 Jan 2007 13:37 On Jan 28, 12:32 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > William Hughes <wpihug...(a)hotmail.com> writes > >No, the fact that one form of Cantor's diagonalization cannot be used > >in one particular case does not mean that Cantor's diagonalization > >is dead and buried. Cantor's diagonalization can be used in > >many other cases and can be used to show that the reals > >are not countable. > Thanks for that. Understood. But actually when Randy said "I do know > that the set of infinite binary strings is uncountable, but I know that > because it is easily proven" I assumed that he meant Cantor's > diagonalisation - I had a mental equivalence set of real numbers <=> set > of infinite binary strings, clearly too primitive a perspective. Cantor's diagonal argument works for binary strings because you don't have the problem of two different strings representing the same number.
From: davidmarcus on 28 Jan 2007 13:40 On Jan 28, 10:11 am, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > Nobody has actually said that I can't attempt to enumerate the reals in > the suggested fashion (on an assumption that they were countable) - > although it is clear that doing so as I proposed can only ever index the > (subset of) the rationals with a terminating sequence of zeroes in their > expansion ... which (as I previously said) is prima facie evidence for > the reals being uncountable. 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.
From: G. Frege on 28 Jan 2007 13:54 On Sun, 28 Jan 2007 15:11:31 GMT, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: >>> >>> [...] most cranks are probably convinced that they are right >>> >> That's for sure! :-) >> > I will say it again - I don't have the knowledge to be a crank. > I don't think so. Remember: "Cranks who contradict some mainstream opinion in some highly technical field, such as mathematics or physics, almost always 1. exhibit a marked lack of technical ability, 2. misunderstand or fail to use standard notation and terminology, 3. ignore fine distinctions which are essential to correctly understanding mainstream belief." Source: http://en.wikipedia.org/wiki/Crank_(person)#Common_characteristics_of_cranks ;-) Of course the difference between a beginner and a crank is that the beginner is able to (listen and) learn. > > Please explain why my post was nonsense, if you have the time, or unless > it was so misguided that you don't know where to begin. I am actually > rational (most of the time anyway). > I'm sorry, I just have the flu, hence... F. -- E-mail: info<at>simple-line<dot>de
From: Virgil on 28 Jan 2007 14:18 In article <1169976214.817714.231970(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Since there are an infinite number of natural numbers to serve as path lengths, and path lengths for all those naturals, why not? > 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 now WM can't tell whether a path staring at 0 end ending at n is finite or not? > > > > > > So a union of different > > > natural numbers must contain an infinite number should it ex�st? > > > > > > 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. If p is the union of all paths with finite lengths it must have a length at least as great as any natural, so which natural does WM claim is as great as every natural? > > > > 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). There IS an infinite ordinal number, but it is not a natural number. There IS an infinite path but it is not a path of finite length. > > Yes, that is the implication of finite infinity: infinite finity. WM's quantifier dyslexia strikes again.
From: Virgil on 28 Jan 2007 14:20
In article <1169976441.727208.225560(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Then why does WM keep claiming that N is finite? |