Cantor Confusion
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From: mueckenh on 25 Jan 2007 11:02 On 25 Jan., 16:12, Carsten Schultz <cars...(a)codimi.de> wrote: > mueck...(a)rz.fh-augsburg.de schrieb: > > > > > On 25 Jan., 13:39, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > >>> 1) If we identify path-lengths with natural numbers, then there is no > >>> infinite path in any of the finite trees, because there is no infinite > >>> natural number. Therefore, also the union of all finite trees cannot > >>> contain an infinite path. > > >> Clearly the tha path-length of an infinite path cannot be a > >> natural number. > > > The path length is the number of nodes of the path. As long as the > > nodes are enumerated by natural numbers, the length (distance from the > > root node) is a natural number.This is just great. > > Of course this is true for WM, because he does not believe in infinite > sets anyway, so in particular every set of natural numbers is finite. > Good luck for those who think they can convince him of anything using > mathematical arguments. Consider a set of paths p(n) of lengths |p(n)| = n for all n in N. Each path is finite. Define the union as we defined the union of paths in a binary tree. Is it not true for you that the union path is infinite? Regards, WM
From: Dik T. Winter on 25 Jan 2007 11:03 In article <1169736539.298735.176670(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1169728742.271970.13430(a)v33g2000cwv.googlegroups.com> "William Hughes" <wpihughes(a)hotmail.com> writes: > > > On Jan 25, 3:19 am, mueck...(a)rz.fh-augsburg.de wrote: > > ... > > > > and S is the set of corresponding sequences > > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > > > then L cannot be larger than S, because there cannot be more limits > > > > than sequences. > > > > > > Note that S contains only infinite sequences. > > > > S is not a set of sequences. S is an ordered multi-set of the digits > > 0 and 1. Or, alternatively, S is one of the sets {0, 1}, {0} or {1}. > > Sorry. S contains for instance the paths 0.111... and 0,010101... the > bits (nodes) of which can be indexed by natural numbers. Yes? The elements of S as written above are the a_ik for i and k natural numbers. So the elements are elements from {0, 1}. S *does* contain those paths, but not as elements, but as subsets. S is not the set of "corresponding sequences". You specifically wrote: > However, if L is a set of > limits L_k > L = { L_k | k in N } > and S is the set of corresponding sequences > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > then L cannot be larger than S, because there cannot be more limits > than sequences. this makes no sense if S is *not* a set of sequences. Consider the following: L = { L_k | k in {0, 1, 2}} S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in {0, 1, 2} } If S is a set the cardinality of S is at most 2. If S is an ordered *multi-set*, the ordinality is omega. The cardinality of L is 3. S is certainly not an ordered set because it contains multiple identical elements. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 25 Jan 2007 11:16 On 25 Jan., 17:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1169736539.298735.176...(a)v33g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > In article <1169728742.271970.13...(a)v33g2000cwv.googlegroups.com> "William Hughes" <wpihug...(a)hotmail.com> writes: > > > > On Jan 25, 3:19 am, mueck...(a)rz.fh-augsburg.de wrote: > > > ... > > > > > and S is the set of corresponding sequences > > > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > > > > then L cannot be larger than S, because there cannot be more limits > > > > > than sequences. > > > > > > > > Note that S contains only infinite sequences. > > > > > > S is not a set of sequences. S is an ordered multi-set of the digits > > > 0 and 1. Or, alternatively, S is one of the sets {0, 1}, {0} or {1}. > > > > Sorry. S contains for instance the paths 0.111... and 0,010101... the > > bits (nodes) of which can be indexed by natural numbers. > > Yes? The elements of S as written above are the a_ik for i and k natural > numbers. Look closer: S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } The elements are sequences S_k like a_1k, a_2k, a_3k, ... > > However, if L is a set of > > limits L_k > > L = { L_k | k in N } > > and S is the set of corresponding sequences > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > then L cannot be larger than S, because there cannot be more limits > > than sequences. > this makes no sense if S is *not* a set of sequences. S is a set of sequences S_k = a_1k, a_2k, a_3k, ... > Consider the > following: > L = { L_k | k in {0, 1, 2}} > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in {0, 1, 2} } > If S is a set the cardinality of S is at most 2. S contains all paths in the union of all finite trees T(oo). > If S is an ordered > *multi-set*, the ordinality is omega. The cardinality of L is 3. > S is certainly not an ordered set because it contains multiple identical > elements. S is a set of sequences S_k = a_1k, a_2k, a_3k, ... Regards, WM
From: Andy Smith on 25 Jan 2007 11:47 G. Frege <nomail(a)invalid.?.invalid> writes >On Thu, 25 Jan 2007 08:38:08 GMT, Andy Smith ><Andy(a)phoenixsystems.co.uk> wrote: > >> >> [that] doesn't mean that everything I say is rubbish... >> > >Concerning math, it is. Sorry. > Maybe. I don't think that you read what I say, or cut any slack to try to understand what I am trying to convey. Cantor says, you say, almost everybody says (apart from a few denizens of this NG) that the reals are uncountable. I think that is clearly right, but am not wholly convinced by Cantor's diagonalisation proof, not least because I need to take on faith the validity of some of the reasoning and defer my judgement to people more expert in these matters. In any event, while Cantor's proof may show that the reals are uncountable, it doesn't explain why - other than in terms of the proof. So I thought that it would be useful to consider a systematic scheme that would count all the reals if they were countable i.e. for a given finite integer n one can read out the corresponding real, and for a given real determine its integer index. I suggested such a systematic counting scheme, which, of course, fails - it fails because the reals are intrinsically infinite, whereas the natural numbers just have no greatest number. I don't think this deserves as much scorn as you dump on it. Regards -- Andy Smith
From: David Marcus on 25 Jan 2007 12:26
Andy Smith wrote: > I said (and this ought not to be controversial) that if you have n > objects you need an address space defined by log(no of items, base 2) to > index(i.e. count) them. > > In your example, A n e N you need an address space of log(n,2) bits. The > address space is finite for all finite n. Fine. > When I talked about the reals, I will say it again: > > 1) Any systematic method for counting the reals is equivalent (you would > probably say isomorphic under permutation). This is wrong or at least too vague to say whether it is right or wrong. What do "systematic method" and "equivalent" mean? And, I don't know what "isomorphic under permutation" means, so I wouldn't say that. Does the following qualify as a "systematic method" for enumerating the reals? 0 0.1 0.2 .... 0.9 0.01 0.02 .... 0.09 0.11 0.12 .... 0.19 .... > 2) I proposed to try to count the reals by progressively increasing the > degree of precision of their representation. To n bits precision, there > are 2^n distinct numbers, requiring 2^n indices. If I wish I can > represent a unique set of indices covering this space by e.g. reflecting > the bit pattern of the reals to the first n bits about the binary point. > > 3) For all finite n, no problem. The set of indices are finite, and this > basically is just indexing (a subset of) the rationals. However the > reals require an infinite n and cannot be represented with indices with > a finite number of bits. (Aristotle would say that the natural numbers > are potentially infinite, but the reals are actually infinite). Is the following what you are saying? We can enumerate the reals by first enumerating the reals that take only 1 bit to write in binary, then the ones that 2 bits, then the ones that take 3 bits, etc. This doesn't work because we never get to the ones that take an infinite number of bits. If so, that's fine as far as it goes, but it doesn't tell us whether some other method of enumerating the reals will work. > Incidentally, if you will excuse the cross-threading, re the > space-filling curve, yes it fills the space, but with my variant, are > the points in the plane painted white or black (at each stage of > iteration by construction the white area = black area).? I'm sorry, but I didn't understand your variant well enough when I first read it to answer your question. Please explain it again. -- David Marcus |