Cantor Confusion
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From: Dik T. Winter on 22 Jan 2007 10:50 In article <uC3ofBOf0MtFFwCD(a)phoenixsystems.demon.co.uk> Andy Smith <Andy(a)phoenixsystems.co.uk> writes: .... > But with a finite list of n entries we can always define the number e.g. > 2^n which is not in the list. So for an infinite list of natural numbers > of size n we can always construct a number not in the list (trivially true > anyway because any natural number has a successor). This is wrong. With an infinite list you get a string of digits of an infinite length. But each natural number has a finite number of digits, and so the constructed string does not represent a natural number. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dave Seaman on 22 Jan 2007 12:07 On Mon, 22 Jan 2007 15:28:19 GMT, Andy Smith wrote: > Dave Seaman writes >>> But why is Cantor's argument valid but such a construction (or variants >>> of it) invalid? Cantor's argument goes, as I understand it : construct >>> a hypothetical table of all the reals (in [0,1) , defined for the sake >>> of argument as a infinite bit patterns from .000.. to .111.. . Assume >>> that this is a complete list - then show that E a number not in the >>> list, therefore you cannot have such a list. >>The difference is that when you apply the Cantor diagonal argument to a >>list of real numbers, you get a real number not in the list. When you >>apply a similar argument to a list of integers, you do not get an integer >>at all. > Even if you truncate the exercise at the nth bit? If you do that, you get a bit string that is already present in the list. That destroys the argument. >>> But it isn't clear to me that such a line of argument is valid ( "do not >>> extrapolate reasoning about finite sets to infinite sets") ? >>In this case, it would be the reasoning that every finite bit string >>represents an integer. We can show that it works for the reals by >>invoking the fact that every Cauchy sequence of rationals converges to >>some real number, but there is no corresponding fact for the integers. > OK, understand that re the reals, but not necessarily about validity of > reasoning over a hypothetically infinite list, nor about the integer > case - you don't have to try to generate numbers like ...111 not to be > on the list of n integers (indeed , n+ 1 will do as number not in the > list). What is n? > So for integers, the argument goes - let there be a 1:1 > correspondence between the set of all integers and itself. Construct a > numbered table of all integers that is complete. But we can create a new > number not in the list, so such a list cannot be constructed, therefore > there is no 1:1 correspondence between the integers and themselves? No, that doesn't work, for the reason that I have explained. An infinite bit string does not denote an integer, and every finite bit string is already accounted for in the table. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: mueckenh on 22 Jan 2007 13:06 Franziska Neugebauer schrieb: > > Max n,m is not defined either, > > Aha > > ,----[ WM in <1169111380.377993.67320(a)l53g2000cwa.googlegroups.com> ] > | The union of two finite trees T(m) and T(n) with m and n levels, > | respectively, where m < n, is the tree with n levels. > `---- > > So you mean m < n is not defined? Then it makes no sense at all to write > about trees? > Sorry, this should read: Max (n,m) is not defined *other* (than for finite m and n). The union of m and n is the maximum of both. Nevertheless the union of all natural numbers exists as well as the union of segments {1,...,n} and {1,..., m} and the infinite union of all segments. Regards, WM
From: mueckenh on 22 Jan 2007 13:14 Virgil schrieb: > > The directions of the paths are the same in T1 and T2. If you insist on > > a difference, then it can only result from the length of the paths. > > Two paths will be different it there is some node from which one > branches in one direction and the other does not branch in the same > direction For a given finite length all segments of paths are there, no? So all finite segments of 0.111... as well as of 0.010101... are there. > (it may branch in another direction or it may have that as a > terminal node). There is no terminal node in T1 and T2. > WM has been working in vain for years regardless of the outcome. I know. It was certainly in vain - as far as you are concerned. But we see at least the following result: You think to have learned that identical trees can have different paths. This idea is so absurd that it may serve as a good example to show how dangerous set theoric influences are for the mind. > > > If you are wrong, then thousands > > of mathematician have been working in vain for 130 years. > > WM is not competent to pass judgement on mathematicians. That could be decided by mathematicians only. > > > Two trees which are identical by all nodes and edges are identical by > > all paths. > > For finite sets, a tree is uniquely determined by its set of nodes and > set of edges. > > This is false for infinite trees This assertion alone is capable of showing the non-existence of infinity. > One can easily see that the sets of nodes are the same for both trees > and the sets of edges is also the same for both trees That is easily to be seen. > but the sets of > paths are quite different, That is, ae, silly to be seen. > Regards, WM
From: Franziska Neugebauer on 22 Jan 2007 13:34
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> > Max n,m is not defined either, >> Aha >> >> ,----[ WM in <1169111380.377993.67320(a)l53g2000cwa.googlegroups.com> ] >> | The union of two finite trees T(m) and T(n) with m and n levels, >> | respectively, where m < n, is the tree with n levels. >> `---- >> >> So you mean m < n is not defined? Then it makes no sense at all to >> write about trees? >> > Sorry, this should read: Max (n,m) is not defined *other* (than for > finite m and n). The problem with defining the tree-union with the max-Function is that max N is not defined. (N has no maximum). (If your N has a maximum, i.e. is finite, you leave the contemporary set theory.) > The union of m and n is the maximum of both. > Nevertheless the union of all natural numbers exists Your tree-union of two trees is _defined_ to be the tree having the greatest of both trees depths. This is not a set-theoretical union. The only thing it has in common is the name "union" (equivocation). The extension of your tree-union to V* = { T(n) | n e N } fails as I have pointed out already three times (or even more) because max (N) does not exist. > as well as the > union of segments {1,...,n} and {1,..., m} and the infinite union of > all segments. These unions are standard-unions of set theory, your tree-union is not. The latter is an equivocation. http://en.wikipedia.org/wiki/Equivocation Again. U V* is not defined. A recap of the definitory problems with your and other non-standard notions of trees you will find in. <45b489e1$0$97253$892e7fe2(a)authen.yellow.readfreenews.net> F. N. -- xyz |