Cantor Confusion
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From: Dik T. Winter on 16 Jan 2007 20:44 In article <1168962527.282591.33370(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1168955262.910310.48680(a)v45g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > Please look at the paths in original definition: I'll try to draw it > again somewhat more suggestive: The finite tree from level 0 to level 1 > is: > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > | | | | > 0 1 0 1 > | | | | > 0 1 0 1 > ........................... So it contains infinitely many nodes? You are again shifting your paradigm. And I would not think about that as a finite tree, only as a tree that splits at finitely many places. > > You do when you want to prove that the set of paths is countable. > > Remember what you wrote: > > (1) In each finite tree the set of paths is finite. > > (2) A countable union of finite sets is countable. > > (3) So the set of paths in the complete tree is countable. > > (3) is not justified because (2) is not about the set of paths in the > > complete tree. Now try to do this proof *without* using sets of paths. No answer to this? > > > In the union of all finite squares > > > > > > (11) > > > > > > (11), (12) > > > (21), (22) > > > > > > etc. > > > > > > there is no infinite diagonal. > > > > Why not? I only state that that diagonal is not in the union of the sets > > of diagonals of the finite squares. > > Which digit is missing? None. But all sets of diagonals of finite squares contain only finite diagonals, so there can not be an infinite diagonal in their union (which is a set of infinite size). > > > Why do you refuse to talk about the limit in case of paths? > > > > In that case *you* have to define what you understand under limit. But > > lets suppose that s_n is the set of paths in the finite tree of level n. > > I concede that you might be able define a limit such that > > lim{n -> oo} s_n = s > > where s is the set of paths in the complete tree. But with this you > > can show nothing about the cardinality of s because > > lim{n -> oo} | s_n | = | s | > > is not necessarily valid. > > There are more limits than sequences? Eh? Why do you think so? Each sequence has only a single limit (when the limit is properly defined and when it exists). But you expect some form of continuity (which you do not express). In order to show that the above limit equality is valid you have to prove that. And in earlier articles I have already shown that interchanging "taking the limit" and "taking the cardinality" was not a valid operation. > What we need is: All paths > in the union tree are countable. All possible nodes are in the union > tree. Yes, you desperately need it. I do not need it, but you do. > A path can only then differ from all paths of the union tree if > there is another node (or if it is running in other directions --- try > that argument?). What is the argument here? I think you are back to the old argument, that is not valid either. (1) For each pair of paths there is an edge were they differ. (2) For each path there is no edge where it differs from all other paths. You think they are in contradiction (but have not been able to prove that). Mathematicians think there is no contradiction. Consider the set of natural number when written out in decimal: (1) For each pair of numbers there is a digit where they differ. (2) For each number there is no digit that is different from all other numbers. (And do not come up with the finitistic view of 10^100 * pi. You are arguing within the axioms of ZF. If you want to prove inconsistency you need to proof that with those axioms you can show for some proposition P that both P and not P hold. Always using arguments that are within the logic.) > > > The union of finite trees contains all (initial) sequences (of paths). > > > Their number is countable. > > > > As long as you allow only the *finite* initial sequences, they are > > countable. Not if you allow more. That is because the union of the > > sets of paths in finite trees does exist just of the set of finite > > initial sequences. And you proof about sets of paths works for this. > > But that set is not the set of all paths. > > But if there are other paths, then they should prove their being > different by showing a node not in the union. Different sets differ by > at least one element. Or not? Yes. This does no mean that there is an element that is particular to a single set. This can already be shown with a finite collection of finite sets. All sets can be different, but there is no single element that distinguishes between the sets. > Did you receive and enjoy my book which I sent off on 10. Jan.? Not yet. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Marcus on 16 Jan 2007 20:57 MoeBlee wrote: > Andy Smith wrote: > > > Doubtless you lot have some legally watertight definition. As far as I > > am concerned, a function is a formula that provides a value for a given > > input (or inputs, in a multidimensional situation). > > Dump that definition. It's picture language. Use mathematical > definitions instead: > > {x y} = z <-> At(t in z <-> (t=x v t=y)) > > {x} = {x x} > > <x y> = {{x} {x y}} > > p is an ordered pair <-> Exy p = <x y> > > r is a relation <-> Ax(x in r -> r is an ordered pair) > > f is a function <-> (f is a relation & Axyz((<x y> in f & <x z> in f) > -> y = z)) I think this is a bit (actually quite a bit) too formal considering Andy's mathematical background. You have to walk before you can run. -- David Marcus
From: Dik T. Winter on 16 Jan 2007 20:51 In article <10376996.1168964404012.JavaMail.jakarta(a)nitrogen.mathforum.org> Andy Smith <andy(a)phoenixsystems.co.uk> writes: .... > Sometimes pi is estimated by computing the circumference > of a regular convex polygon with N vertices, and letting > N->oo . But actually such a limit does not exist? The limit of the value does exist. That does not mean that the limit of the polygon is something sensible. > All I meant was that if you have the ordered finite set > e.g. {0,1,2,3,..,N} you could partition it into exclusive subsets > {0,1,2,3,..,N-1} and {N}, and then, if you wished, > recreate the original set by appending {N} to > {0,1,2,3,..,N-1}. But when you form the open subset > of [0,1) by e.g. excluding the point {1}, you could > not form the ordered closed set [0,1] from > [0,1) and {1} - because [0,1) does not have a last member, > so you would not know where to append {1}. When you are talking about ordered sets, it is clear where you have to put {1}. For all elements of [0,1) the element is smaller than 1, so you should put it at the right end. It does not matter whether there is a last member or not. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Marcus on 16 Jan 2007 21:03 Dik T. Winter wrote: > In article <1168962527.282591.33370(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Did you receive and enjoy my book which I sent off on 10. Jan.? > > Not yet. You didn't receive it or you didn't enjoy it? -- David Marcus
From: MoeBlee on 16 Jan 2007 21:12
David Marcus wrote: > MoeBlee wrote: > > Andy Smith wrote: > > > > > Doubtless you lot have some legally watertight definition. As far as I > > > am concerned, a function is a formula that provides a value for a given > > > input (or inputs, in a multidimensional situation). > > > > Dump that definition. It's picture language. Use mathematical > > definitions instead: > > > > {x y} = z <-> At(t in z <-> (t=x v t=y)) > > > > {x} = {x x} > > > > <x y> = {{x} {x y}} > > > > p is an ordered pair <-> Exy p = <x y> > > > > r is a relation <-> Ax(x in r -> r is an ordered pair) > > > > f is a function <-> (f is a relation & Axyz((<x y> in f & <x z> in f) > > -> y = z)) > > I think this is a bit (actually quite a bit) too formal considering > Andy's mathematical background. You have to walk before you can run. > > -- > David Marcus Gulp. That IS walking. MoeBlee |