Cantor Confusion
in [Math]
|
From: mueckenh on 12 Apr 2007 12:59 On 11 Apr., 19:42, Virgil <vir...(a)comcast.net> wrote: > In article <1176298903.609533.227...(a)d57g2000hsg.googlegroups.com>, > > > > > > Are the > > > > numbers represented by paths which are completely within the > > > > infinitely many levels of the tree or not? > > > > This sentence is quite difficult to pars. Yes, each real number is > > > represented by a path which is completely within the infinitely many > > > levels of the tree. > > > But not all real numbers coexist in the tree? > > Yes, all of them do at least once, and some twice. But not side by side? More above or below each other? > > > Only a countable number > > of them is admitted simultaneously? > > All of them simultaneoulsy, and some twice. But not side by side? Regards, WM
From: mueckenh on 12 Apr 2007 13:02 On 11 Apr., 20:00, Virgil <vir...(a)comcast.net> wrote: > In article <1176300393.408149.165...(a)w1g2000hsg.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 11 Apr., 03:56, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1175968689.240804.64...(a)e65g2000hsc.googlegroups.com> > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 4 Apr., 16:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > ... > > > > > > You can mirror the series without change of value. You can decide > > > > > > to > > > > > > write from right to left. > > > > > > Yes. In that case the mirroring is just virtual and the first number > > > > > is > > > > > on the right. But that means that the '1' applies to the *first* node > > > > > in the path. I do not think you want that... > > > > > The mirror is not "just virtual" unless you claim that the infinite > > > > paths exist only virtually. > > > > That is not what I state. In your mirrored sequence you have a first > > > element > > > on the right. Well if you wish to go with the right-left writing people, > > > do so. But that does not change the sequence at all, Only the way you > > > look at it. > > > In the geometric series there is no last element. Nevertheless you > > think the series has a limit value. In a geometric series with last > > but not first element, the things are different? Why is there not a > > limit value for > > > lim[n --> oo] SUM [1/2^n + ... + 1/4 + 1/2 + 1]? > > The limit of a series, if it exists, is by definition the limit of its > sequence of (finite) partial sums. What are the partial sums of your > series? If they are, as I suspect, 1, 1/2 + 1, 1/4 + 1/2 + 1, ..., then > the limit is the same as for 1 + 1/2 + 1/4 + ... . > > > > > > > For every path there is a node assumed to contribute 1. If it turns > > > > out that this node is shared by another path, then take the next node > > > > and continue to go on. > > > > And you never will terminate searching because there is no last node. Now > > > what? > > > That is the same for searching the digits of pi. Nevertheless you say > > pi exists. > > One can prove that the series for pi (decimal base or otherwise) > converges, so that the limit exists. > > > > > > You think so. In that case you should throw away *all* of mathematics, > > > because there is (for instance) no direct proof that sqrt(n) is rational > > > only if n is a square. When something is proven for an arbitrary element > > > that means that is has been proven for all elements. > > > Yes, it has been proven for all elements (which exist). > > Name one which doesn't. If I could name it, it would exist (at least as an idea). But we cannot name more than countably many. > > > But it has not > > been proven for the sequence eventually constructed by these elements. > > On the contrary, that is precisely what has been proved. > > If a binary sequence has for each binary sequence in a set of such > sequences at least one term different from that sequence, then it is > different from every sequence in that set. > > Why WM keeps trying to claim this to be false is incomprehensible. > > > > > > > You cannot apply Cantor's proof to the whole diagonal. In particular > > > > because the diagonal does not exist. (It does not exist as a path > > > > separated from all other paths in the binary tree.) > > This fixation on "separation" is puerile. Every path in any binary tree > is separated from any other path from the children of some node onwards, > and that is as much separation as is needed to give each path a unique > identity. But there are only countably many nodes! > > > > > > Again, your mantra. Proof, please. > > > If all paths exist simultaneously, then there must exist uncountably > > many in the infinite tree. > > Which is precisely the case! At least in mathematics.- But the tree is not in mathematics? Regards, WM
From: Virgil on 12 Apr 2007 13:53 In article <1176397196.082457.230040(a)y5g2000hsa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 11 Apr., 19:42, Virgil <vir...(a)comcast.net> wrote: > > In article <1176298903.609533.227...(a)d57g2000hsg.googlegroups.com>, > > > > > > > > > Are the > > > > > numbers represented by paths which are completely within the > > > > > infinitely many levels of the tree or not? > > > > > > This sentence is quite difficult to pars. Yes, each real number is > > > > represented by a path which is completely within the infinitely many > > > > levels of the tree. > > > > > But not all real numbers coexist in the tree? > > > > Yes, all of them do at least once, and some twice. > > But not side by side? More above or below each other? If WM visualizes paths as running from the root node downward, then sort of side by side. When two paths represent the same binary rational (the only reals having dual representation) there will be no other paths "between" them. From the last node they have in common, one will branch left once then forever branch right and the other will branch right once then forever branch left, like 00111... and 0.1000... in binary both represent the rational number 1/2. > > > > > Only a countable number > > > of them is admitted simultaneously? > > > > All of them simultaneoulsy, and some twice. > > But not side by side? Whyever not, at least for those represented twice? > > Regards, WM
From: Virgil on 12 Apr 2007 14:02 In article <1176397336.310678.13030(a)l77g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 11 Apr., 20:00, Virgil <vir...(a)comcast.net> wrote: > > In article <1176300393.408149.165...(a)w1g2000hsg.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 11 Apr., 03:56, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > You think so. In that case you should throw away *all* of mathematics, > > > > because there is (for instance) no direct proof that sqrt(n) is > > > > rational > > > > only if n is a square. When something is proven for an arbitrary > > > > element > > > > that means that is has been proven for all elements. > > > > > Yes, it has been proven for all elements (which exist). > > > > Name one which doesn't. > > If I could name it, it would exist (at least as an idea). It follows that if you cannot name it , it does not exist, even as an idea. > But we cannot name more than countably many. That is not because there cannot be more than countably many things but because there cannot be more than countably many names. So that just as there can be more things than can be numbered by the naturals, there can be more things than can be given all different names. > > > > > But it has not > > > been proven for the sequence eventually constructed by these elements. > > > > On the contrary, that is precisely what has been proved. > > > > If a binary sequence has for each binary sequence in a set of such > > sequences at least one term different from that sequence, then it is > > different from every sequence in that set. > > > > Why WM keeps trying to claim this to be false is incomprehensible. > > > > > > > > > > > You cannot apply Cantor's proof to the whole diagonal. In particular > > > > > because the diagonal does not exist. (It does not exist as a path > > > > > separated from all other paths in the binary tree.) > > > > This fixation on "separation" is puerile. Every path in any binary tree > > is separated from any other path from the children of some node onwards, > > and that is as much separation as is needed to give each path a unique > > identity. > > But there are only countably many nodes! There are only countably many binary digit positions, but they are enough to create uncountably many binary sequences. > > > > > > > > > > Again, your mantra. Proof, please. > > > > > If all paths exist simultaneously, then there must exist uncountably > > > many in the infinite tree. > > > > Which is precisely the case! At least in mathematics. > > But the tree is not in mathematics? If mathematically defined trees, finite or infinite, do not exist within mathematics, why does mathematics bother to define them? And it does bother to!
From: MoeBlee on 12 Apr 2007 14:15
P.S. I forgot to mention one that Daryl and Rupert were helping me with: x is finite <-> ER(R well orders x & converse(R) well orders x). And one I thought of the other day but I need to check that my proof is okay: x is finite <-> E!s(s is an ordinal & x is 1-1 with s) MoeBlee |