Cantor Confusion
in [Math]
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From: mueckenh on 11 Apr 2007 10:20 On 10 Apr., 20:52, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Apr 10, 2:32 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > Do all paths exist separated from one another in the infinite binary > > tree or not? > > Yes, but no level separates an uncountable number of paths. Why then do you answer "yes" while the answer is "no"? If no level of infinitely many levels separates all paths, then not all separated paths are separated in any level of the tree. Where are they separated? > > i. No level of the infinite binary tree separates an > uncountable number of paths. > > ii. The union of all countable levels is the union of > countable levels. > > And we get the usual logical jump > > iii. The union of all countable levels does not separate > an uncountable number of paths > > Which bit of "i and ii do not imply iii" do you fail to > understand? I do not understand the following. If, as you seem to suggest, iii is correctly spelled out: The union of all countably many levels separates an uncountable number of paths. Where are the uncountably many separation points = nodes? The union of all levels contains only countably many nodes? Regards, WM
From: mueckenh on 11 Apr 2007 10:36 On 10 Apr., 22:02, Virgil <vir...(a)comcast.net> wrote: > In article <1176230527.931928.12...(a)y5g2000hsa.googlegroups.com>, > > > All mathematics is equally virtual. Finite trees and their finite paths > > > are as virtual as infinite trees and their infinite paths, as they exist > > > only in imaginations. > > > Wrong. The finite tree > > 0. > > /\ > > 0 1 > > > does exist here, on your screen. > > That is only a picture of a tree, and is no more an actual tree than a > picture of a triangle is an actual triangle. > > 'Actual' trees, like 'actual' triangles are only actual in one's > imagination. Without pictures or words, there would be no imagination of mathematics. The finite tree shown above is the best image one can give, mental or electronical. > > Cantor's proof does not cover the number behind the last one proved. > > Since the proof covers all members of N simultaneously, there is no > "last one" There is a last position to be recognized so far. To imagine all numbers simultaneously can only mean that your are incapable of imagine anything at all. > > > Same as: > > The binary tree does not contain all paths, because otherwise it must > > contain a level with uncountably many nodes. > > Not in my world: > (1) A tree without all paths is not actually a tree at all. > > (2) each partition of WM's set of countably many node levels into an > ordered pair of two sets determines a path in which the first set of the > pair contains those nodes levels from which the path branches left and > the other contains the set of node levels from which the path branches > right. This works equally well for finite or infinite trees. > > For a finite tree with n levels, excluding leaves, > this produces 2^n paths. > > For an infinite tree with aleph_0 levels, there being no leaves, > this produces 2^aleph_0 paths. How does the production produce separated paths without separation points? > > That's how it works in mathematics, including ZF and NBG and most other > systems. Deplorable mathematics. > > If that is not how it works in WM's system, then WM's system is no part > of any standard mathematics.- No these are unavoidable considerations the logic of which forces upon us. Regards, WM
From: Virgil on 11 Apr 2007 13:42 In article <1176298903.609533.227830(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 11 Apr., 03:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1175967715.832820.167...(a)q75g2000hsh.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 4 Apr., 15:47, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > Such laws show that there is no separated path without a separation > > > > > point. > > > > > > > > Indeed. But that does *not* prove your statement. As I have said > > > > time > > > > and time again, for every two paths there is a single node where they > > > > separate. On the other hand, there is not a single level where all > > > > the > > > > paths are separated. > > > > > > Do you claim that all paths exist as separated entities? If so: how or > > > where do they exist? > > > > Philosophy? I would expect mathematics in this newsgroup. > > I think that numbers belong to mathematics. Issues of how or where numbers exist are philosophical, not mathematical. > > > > > Can you explain what you mean by existence of > > > real numbers at all? > > > > Dedekind-cuts, Cauchy-sequences, whatever you want. > > > > > How are they related to separated paths? > > > > Each path is (in principle) a Cauchy-sequence. > > To be or not to be, that is the question. If yes, then the tree must > have a width of 2^aleph0. If not, then there are no irrational numbers > - nowhere, neither in the tree nor else in mathematics. Perhaps not in WM's philosophy, but there are more things in heaven and earth, WM, than are dreampt of in your philosophy. > > > > > 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. > Only a countable number > of them is admitted simultaneously? All if them simultaneoulsy, and some twice.
From: Virgil on 11 Apr 2007 14:00 In article <1176300393.408149.165530(a)w1g2000hsg.googlegroups.com>, mueckenh(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. > 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. > > > > 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.
From: Virgil on 11 Apr 2007 14:05
In article <1176300643.280458.120840(a)e65g2000hsc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 11 Apr., 03:29, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1175967792.541552.222...(a)q75g2000hsh.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 4 Apr., 15:48, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > > > If you unite natural numbers, you cannot get an > > > > > > > infinite number but only an infinite set of natural > > > > > > > numbers. Why is this different for paths? > > > > > > > > > > > > Careless again. You get an infinite number, but not an > > > > > > infinite natural number. The reason is simply that > > > > > > natural numbers are finite by definition. > > > > > > > > > > Not by definition, but by nature. > > > > > > > > Natural numbers have nothing to do with nature. From the > > > > definition it immediately follows that all natural numbers are > > > > finite. > > > > > > Without nature (or reality, as we say today) there would not be > > > any natural number. > > > > And, so what? If I now define that the natural numbers are the > > ordinal numbers, there is nothing wrong with that. It is just > > contrary to common nomenclature. So, no, finiteness of the natural > > numbers is *not* because of nature, but it is because of > > definition. > > You could define a triangle with four corners. It would be as > meaningful and as possible. Note that in mathematics one defines finiteness (of sets) in terms of naturals, not the other way round. A set is finite if and only if it can be put into bijection with the set if naturals less than some fixed natural. > > > I may put in your mind another system where '0' is also called a > > natural number. Mathematicians use terms as they see fit. It is > > the same with physicians. How red is a red quark? And in what way > > is it red? > > Physicists will never look for a red quark because of wavelength > problems. Irrelevant as usual! |