Cantor Confusion
in [Math]
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From: WM on 16 May 2007 16:16 On 16 Mai, 04:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178808468.554108.46...(a)y80g2000hsf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 10 Mai, 03:59, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > There is a slight difference. Finite definitions do not imply finitely > > > definable. Consider the computable numbers as they are presented by halting > > > Turing machines. The definition is indeed finite. > > > > The definition of the set is finite, not of the numbers. > > That is your interpretation. I would consider each of the numbers in the > set to be finitely defined, but the set not finitely defined. > > > > Nevertheless they are > > > considered not to be finitely defined because it is impossible to find out > > > whether a particular Turing machine indeed *does* halt. > > > > Therefore only few of the numbers are finitely definable. > > Each of the numbers is finitely definable, because each is a halting Turing > machine, and Turing machines are finitely definable. It is the set that is > not finitely definable. If each number is finitely defined, then there are only finitely many numbers, because there are only finitely many Turing machines. Of course, this finite set then is finitely defined too. > > > > > Yes, of the set. But it is only a definition of those real numbers for > > > > which Cauchy sequences can be finitely defined. > > > > > > No. It is a definition of the set. It is not a definition of the real > > > numbers at all. > > > > A well defined Cauchy sequence, for pi, for instance, makes pi well > > defined - not as a number in the sense of MatheRealism, but as a > > number in the sense of mathematics. > > Now you come up again with a new term. 'Well defined'. What is > 'well defined'? A number is well defined if it is defined by a finite number of words such that, in principle (i.e., given an infinite amount ressources) the Cauchy epsilon can be made arbitrary small. > > > > > I gave a bijection between the set of nodes and the set of branching- > > > > offs of paths bunches. > > > > > > You did not. What node is bijected with 0.1010101010...? > > > > Please read: > > I gave a bijection between the set of nodes and the set of branching- > > offs of paths bunches. > > Yes. I do not understand. 0.1010101010... is a path bunch. So there is > a branching-off of this path bunch. And you claim a bijection with the > nodes. I showed that there are countable sets which cannot be bijected with the set of natural numbers, for instance the set of finitely definable numbers. > What node is bijected with the branch-off of 0.101010101010...? I claim that there are no more branching-offs than odes and that there are no more real numbers represented in the tree than are branching- offs. There is no path ever finished, but it is only branching off from other paths in infinity. But the number of paths separated from other paths cannot surpass the number of branching-offs. > > > > > As there can be not more results of branching- > > > > offs than branching-offs, the task is done. > > > > > > As in each node the number of path bunches decreases I wonder about this > > > statement. > > > > In each node the number of path bunches increases. > > This is utter nonsense. *All* path bunches start at the root. Or are you > now considering bunches that do *not* start at the root? > > > If we add one bunch > > going into the first node, we an say: In all nodes one bunch goes in > > and two come out. > > Makes no sense. As all bunches start at the root, there cannot be more > bunches that come out of a node than go in. By your own definitions a > bunch that comes out of a node, also comes in it (when it arrives there > from the root). One bunch goes in, because the two which come out have not yet been separated when the paths which they consist of, went in. The only question is how man *separated* bunches are present. And this number is not larger than the number of nodes. Regards, WM
From: WM on 16 May 2007 16:19 On 16 Mai, 04:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178808886.504395.245...(a)w5g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 10 Mai, 04:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > I will answer there. > > > > > > I will not see it, so you clearly cut short the discussion. > > > > Please try, we could reduce the noise significantly and could kill any > > polemics and insults. If you are not able to see it, I will repeat it > > here. But I wonder why Google introduced this technique if it remains > > inaccessible to most users and even to experts. > > It is not inaccessible to me. So you saw my answer? This set F of functions can be bijected with the set R of reals. The reals form an intercession with the rationals Q. The rationals form a bijection with the naturals N. So use an intercession of R and Q. Replace every r of R by a function f(n) (by the first bijection F <--> R). Replace every q of Q by a natural n (by the second bijection Q <--> N). Done. > But I do not want to sign up with google > just for the privilege of accessing it. And I do not want to use a web > browser to access discussions. Why google introduced it beats me. I > prefer my keyboard based access to newsgroups, no mouse for me. If you are interested in the discussion which we had about Cantor's mistake or not mistake concerning his second diagonal proof, you may enter http://groups.google.com/group/sci.math.research/browse_frm/thread/5d9926c33ee6e40f?scoring=d&hl=de Regards, WM PS: What about the critique of ma chapter 10?
From: Virgil on 16 May 2007 19:42 In article <1179345320.393851.200920(a)u30g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 14 Mai, 22:09, William Hughes <wpihug...(a)hotmail.com> wrote: > > On May 14, 3:17 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > Which of the following statements is not true > > > > Every element of R is different from p. > > > > Every node of p is in some element of R > > > Back to the facts Back FROM the fact, more like. > > 1) No element r =/= p of R is required to cover any node of p. (I can > show of every such element r =/= p which you may name that it is not > required.) > 2) No element of R is sufficient to cove all nodes of p. > Hence your set R is not suitable to demonstrate your claims. Let R be the set of all finite initial subsets of N. No member of R covers N yet R itself does. For every p' with p' =/= p, the set of levels at which they have nodes in common is a finite subset of N, and every finite initial subset of N is represented by many such paths, so taking R as any infinite set of paths such that for any n in N there is path with at least n nodes in common satisfies: Every element of R is different from p AND Every node of p is in some element of R SIMULTANEOUSLY. That Wm is willfully blind to the necessity that both to be satisfied by such an R does not make it false, it only makes WM willfully blind to the truth of what transpires in ZF and NBG.
From: Virgil on 16 May 2007 19:53 In article <1179346082.257799.291260(a)k79g2000hse.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 16 Mai, 03:55, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1178795907.957410.94...(a)o5g2000hsb.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > > On 10 Mai, 03:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > So in your opinion there are only finitely many paths in the > > > > > > infinite tree. > > > > > > > > > > I did not say that. > > > > > > > > Quote: "Only from those few you can ask for". I can ask only for > > > > finitely > > > > many paths. > > > > > > That is correct. But that does not mean that I said that there are > > > only finitely many paths in the tree, as you implied. > > > > Why not? I can ask only for finitely many paths, and (if I interprete your > > thinking correctly) path which I can not ask for do not exist. > > That MatheRealism. Which is Monumentally Misnamed. WMRealism has nothing whatever to do with mathematics, nor much to do with reality, as far as that goes. > > This sentence makes no sense. I do not ask it for "all natural numbers > > which > > obey this formula", I ask it for "all natural numbers". And what natural > > numbers that obey that formula do not obey that formula? > > Up to each one they obey it. *All* natural numbers obviously do not > obey the formula. The only way in which a formula can fail to be true for "all naturals" is if for at least one of them it is false. The negation of "For all x, f(x)" is "for at least one x, not f(x)". So that WM is specifically claiming that there is at least one natural, n, for which the formula fails, but cannot name one.
From: Virgil on 16 May 2007 20:19
In article <1179346617.111313.267350(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 16 Mai, 04:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > Each of the numbers is finitely definable, because each is a halting Turing > > machine, and Turing machines are finitely definable. It is the set that is > > not finitely definable. > > If each number is finitely defined, then there are only finitely many > numbers, because there are only finitely many Turing machines. If there are only finitely many then you should be able to say how many. In fact there is no finite limit on the number of Turing machines. Of > I showed that there are countable sets which cannot be bijected with > the set of natural numbers, for instance the set of finitely definable > numbers. Then there must exist an injection from the set of finitely define]able numbers to the set of naturals or a surjection from the set of naturals to the set of finitely definable numbers, as absent both of these, there is no way to show that the set of finitely definable numbers is actually countable. > > > What node is bijected with the branch-off of 0.101010101010...? > > I claim that there are no more branching-offs than odes Getting poetic? But note that each 'branching off' contains as many paths as in the entire tree. > and that there > are no more real numbers represented in the tree than are branching- > offs. That mistakenly implies that branchings off are all single paths, or at least finite sets of paths, which is trivially false. The set of paths branching off from some p at any node of p is as large as the set of paths of the entire CIBT. > There is no path ever finished, but it is only branching off > from other paths in infinity. But the number of paths separated from > other paths cannot surpass the number of branching-offs. The number of paths in EACH branching off is uncountably infinite .. Through every node in the entire tree pass as many paths as issue from the root node. > > > > > > > > As there can be not more results of > > > > > branching- > > > > > offs than branching-offs, the task is done. False claims finish up nothing. > > > > > If we add one bunch > > > going into the first node, we an say: In all nodes one bunch goes in > > > and two come out. Nothing "goes into" the root node. But in any case, bunches are irrelevant. Every "bunch" in a CIBT contains contains uncountably many paths. There are no "bunches" of any other size. > > > > Makes no sense. As all bunches start at the root, there cannot be more > > bunches that come out of a node than go in. By your own definitions a > > bunch that comes out of a node, also comes in it (when it arrives there > > from the root). > > One bunch goes in, because the two which come out have not yet been > separated when the paths which they consist of, went in. The only > question is how man *separated* bunches are present. And this number > is not larger than the number of nodes. But the number of bunches coming out of a node is irrelevant when each of them contains uncountably many paths, which they all do. |