Why can no one in sci.math understand my simple point?
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From: Virgil on 19 Jun 2010 18:25 In article <f91b4fc9-7aa9-4459-81e5-a9a9dbc70ffe(a)d37g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > A crank suffers from selective perception of reality, doesn't he? WM certainly does. If he thinks that makes him a crank, I, for one, will not try to change his mind on that point.
From: Virgil on 19 Jun 2010 18:27 In article <f9ecf9b6-02a1-4f84-9fac-f5ca6c57b844(a)x27g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 19 Jun., 08:37, "Peter Webb" > > In Cantor's diagonal proof, the list of Reals is provided in advance, such > > that the nth digit of the nth item is known. > > Where can I see such a list? Is it available for cheap money in the > net? Such lists abound. The identity function on N is one of them. In fact any real function with domain N is one.
From: Virgil on 19 Jun 2010 18:31 In article <492b178b-0e94-4c93-b5f7-749ac339d378(a)h13g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 19 Jun., 10:06, Sylvia Else <syl...(a)not.here.invalid> wrote: > > On 19/06/2010 4:11 PM, |-|ercules wrote: > > > > > To support your argument you should at least show that you've formed a > > > new sequence of digits. > > > > I'll explain it simply then. The first digit of the created number > > differs from the first digit of the first number in the list. The second > > digit differs from the second digit of the second number in the list. > > > > In general, digit n differs from digit n of the nth number in the list. > > > > So for all n, the created number differs from number n. Therefore the > > created number is not in the list - it is a new sequence of digits. > > Who constructed your list? Has it been constructed in an infinite > process? Or has it been defined by a finite definition? In FOL+ZFC things exist without being "constructed" in WM's sense.
From: Virgil on 19 Jun 2010 18:59 In article <10c118d2-4751-4fd9-ade8-30c2f26afe7f(a)i31g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 19 Jun., 11:07, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > > The list of computable reals contains every digit (in order) of all > > possible infinite sequences. > > Hi Herc, > > why not instead of a list of all reals produce a Binary Tree. This > tree can be shown to produce every infinite binary sequence that can > be produced by the following step-by-step construction. This > construction is possible, because the set of all nodes is a countable > set and all paths exist among the nodes and nowhere else. The > construction is as follows: > > The Binary Tree contains all real numbers of the interval [0, 1] as > infinite paths. > > 0, > / \ > 0 1 > / \ / \ > 0 1 0 1 > / > 0 ... > > The nodes K_i with numerical values 0 or 1 are countable: > > K_0 > / \ > K_1 K_2 > / \ / \ > K_3 K_4 K_5 K_6 > / > K_7 ... > > Everey step adds one node to the configuration B_i and yields > configuration B_(i+1) > > _________________ > B_0 = > > K_0 > _________________ > B_1 = > > K_0 > / > K_1 > _________________ > B_2 = > > K_0 > / \ > K_1 K_2 > _________________ > B_3 = > > K_0 > / \ > K_1 K_2 > / > K_3 > _________________ > > B_4 = > > K_0 > / \ > K_1 K_2 > / \ > K_3 K_4 > _________________ > ... > _________________ > B_j = > > K_0 > / \ > K_1 K_2 > / \ > K_3 K_4 ... > ... > ... K_j > _________________ > ... > _________________ > > There is no end, hence there is no node that is not constructed. If > there is no infinite path constructed at all, this means either that > infinite paths consist not only of nodes (but of phantasy-products of > matheologicians) or they do not exist at all. > > The latter is true. There is no completed infinite path but there is > merely the possibility to add a node to any path of any finite length. > But that does not yield an uncountable set of paths. Note than in FOL+ZFC, the set N necessarily exists, and, with suitable definition of left-child and right-child is already a complete infinite binary tree with suitable subsets denoting paths. E.g., for n = {0,1,2,3,...} let the left child of n be 2*n+1 and the right child of n be 2*n+2 Then one has immediately a complete infinite binary tree with infinite paths. E.g., the path having only left branchings is {2^n-1: n in N} So that what WM claims abut binary trees does not hold in FOL+ZFC, nor in any system in which a set like N exists.
From: Virgil on 19 Jun 2010 19:05
In article <dd3dc65c-aa76-4921-9696-aead077ab221(a)k39g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 19 Jun., 16:00, Sylvia Else <syl...(a)not.here.invalid> wrote: > > > ZFC makes claims in the context of ZFC. You can't take it down using a > > different set of axioms, because ZFC doesn't make statements under those > > other axioms. If you want to attack ZFC, as distinct from inventing > > competing sets of axioms, your only viable course is to seek to show > > that it is inconsistent. > > There are things more elementary than ZFC. Induction for instance: If > a list is constructed as follows: > > 0.0 > 0.1 > 0.11 > 0.111 > ... > > then the anti-diagonal p = 0.111... does not exist or it is in one and > the same line of the list. Until WM presents a complete system, such as FOL+ZFC represents, there is no way to tell if FOL+ZFC can be imedded in WM's system, and unless it can be embedded, what happens in WM's system is irrelevant to what can happen in ZFC. A long and irrelevant argument about what happens in WM's system but outside FOL+ZFC deleted. |