Why can no one in sci.math understand my simple point?
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From: WM on 19 Jun 2010 10:52 On 19 Jun., 08:12, Virgil <Vir...(a)home.esc> wrote: > > No, for the fifth time now. The antidiagonal of a list of all > > computable real is not computable. How many more times would you like > > me to repeat this simple and mathematically obvious statement? Please stop to repeat that. It is completely irrelevant. There is a list that contains all finite words. It includes all definable, constructible, computable numbers. > > There is, however, some question in my mind about the existence of a > list of all and ONLY computable reals. Completely irrelevant. > > For countability of a set it is certainly sufficient to have a list > containing all its members even if that list is allowed to contain other > things as well.- Correct. All numbers that somehow can be identified belong to a countable set. Therefore Cantor either proves the uncountability of a countable set (by producing a diagonal number that can be identified) or he identifies an unidentifiable disgonal number. Both is a contradiction. And I am very sorry for those "logicians" who try to escape this simple truth by sophisticated definitions that yield unidentifiable identities and other foolish notions. A crank suffers from selective perception of reality, doesn't he? Regards, WM
From: WM on 19 Jun 2010 10:56 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? There has never been an infinite Cantor-list other than definied by a finite definition. And that means there are at most counatble many lists and countably many diagonals (which are necessarily also defined by finite definitions). Regards, WM
From: WM on 19 Jun 2010 11:00 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? There are only countable many such definitions. And there are only countable man rules to construct the anti diagonal. What is your conclusion? Regards, WM
From: WM on 19 Jun 2010 12:27 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. Regards, WM
From: |-|ercules on 19 Jun 2010 16:03
Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote ... > Newberry <newberryxy(a)gmail.com> writes: > >>> Because every infinite sequence of digits represents a real number? And >>> the antidiagonal is one such sequence? >> >> If it does not exist then it does not represent anything let alone a >> number. >> >> Now it is clear that it does not exist. Since all the reals are on the >> list and the anti-diagonal would differ from any of them. This >> violates the assumption. Hence the anti-diagonal does not exist. > > Wow. Are you saying that the *sequence of digits* specified by the > anti-diagonal does not exist? > > Anyway, in a sense, you're right. If we assume that every real is > represented by a sequence on the list, then we can prove that every > sequence occurs on the list (ignoring the issue of multiple > representations). And yet, we can also show that a particular sequence > is not on the list. This is a contradiction and hence *our assumption* > that every real is represented by a sequence on the list must be false. > > You've never seen proof by contradiction in your life? How remarkable. > You see the words but you don't read. Cantor's proof should work if you give a definition for ALL antidiagonals instead of a SPECIFIC antidiagonal. CONSTRUCT A REAL An AD(n) =/= L(n,n) PROOF THAT IT'S NEW An AD(n) =/= L(n,n) THEREFORE [ An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) ] -> Higher Infinity Do you agree with the above proof? Herc |