Why can no one in sci.math understand my simple point?
in [Logic]
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From: Virgil on 24 Jun 2010 15:18 In article <09e3f5b1-ae8f-4ec9-8f77-278914d90053(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 24 Jun., 11:04, Virgil <Vir...(a)home.esc> wrote: > > > > Thise prodedure is abbreviated by > > > (..., An, ... A2, A1, A0, L0) > > > but of course there cannot appear at any stage a list without first > > > line. > > > > But that list can be rearranged into a more standard form, for example > > with the An and the members of L0 listed alternatingly, and from such a > > rearrangement a non-member anti-diagonal can be constructed. > > What should that be good for? The countable set under investigation > contains the antidiagonal of the arragement that I prescribed. The "arrangement" you prescribe does not suport antidiagonal construction without rearrangement into standard list form. You claim a function from N to R (the reals) or B (the set of endless binary sequences) which contains its own antidiagonal. Nonsense! Cantor proved no such containment possible, and nothng WM has done has shown that proof flawed. > > Your arguing reminds of another argument: I can eat cherries, > therefore the list is neither complete nor incomplete. That certainly reminds me of many of WM's arguments. > > I for my person would not accept that. Perhaps set theorists have > another taste. Mathematicians accept that in certain set theories there re uncountable sets. > > > > > > It is as simple as that. > > > > > It is not a big deal. Set theory shows many contradictions arising > > > from the idea, that from "every n in N can be constructed" it is > > > implied that "N can be constructed". > > In, for example, FOL+ZFC, there is no assumption that every n in N "can > > be constructed". What is assumed is that there exists a set, along with > > all its elements, having the properties that we want for N and its > > elements , which we then call N. > > And being ready to be used for the construction of a bijection. Such bijections from to suitable other sets can often be "constructed", though the more common term is "defined". > > > > �This is wrong, because for every > > > > > n, there is an infinite set of unconstructed elements of N. > > > > But we do not "construct" any of them. Though we do construct various > > naming conventions for elements of N. > > Bijections from M to N are constructed. Sequences (having natural > indices) are constructed. Many constructivists do not believe in > uncountability but in N because N can be constructed. If your "constructed" means "defined" then yes. But in the Cantor theorem, the "antidiagonal" is defined (but not necessarily constructed) and since it defines a binary sequence not listed in the list from which it is defined, it shows that no list can include all such binary sequences, QED.
From: Virgil on 24 Jun 2010 15:20 In article <53058a2b-9675-469e-b3b2-095661656ef3(a)e5g2000yqn.googlegroups.com>, Newberry <newberryxy(a)gmail.com> wrote: > On Jun 23, 9:09�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <b0258a2f-1679-492d-99e9-3397093bb...(a)t10g2000yqg.googlegroups.com>, > > > > �Newberry <newberr...(a)gmail.com> wrote: > > > On Jun 23, 1:49 pm, Virgil <Vir...(a)home.esc> wrote: > > > > In article > > > > <8cd5db0d-46a5-4deb-8fe5-9a28acad5...(a)k39g2000yqb.googlegroups.com>, > > > > > > Newberry <newberr...(a)gmail.com> wrote: > > > > > Cantor's proof starts with the assumption that a bijection EXISTS, not > > > > > that it is effective. > > > > > > Actually, a careful reading shows Cantor's proof merely assumes an > > > > arbitrary INJECTION from N to R (originally from N to the set of all > > > > binary sequences, B) which is NOT presumed initially to be surjective, > > > > and then directly proves it not to be surjective by constructing > > > > > OK, so construct it assuming injection. > > > > It is simplest to do for an injection from N to B. > > A member of B, a binary sequence, �is itself a function from N to an > > arbitrary two element set which, without loss of generalization, we may > > take to be S = {0,1}. > > A list of such functions is then equivalent to a single function from > > NxN, the Cartesian product to {0,1}, say F(-,-), so that for each m in > > N, the function F(m,-): N -> {0,1}: m |--> F(n,m) is one of the binary > > functions in the list. > > For any such list of binary functions, define a new function > > � �g(-):N -> {0,1}:n |--> 1 - F(n,n), or more briefly, g(n) = 1-F(n,n). > > > > This new function is the desired "antidiagonal" for the list of lists > > F(-,-) and g(-) is not in the original list since g(-) differers from > > each F(n,-) at n. > > Now costruct it when F is not effectvely computable. I don't have to construct it, merely define it. > > > > > > > > > > > > > > > the > > > > "antidiagonal"as a member of the codomain not in the image. > > > > > > Thus proving that ANY injection from N to R (or B) fails to be > > > > surjective. > > > > > > For some unknown reason, the DIRECT "anti-diagonal" proofs given by > > > > Cantor, and his followers, are often misrepresented as being proofs by > > > > contradiction, but they never were.
From: Mike Terry on 24 Jun 2010 15:26 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:41a66545-c0bb-44e3-bb41-1fdaf0c83d71(a)j4g2000yqh.googlegroups.com... > On 24 Jun., 01:02, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > > > a) Does this list contain the anti-diagonal of > > > (..., An, ... A2, A1, A0, L0)? > > > > This is not a list of numbers. L0 is not a number, it is a list. > > > > Therefore (..., An, ... A2, A1, A0, L0) does not have an anti-diagonal. > > You are wrong. I am obviously right. What you have written is not a list of real numbers. Anti-diagonals (as used in Cantor's proof) require as "input" a list of real numbers. Therefore, what you've written does not have an anti-diagonal in the sense of Cantor's proof. > > The symbols above abbreviate the sequence of lists > > An > ... > A0 > L0 Which I'll take that you're trying to say that that (..., An, ... A2, A1, A0, L0) abbreviates the SEQUENCE of lists (L0, L1, L2...) (since by your construction, (An, ...A0, L0(0), L0(1),...) is the list L_(n-1).) NOTE: a sequence of lists does not have an antidiagonal, so your earlier question: > > > a) Does this list contain the anti-diagonal of > > > (..., An, ... A2, A1, A0, L0)? is nonsense, as I pointed out. > > Each of them has an antidiagonal .... yes, the antidiagonal of Ln is An.. > that either is not in the list .... sorry, what list? Do you mean the list of sequences (L0, L1, ...) or the list Ln, or something else? > (then > the set of all of them is unlistable) or is in a list. Then Cantors > argument is wrong. > > Regards, WM
From: Newberry on 24 Jun 2010 23:34 On Jun 24, 12:20 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <53058a2b-9675-469e-b3b2-095661656...(a)e5g2000yqn.googlegroups.com>, > > > > > > Newberry <newberr...(a)gmail.com> wrote: > > On Jun 23, 9:09 pm, Virgil <Vir...(a)home.esc> wrote: > > > In article > > > <b0258a2f-1679-492d-99e9-3397093bb...(a)t10g2000yqg.googlegroups.com>, > > > > Newberry <newberr...(a)gmail.com> wrote: > > > > On Jun 23, 1:49 pm, Virgil <Vir...(a)home.esc> wrote: > > > > > In article > > > > > <8cd5db0d-46a5-4deb-8fe5-9a28acad5...(a)k39g2000yqb.googlegroups.com>, > > > > > > Newberry <newberr...(a)gmail.com> wrote: > > > > > > Cantor's proof starts with the assumption that a bijection EXISTS, not > > > > > > that it is effective. > > > > > > Actually, a careful reading shows Cantor's proof merely assumes an > > > > > arbitrary INJECTION from N to R (originally from N to the set of all > > > > > binary sequences, B) which is NOT presumed initially to be surjective, > > > > > and then directly proves it not to be surjective by constructing > > > > > OK, so construct it assuming injection. > > > > It is simplest to do for an injection from N to B. > > > A member of B, a binary sequence, is itself a function from N to an > > > arbitrary two element set which, without loss of generalization, we may > > > take to be S = {0,1}. > > > A list of such functions is then equivalent to a single function from > > > NxN, the Cartesian product to {0,1}, say F(-,-), so that for each m in > > > N, the function F(m,-): N -> {0,1}: m |--> F(n,m) is one of the binary > > > functions in the list. > > > For any such list of binary functions, define a new function > > > g(-):N -> {0,1}:n |--> 1 - F(n,n), or more briefly, g(n) = 1-F(n,n).. > > > > This new function is the desired "antidiagonal" for the list of lists > > > F(-,-) and g(-) is not in the original list since g(-) differers from > > > each F(n,-) at n. > > > Now costruct it when F is not effectvely computable. > > I don't have to construct it, merely define it. In this sub-thread we are debating if it makes sense to say that the anti-digonal does not exist. The objection was that it must exist because we can construct it. If that is what you contend then you do have to construct it. > > > > > the > > > > > "antidiagonal"as a member of the codomain not in the image. > > > > > > Thus proving that ANY injection from N to R (or B) fails to be > > > > > surjective. > > > > > > For some unknown reason, the DIRECT "anti-diagonal" proofs given by > > > > > Cantor, and his followers, are often misrepresented as being proofs by > > > > > contradiction, but they never were.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: WM on 25 Jun 2010 06:13
On 24 Jun., 21:26, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: There is a starting list L0 and many following lists. During construction step n+1 the list is An .... L0 and has an antidiagonal. The antidiagonals of all these lists are countable but canot be written in form of a list (it does not matter if in a separate list or in a list with spaces or else), because, if written in form of some list, they would yield an antidiagonal that was not in the list. Hence all antidiagonals of this process, though countable, cannot be written in a list. Regards, WM |