Why can no one in sci.math understand my simple point?
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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
From: Charlie-Boo on 25 Jun 2010 08:26
On Jun 24, 8:49 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > On Jun 15, 2:15 am, "Peter Webb" > > >> No. You cannot form a list of all computable Reals. > > > Of course you can - it's just the list of Turing Machines. > > No, it's not. I asked for a counterexample, to no avail. Don't you think you should substantiate your statement or retract it? Each Turing Machine represents some computable real (all computable reals are included) and you can list those Turing Machines. The Turing Machine represents it as well as any other system of representation. C-B > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, dar ber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |