Why can no one in sci.math understand my simple point?
in [Logic]
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From: WM on 24 Jun 2010 05:52 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. Your arguing reminds of another argument: I can eat cherries, therefore the list is neither complete nor incomplete. I for my person would not accept that. Perhaps set theorists have another taste. > > > 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. > > 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. Regards, WM
From: Charlie-Boo on 24 Jun 2010 08:03 On Jun 15, 2:15 am, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > "|-|ercules" <radgray...(a)yahoo.com> wrote in message > > news:87ocucFrn3U1(a)mid.individual.net... > > > Consider the list of increasing lengths of finite prefixes of pi > > > 3 > > 31 > > 314 > > 3141 > > .... > > > Everyone agrees that: > > this list contains every digit of pi (1) > > Sloppy terminology, but I agree with what I think you are trying to say. > > > as pi is an infinite digit sequence, this means > > > this list contains every digit of an infinite digit sequence (2) > > Again sloppy, but basically true. > > > similarly, as computable digit sequences contain increasing lengths of ALL > > possible finite prefixes > > Not "similarly", but if you are claiming that all Reals which have finite > decimal expansions can be listed, this is correct. > > > the list of computable reals contain every digit of ALL possible infinite > > sequences (3) > > No. You cannot form a list of all computable Reals. Of course you can - it's just the list of Turing Machines. His mistake is simply that he is saying that a finite sequence of digits contains an infinite sequence of digits. C-B > If you could do this, > then you could use a diagonal argument to construct a computable Real not in > the list. > > > > > OK does everyone get (1) (2) and (3). > > No. (3) is not true, as it is based on a false premise (that the computable > Reals can be listed). > > > > > There's no need for bullying (George), it's just a maths theory. Address > > the statements and questions and add your own. > > > Herc > > -- > > If you ever rob someone, even to get your own stuff back, don't use the > > phrase > > "Nobody leave the room!" ~ OJ Simpson- Hide quoted text - > > - Show quoted text -
From: Aatu Koskensilta on 24 Jun 2010 08:49 Charlie-Boo <shymathguy(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. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Newberry on 24 Jun 2010 10:45 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. > > > > > > > > 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 -
From: Charlie-Boo on 24 Jun 2010 12:02
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. Witness? > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, dar ber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |