Why can no one in sci.math understand my simple point?
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From: WM on 16 Jun 2010 13:53 On 15 Jun., 21:11, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > 1) What you write is sloppy (mathematically) and so it's not > clear what you intend to say with your individual statements. But actually everybody can easily imagine what is meant. You prove it for your person below. > > > 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) > > Literally this just doesn't make sense. Obviously he means that this list contains every digit of pi with the correct index enumerating its position. In discussions like the present one it is not usual to state everything explicitly that can be understood by an experienced reader. Perhaps you mean to say: > > this list contains every finite prefix of the infinite > digit sequence for pi (1)? > > I would agree with that... > > (What you actually said is gibberish because the list is not a list of > digits. If we try to treat it as such, then the only digit in the list is > 3). A list of people can contain and usually does contain letters. Thiks list is not a set in orthodox set theory. > > [My suspicion at this point is that your gibberish wording is actually the > *key* in some way to how you want to introduce some incorrect conclusion, > but time will tell on that. If I'm right, you won't like my clarification, > because it will make it harder for you to express your mistake...] This suspicion is wrong. Every initial segment of the decimal expansion of pi is in at least one line of your list 3. 3.1 3.14 3.141 .... What we can find in the diagonal, namely 3.141 and so on, exactly that can be found in one line. This is obvious by construction of the list. > > Same again :) This is harder to guess what you're trying to say though... So consider this: Every part of the diagonal is in at least one line. That means, every part is in one single line, or there are parts that are in different lines but not in one and the same. The latter proposition can be excluded. If there are more than one lines that contain parts of pi, then it can be proved, be induction, that two of them contain the same as one of them. This can be extended to three lines and four lines and so on for every initial segment of n lines. Hence we prove that all of pi, that is contained in at least one of the finite lines of this list, is contained in one single line. Conclusion: Either the complete diagonal pi does not exist, or it exists also in one and the same single line (because every line that can contain a digit sequence, is a finite line). As the latter is wrong, so is the former. There does not exist an actually infinite sequence. Actually infinite mean finished infinite. That is nonsense. (The proof has the same status as Cantor's diagonal proof. Also his proof is valid only for all finite n. In a similar way we find above: For all finite n: All of pi that is in the list up to line n, is in a single line of the list.) > > There is no unique list of computable reals - however, people would agree > that the computable reals are countable, and so can certainly be enumerated. > Perhaps the actual list doesn't matter and we should just choose one? Choose this one for example: The proof of countability of S does not require a bijection of S with N but only an injection of some superset of S into N. This is established by my list 0 1 00 01 .... where every line may be enumerated by an element of the countable set omega^omega^omega (and, if required, finitely many more exponents for alphabets, languages, dictionaries, thesauruses, and further properties) An obvious enumeration of the lines is 1, 2, 3, ... where every line n can have many sub-enumerations n n.1.a n.11.a n.111.a .... where every section of 1's and a's contains enuogh symbols to enumerate every single line as often as required to cover all its meanings in all possible languages. If necessary we can add 2's and b's and so on and remain in the countable domain. > > > Also, now your talking about reals rather than digit sequences. These are > distinct objects, but there are obvious correspondences. In particular users of Cantor-lists do intermingle them. Digit sequences do not converge at all. Therefore infinite digit sequences cannot be used to express something meaningful. Regards, WM
From: WM on 16 Jun 2010 14:04 On 16 Jun., 19:45, MoeBlee <jazzm...(a)hotmail.com> wrote: > But, every countable set does have a bijection with N or has a > bijection with some member of w (the set of natural numbers). That would be true if countability and aleph_0 were not self- contradicting concepts. But there is a set that is less than uncountable but has no bijection with N or a definasble subset of N. This set is the set of all finite definitions (in binary representation). 0 1 00 01 .... where every line may be enumerated by an element of the countable set omega^omega^omega (and, if required, finitely many more exponents for alphabets, languages, dictionaries, thesauruses, and further properties) An obvious enumeration of the lines is 1, 2, 3, ... where every line n can have many sub-enumerations n n.1.a n.11.a n.111.a .... Every section of 1's and a's contains enough symbols to enumerate every single line as often as required to cover all its meanings in all possible languages. If necessary we can add 2's and b's and so on and remain in the countable domain. This set contains all possible finite words in all possible languages over all possible alphabets over all possible whatever may be required in addition. This set is countable. Hence the set of all definable, computable, and "whatever may be invented by clever set theorists" real numbers is countable as a subset. Cantor shows that a countable set is uncountable or that a constructed real is unconstructable. Regards, WM
From: MoeBlee on 16 Jun 2010 15:21 On Jun 16, 1:04 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 16 Jun., 19:45, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > But, every countable set does have a bijection with N or has a > > bijection with some member of w (the set of natural numbers). > > That would be true if countability and aleph_0 were not self- > contradicting concepts. What I mentioned is pretty much just a definition of 'countable'. x is countable <-> Ef(f is a bijection from x onto w or Enew x is a bijection from x onto n) Meanwhile, I don't have a thousand lifetimes to wait for you to show a formula P such that both P and ~P are derivable in ZFC from the above definition. MoeBlee
From: Virgil on 16 Jun 2010 15:37 In article <91a57f95-54ee-4f0a-8341-b2a7dc2f11de(a)h13g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 16 Jun., 05:37, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > Virgil <Vir...(a)home.esc> writes: > > > But until you can determine which of those 10 cases, how can you > > > compute the number? > > > > You can't. The number is computable nonetheless, in the sense that there > > exists an effective procedure for churning out its decimal expansion. > > > > As noted, computability is a purely extensional notion. Recall the > > classical recursion theory exercise, which we find, in some form or > > other, in pretty much any text on the subject: > > > > �Let f : N --> N be a function such that > > > > � �f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. > > > > �Is f computable? > > All computable, Turing-computable, nonetheless computable, definable > (in any useful, i.e., finite language), and by other means > determinable numbers form a countable set. I am not sure that in standard mathematics that that description can "form a set" at all, much less a countable one. > > If Cantor's diagonal proof results in any such a number, then it > proves in effect the uncountability of a countable set. If WM claims that every set of 'real' numbers is countable, meaning that one can construct a surjection from N to it, then the very constructability of such a surjection proves that set of numbers to be incomplete by the Cantor "anti-diagonal" construction.
From: MoeBlee on 16 Jun 2010 15:40
On Jun 16, 2:21 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > I don't have a thousand lifetimes to wait for you to show a > formula P such that both P and ~P are derivable in ZFC from the above > definition. CORRECTION: I should have said: [...] to show, for some formula P, a proof in ZFC (with said definition included) of P and a proof in ZFC (with said definition included)of ~P. MoeBlee |