Why can no one in sci.math understand my simple point?
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From: Transfer Principle on 17 Jun 2010 20:57 On Jun 17, 12:59 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 17, 1:54 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > Does ZFC not prove that all constructible numbers are countable? > I don't know. What is the definition IN THE LANGUAGE of ZFC of > 'constructible number'? I know that the word "constructible" occurs in the context of V=L, where L is called the "constructible universe." So it is possible to call the elements of L intersect R "constructible reals"? (Of course, even if we can, I'm not sure what effect, if any, this would have on the cardinality of R.)
From: |-|ercules on 17 Jun 2010 21:42 "Transfer Principle" <lwalke3(a)lausd.net> wrote > On Jun 17, 6:56 am, Sylvia Else <syl...(a)not.here.invalid> wrote: >> On 15/06/2010 2:13 PM, |-|ercules wrote: >> > the list of computable reals contain every digit of ALL possible >> > infinite sequences (3) >> Obviously not - the diagonal argument shows that it doesn't. > > But Herc doesn't accept the diagonal argument. Just because > Else accepts the diagonal argument, it doesn't mean that > Herc is required to accept it. > > Sure, Cantor's Theorem is a theorem of ZFC. But Herc said > nothing about working in ZFC. To Herc, ZFC is a "religion" > in which he doesn't believe. > > Else's post, therefore, is typical of the posts which seek > to use ZFC to prove Herc wrong. To say the list of computable reals DOES NOT contain every digit (in order) of ALL possible infinite sequences is to say this list does not contain every digit (in order) of PI. 3 31 314 .... Herc
From: Ross A. Finlayson on 17 Jun 2010 21:45 On Jun 17, 1:49 am, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-17, Peter Webb <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > "Virgil" <Vir...(a)home.esc> wrote in message > >> An infinite set is defined to bee countable if and only if there is a > >> surjection from the set of natural numbers to that set. When such a > >> function is a bijection, it is commonly called a list. > > > Only if the bijection can be explicitly created. > > You apparently have some bizarre private definition of "list". > > Explicitness has nothing to do with it. > > Though even if it did, you are incorrect. Given any numbering of > Turing machines, a list of computable reals ordered by the > least-numbered Turing machines that compute them is quite explicit. > > The practical difficulties of establishing which Turing machines halt, > which are equivalent and so on are just that: practical difficulties > which have nothing to do with mathematical theory. > > - Tim Uncountability of the rationals? The rationals are well known to be countable, and things aren't both countable and uncountable, so to have a reason to think that arguments about the real numbers that are used to establish that they are uncountable apply also to the rationals, the integer fractions, has for an example in Cantor's first argument, about the nested intervals, that the rationals are dense in the reals, so even though they aren't gapless or complete, they are no- where non-dense, they are everywhere dense on the real number line. So, even though the limit of the sequences that are alternatively sampled to bound the minimum might not be in the rationals, to be an irrational number in the reals, still at every step there are more rationals then between the limiting value defined by the function from the naturals to the real numbers (or rational number). Between them are some non-rational numbers, the irrational numbers, not integer fractions although everywhere sums of them with infinite sums like Cauchy sequences besides the finite, the rationals are not continuous anywhere, although the normal definition of continuity as is presented includes the rationals. Many of the arguments that result from normal definitions of continuity apply to the rational numbers which in analysis are generally ignored in deference to the real numbers. That's still the general result of approximation, here the function that defines the evolution of the minimum bounds defined upper and lower by the two sequences of the evens and odds of F(n) as defined preserves for example Markov sequence, error term, ergodicity, of course any results. Also preserves there, the function, looking at all the functions from the natural integers to the real numbers, besides of course any results, neat translations from [0,1] to for example [-1/2,1/2], or, [-1,1], i.e., average zero. Warm regards, Ross Finlayson
From: Virgil on 17 Jun 2010 22:03 In article <5400048f-9123-4823-bf4a-06769640ef87(a)q12g2000yqj.googlegroups.com>, Transfer Principle <lwalke3(a)lausd.net> wrote: > On Jun 17, 3:16�am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > Virgil <Vir...(a)home.esc> writes: > > > In article > > > <995d761a-f70f-4bca-b961-8db8e1663...(a)d37g2000yqm.googlegroups.com>, > > > �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > >> On 15 Jun., 22:24, Virgil <Vir...(a)home.esc> wrote: > > >> > Note that it is possible to have an uncomputable number whose > > >> > decimal expansion has infinitely many known places, so long as it > > >> > has at least one unknown place. > > >> That is mathematically wrong. > > > It may not match every definition of 'uncomputable', but otherwise it is > > > right. > > It doesn't really match any definition of 'computable'. > > It's a rare day indeed when Aatu and WM actually agree! > > They both agree that Virgil's definition of "uncomputable" > is wrong. Let's see what's at stake here. > > We all agree that Chaitin's omega is uncomputable, so let > us define another real number as follows. > > Suppose we regard Chaitin's omega in let the number x equal > 1 if there exists a natural number N such that for all > natural numbers n>N, we have that a majority of the first n > digits of omega are 1's, and let x equal 0 othewise. So now > we ask, is x a computable real? > > By Virgil's definition, it isn't, since there's no Turing > machine that can compute whether x=0 or x=1. > > But by Aatu and WM, it is computable, since after all, x > must be either 0 (which is computable), or 1 (which is also > computable), so in either case, x is computable. > > We recall that in classical logic, we know that from P->R > and Q->R, we conclude (PvQ)->R. So we have: > > P <-> "x=0" > Q <-> "x=1" > R <-> "x is computable" > > So P->R is "if x=0, then x is computable" (which is true, > since 0 is computable). > > So Q->R is "if x=1, then x is computable" (which is true, > since 1 is computable). > > So we conclude (PvQ)->R, which is "if x=0 or x=1, then x > is computable." And we know that x must be 0 or 1. Thus, > by Modus Ponens, x is computable. QED > > So which definition of computable, Virgil's or Aatu-WM's, > should we use? I'm not sure which of the two definitions > is more prevalent. We could distinguish between them by using "uncomputable" for one of them and "noncomputable" for the other. Since so many seem to think that mine should NOT be called "uncomputable, lets call mine "noncomputable".
From: Ross A. Finlayson on 17 Jun 2010 22:59
On Jun 17, 6:45 pm, "Ross A. Finlayson" <ross.finlay...(a)gmail.com> wrote: > On Jun 17, 1:49 am, Tim Little <t...(a)little-possums.net> wrote: > > > > > On 2010-06-17, Peter Webb <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > > "Virgil" <Vir...(a)home.esc> wrote in message > > >> An infinite set is defined to bee countable if and only if there is a > > >> surjection from the set of natural numbers to that set. When such a > > >> function is a bijection, it is commonly called a list. > > > > Only if the bijection can be explicitly created. > > > You apparently have some bizarre private definition of "list". > > > Explicitness has nothing to do with it. > > > Though even if it did, you are incorrect. Given any numbering of > > Turing machines, a list of computable reals ordered by the > > least-numbered Turing machines that compute them is quite explicit. > > > The practical difficulties of establishing which Turing machines halt, > > which are equivalent and so on are just that: practical difficulties > > which have nothing to do with mathematical theory. > > > - Tim > > Uncountability of the rationals? The rationals are well known to be > countable, and things aren't both countable and uncountable, so to > have a reason to think that arguments about the real numbers that are > used to establish that they are uncountable apply also to the > rationals, the integer fractions, has for an example in Cantor's first > argument, about the nested intervals, that the rationals are dense in > the reals, so even though they aren't gapless or complete, they are no- > where non-dense, they are everywhere dense on the real number line. > So, even though the limit of the sequences that are alternatively > sampled to bound the minimum might not be in the rationals, to be an > irrational number in the reals, still at every step there are more > rationals then between the limiting value defined by the function from > the naturals to the real numbers (or rational number). Between them > are some non-rational numbers, the irrational numbers, not integer > fractions although everywhere sums of them with infinite sums like > Cauchy sequences besides the finite, the rationals are not continuous > anywhere, although the normal definition of continuity as is presented > includes the rationals. Many of the arguments that result from normal > definitions of continuity apply to the rational numbers which in > analysis are generally ignored in deference to the real numbers. > That's still the general result of approximation, here the function > that defines the evolution of the minimum bounds defined upper and > lower by the two sequences of the evens and odds of F(n) as defined > preserves for example Markov sequence, error term, ergodicity, of > course any results. Also preserves there, the function, looking at > all the functions from the natural integers to the real numbers, > besides of course any results, neat translations from [0,1] to for > example [-1/2,1/2], or, [-1,1], i.e., average zero. > > Warm regards, > > Ross Finlayson So, given two functions, how do you tell which is definitely less likely to map to enough of the reals to be like the continuum, compared to one like the mapping to the rationals which would be, for example the fractions in order? That is where the functions define the inductive lines and inputs etcetera. Well, that's where it is the, uh: real number line, along the constructive lines with the, uh, Bishop and Cheng constructive real numbers, rather restricted transfer principle, partially ordered ring and complete ordered field (transferring constructive results like the rationals to continuum results). Simple real numbers, R^bar and R^umlaut, R with a line on it and R with two dots, they work compatibly with fundamental results from standard real analysis, with the at-once line of points from point to point that are the same real numbers, these are the real numbers. This would generally be from the origin. As normal real pure numbers with the geometric identity as points in their normal standard definition, the two functions or n- many functions, in the Cantor approaching functions, simply define a partitioning of the natural integers along inductive lines, into fractions. Then the Cantor pursuit set is the line, basically a function to the rationals from the integers, to the reals. Along standard lines: they have the same range, the line is the points on the line, the Cantor pursuit set and the line segment of the line.. In that way then it's pretty trivial along exact and complete lines. Yeah, that way, too. Technically: it's pretty trivial along exact and complete lines. Warm regards, Ross Finlayson |