Why can no one in sci.math understand my simple point?
in [Logic]
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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
From: Peter Webb on 17 Jun 2010 23:09 "Virgil" <Virgil(a)home.esc> wrote in message news:Virgil-11576A.01595017062010(a)bignews.usenetmonster.com... > In article <4c19cef9$0$17178$afc38c87(a)news.optusnet.com.au>, > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > >> "Virgil" <Virgil(a)home.esc> wrote in message >> news:Virgil-6240F4.21454316062010(a)bignews.usenetmonster.com... >> > In article <4c1995e5$0$26118$afc38c87(a)news.optusnet.com.au>, >> > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> > >> >> > >> >> > It is proof that there is no countable set of all real numbers, >> >> > since >> >> > any alleged such set is provably and constructably incomplete. >> >> > >> >> > Similarly, it is proof that there is no countable set of all >> >> > constructable numbers, since any alleged such set is provably and >> >> > constructably incomplete. >> >> >> >> I hate to disagree with you, because we are on much the same "side", >> >> but >> >> this is not correct. Cantor's proof shows that you cannot form a list >> >> of >> >> all >> >> Reals. This is not the same as the Reals being uncountable. >> > >> > If the reals were countable they would be listable, since such a list >> > would be a "counting" of them, so that NOT being listable implies NOT >> > being countable. >> >> That does not follow, and I have already provided a counter-example. >> Computable numbers are countable, but cannot be listed. >> >> > >> > 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. There are countable sets >> which cannot be listed, such as the countable set of computable Reals. >> >> > Since the set of reals is infinite but cannot be listed in this way, it >> > follows that the reals necessarily are NOT countable. >> >> >> By this (incorrect) logic, the computable numbers must also be >> uncountable. >> But they are not. >> >> >> >> >> >> You can use Cantor's diagonal construction to similarly prove that you >> >> cannot form a list of all computable numbers. However the computable >> >> numbers >> >> are in fact countable. You can't simply equate the two concepts; they >> >> are >> >> not exactly the same thing. >> > >> > For infinite sets, listability and countability are equivalent. >> >> No. Witness the infinite set of computable numbers. Countable but not >> listable. >> >> Cantor's proof shows that the set of Real numbers cannot be listed. It >> does >> not immediately follow that they are uncountable. Plenty of countably >> infinite sets cannot be listed. The set of computable numbers is one. The >> set of halting TMs is another. > > One can create lists which contain all of the computable numbers , > (or all of the halting TMs) but which, of necessity, list some other > things as well. No. Lets imagine you give me a list which is supposed to contain all computable numbers in [0, 1.0] , and possibly some other Reals. You can even have some computable Reals appearing multiple times in the list. I can use the Cantor diagonal construction to create a Real not on the list, which means that it is not a computable number. So the list cannot contain all computable numbers. That you cannot list some set is not the same as the set in uncountable. Computable numbers are countable but cannot be listed.
From: Peter Webb on 17 Jun 2010 23:18 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1jod8.jrj.tim(a)soprano.little-possums.net... > On 2010-06-17, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> "Virgil" <Virgil(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. > There are a countable number of computable Reals. You can apply the Cantor construction to any purported list of all computable Reals to form a computable Real not on the list. This proves that the computable Reals cannot be listed. It does *not* prove the computable Reals are uncountable, and in fact they are not. In exactly the same manner, Cantor proved that the Reals cannot be listed. This does *not* automatically mean they are uncountable, any more than the same proof applied to computable Reals proves they are uncountable. These are different concepts. (Although they were not when Cantor produced his proof).
From: Peter Webb on 17 Jun 2010 23:22 > > Since by definition, "listability" = "countability", Cantor's proof of > unlistability proves uncountability. Really? Where did you get that from? The computable Reals cannot be listed. Therefore according to you they are uncountable. But they aren't. Maybe your definition needs a little work?
From: Sylvia Else on 17 Jun 2010 23:25
On 18/06/2010 4:27 AM, WM wrote: > On 17 Jun., 15:56, Sylvia Else<syl...(a)not.here.invalid> wrote: >> On 15/06/2010 2:13 PM, |-|ercules wrote: >> >> >> >> >> >>> 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) >> >>> as pi is an infinite digit sequence, this means >> >>> this list contains every digit of an infinite digit sequence (2) >> >>> similarly, as computable digit sequences contain increasing lengths of >>> ALL possible finite prefixes >> >>> the list of computable reals contain every digit of ALL possible >>> infinite sequences (3) >> >> Obviously not - the diagonal argument shows that it doesn't. >> > > There is no diagonal element for a list of finite lines. The list of computable reals is not a list of finite lines. Sylvia. |