Why can no one in sci.math understand my simple point?
in [Logic]
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From: Peter Webb on 18 Jun 2010 22:59 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1oa08.jrj.tim(a)soprano.little-possums.net... > On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> That, BTW, is my own interpretation of what is happening. > > It is an incorrect interpretation. > > >> Whether you accept this or not, the simple fact is that Cantor's >> proof can be applied to any purported list of all computable Reals >> and used to generate a computable Real not on the list > > Your "simple fact" is simply wrong. Look up the definition of > "computable real" and get back to us. > Why? If you believe there is an error in what I have said, identify it. Or give me at least a hint. As a start, my argument rests on two observations: 1. Cantor's diagonal construction, when applied to a purported list of all computable Reals, will produce a computable Real not on the list. Therefore you cannot provide a list of all computable numbers. 2. The computable numbers are countable. Which one of these do you believe to be wrong? > > - Tim
From: Virgil on 18 Jun 2010 23:10 In article <79d04ee8-00f7-4aad-a499-2b9241ff3b6d(a)y11g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 18 Jun., 05:25, Sylvia Else <syl...(a)not.here.invalid> wrote: > > 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. > > It is. Every real number that is defined is defined by a finite word > (definition or formula). It is impossible to define a number by an > infinite sequence, because the sequence never ends and the definition > is never known. F:N -> R: n -> 3/10^n is an infinite sequence defining a real number. So as one can easily see it is possible to have an infinite sequence defining a real number. In real mathematics, an infinite sequence is merely a function whose domain is the set of natural numbers, and there are lots of them which define real numbers.
From: Peter Webb on 18 Jun 2010 23:11 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1oal8.jrj.tim(a)soprano.little-possums.net... > On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> I see that you have stopped responding to my arguments, and have >> employed the rather lame tactic of suggesting I look at definitions >> on the web. > > I don't care where you get your definitions from, so long as they are > not from your own orifices. You are using mathematical terminology in > a completely incorrect manner, and further discussion is unlikely to > be fruitful until you learn what the words you are using mean. > > >> In the mean time, how about answering two questions for me: >> >> 1. Do you believe it is possible to create a list of all computable >> Reals? > > What do you mean by "create"? Such an list can be proven to exist, > and I even provided a well-defined mapping from N to the set of > computable reals earlier in this thread. > Hmmm. But you cannot provide me with a list of all Computable numbers, can you? Which is what Cantor's diagonal proof requires. Of course there exists a mapping from N to computable numbers. But Cantor's proof requires more than that; it requires the mapping to be recursively enumerable such that we can also explicitly list them. That a set cannot be listed is not the same as it is uncountable, as the set of Computable numbers clearly illustrates. You cannot give me a list of all Computable numbers, because I can use a diagonal construction to form a computable Real not on the list. Yet the set of computable Reals in countable. Cantor's proof applied to Reals does *not* prove they are uncountable; it merely proves what Cantor said it proved, that they cannot be formed into a list. This is not neccesarily the same as being uncountable, witness the set of computable Reals which also cannot be listed but is in fact countable. > If that doesn't answer your question, you'll have to clarify what you > mean by it. > What I mean is that Cantor proved you cannot provide a list of all Reals. This does not mean that the Reals are uncountable. Exactly the same form of proof can be applied to all Computable Reals, and be used to prove that you cannot provide a list of all Computable Reals, either. Yet these are countable. Clearly the predicate "cannot be listed" is *not* the same as "is uncountable", because computable Reals have the first property but not the second. And to the extent that Cantor's diagonal proof merely proves that the Reals "cannot be listed", this does not automatically prove "they are uncountable". > >> 2. Do you believe the computable Reals are countable? > > Obviously. > > > - Tim
From: Virgil on 18 Jun 2010 23:13 In article <4c8d7871-f386-48b2-8877-bd20b638c1d2(a)d37g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > > Infinity is a mathematical construct. Before you can even being to > > discuss it, you have to have a set of axioms. > > What was the set that Cantor used? Cantor was a pioneer, and pioneering work often needs some modification as the area under consideration becomes more familiar. So now we have, for example, FOL+ZFC, in which WM always loses his way.
From: Peter Webb on 18 Jun 2010 23:14
"K_h" <KHolmes(a)SX729.com> wrote in message news:RcCdndEvyKIquYHRnZ2dnUVZ_hWdnZ2d(a)giganews.com... > > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message > news:4c1b2def$0$14086$afc38c87(a)news.optusnet.com.au... >> >> "Tim Little" <tim(a)little-possums.net> wrote in message >> news:slrni1m68o.jrj.tim(a)soprano.little-possums.net... >>> On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >>>> Because Cantor's proof requires an explicit listing. This is a very >>>> central concept. >>> >>> Cantor's proof works on any list, explicit or not. >>> >> >> Really? >> >> How do you apply Cantor's proof to a list constructed as follows: >> >> "Define a list L such that the n'th entry on the list >> consists of all 1's if the n'th digit of Omega is 1, otherwise it is >> all 0's." >> >> (Your example). >> >>> The rest of your misconception snipped. >>> >>> >>> - Tim >> >> Perhaps if you could point out to me why you believe Cantor's proof that >> not all Reals can be listed (as it appears you do) but you don't believe >> my proof that not all computable Reals can be listed. They appear >> identical to me. > > All computable reals can be listed, but there is no finite algorithm for > doing so. An "infinite algorithm" could list every computable real. An > anti-diagonal, then, could be generated from this list but the algorithm > creating the anti-diagonal is implicitly relying on the "infinite > algorithm" underlying the list. In that sense the anti-diagonal is not > computable. Agreed. > The set of all reals are a different story. Even with an "infinite > algorithm" generating a list of reals, there is no way such a list could > contain every real. For a proof, do a google search on Cantor's theorem. > No, Cantor's diagonal construction does not prove this, and nor is any of this machinery part of his proof. Cantor's proof applied to computable Reals proves that the computable Reals cannot be listed. Cantor's proof applied to all Reals proves all Reals cannot be listed. That a set cannot be listed does not prove it is uncountable; computable Reals cannot be listed, but they are countable. |