"Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote in message news:a_mdnZe4xaoOxIPRnZ2dnUVZ8umdnZ2d@brightview.co.uk... > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message > news:4c1d86f0$0$18229$afc38c87@news.optusnet.com.au... >> >> "Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote in > message >> news:NNqdnTtg2LuExoDRnZ2dnUVZ8lKdnZ2d@brightview.co.uk... >> > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message >> > news:4c1c6be3$0$17177$afc38c87@news.optusnet.com.au... >> >> >> >> "Tim Little" <tim@little-possums.net> wrote in message >> >> news:slrni1om15.jrj.tim@soprano.little-possums.net... >> >> > On 2010-06-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> > wrote: >> >> >> So Cantor's proof when applied to computable Reals proves exactly >> >> >> what in your opinion? >> >> > >> >> > That there is a real not on the list. >> >> > >> >> > >> >> >> I might remind you that as the form of proof is identical to that >> >> >> used by Cantor for all Reals, whatever you believe that Cantor's >> >> >> proof applied to computable Reals proves, his proof applied to all >> >> >> Reals must prove the same thing. >> >> > >> >> > It does prove exactly the same thing: in both cases, all such lists >> >> > omit at least one real. >> >> > >> >> > >> >> >> Its the same proof, after all, except limitting the set to just >> >> >> computable Reals. >> >> > >> >> > No, there is a slight difference: the antidiagonal is a real number. >> >> > It does not necessarily produce a computable real number. >> >> > >> >> >> >> Of course it is computable. Cantor provides a simple construction for > the >> >> number. >> > >> > No, you're misunderstanding the meaning of computable. >> > >> > Hopefully you will be OK with the following definition: >> > >> > A real number r is computable if there is a TM (Turing machine) >> > T which given n as input, will produce as output >> > the n'th digit of r. >> > >> > So let us suppose there is a list L (not a "computable list", just a >> > "list") >> > of computable numbers in [0,1], covering all computerable numbers. >> > >> > [You have your own strange definition of "list", but I am using the >> > term >> > with it's standard mathematical meaning: L is just a function mapping >> > N >> > (natural numbers) to CR (computable reals). Perhaps we should also > insist >> > that lists are injections (no duplicates) - that's fine. YOU seem to > want >> > to define "list" to only include "computable" functions from N to CR, > but >> > that is NOT what everybody else (including Cantor) means by "list"!] >> > >> > L : N --> CR >> > >> > so we have the sequence (=list) of computable reals: >> > >> > ( L(0), >> > L(1), >> > L(2), >> > ....) >> > >> > Each computable real has it's digital representation, so this list can > be >> > written: (using d(m,n) for n'th digit of L(m) >> > >> > L(0) digits : d(0,0), d(0,1), d(0,2), ... >> > L(1) digits : d(1,0), d(1,1), d(1,2), ... >> > L(2) digits : d(2,0), d(2,1), d(2,2), ... >> > ... >> > >> > Now, given m and n, is there necessarily a TM that can take m and n as >> > inputs, and ouput d(m,n)? >> > >> > Well, each L(m) individually is computable: there is a corresponding >> > TM, >> > TM_m such that given n as input it will compute d(m,n). We could say: >> > >> > TM_m (n) = d(m,n) >> > >> > But all the TMs can be different and completely unrelated to each >> > other. >> > >> > Can we say there will always be a *single* TM TM_ALL_m, which >> > "subsumes" >> > all >> > the TM_n, i.e. if given m and n as input, will compute d(m,n)? I.e. >> > >> > for all m,n: TM_ALL_m (m,n) = d(m,n) ????? >> > >> > Unfortunately for your argument, the answer is NO. >> > >> > But that is exactly what you would need to argue that the anti-diagonal > is >> > computable: You need a TM that could calculate d(m,m), but this needs > the >> > TM TM_ALL_m, and we do not have that! >> > >> > I know what you want to say at this point! :) You want to say that I > have >> > "explicitly given you" d(m,n), and I can only do this if d(m,n) is >> > computable. (Right?) And you want to tell me that Cantor's diagonal >> > proof >> > is the same? >> > >> > Well, this is a misreading of Cantor's proof! Look closely: >> > >> > Paraphrasing of Cantor's proof: >> > >> > Given a function L: N --> [0,1], there >> > exists r in [0,1], such that >> > for all n, r != L(n) [i.e. r is not in the list] >> > >> > Proof: >> > 1) Suppose L: N-->R >> > 2) Define LD: (NxN)-->{0,1,2,3,4,5,6,7,8,9} >> > such that LD(m,n) = n'th digit of L(m) >> > 3) Define RD: N-->R >> > such that RD(m) = [...antidiagonal digit m...] >> > 4) Then RD determines a real r in [0,1] >> > 5) Show for all n, r != L(n) from defn. of r above. >> > >> > Where in this proof does Cantor require LD(m,n) to be computable? > (Answer >> > = >> > nowhere.) >> > >> > I think maybe you are thinking it must be computable, due to your >> > misleading >> > use of the phrase "explicitly given", but actually all Cantor does is >> > argue >> > that IF a (any) function L (not necessarily computable) EXISTS mapping >> > N >> > to >> > [0,1], the mere existence of L implies the existence of a corresponding >> > real >> > r not in the list L. There is nothing requiring real computer programs > to >> > actually compute anything here! It's just a mathematical proof of >> > existence >> > of something. >> > >> >> What you are writing above is *not* Cantor's diagonal proof as it appears > on >> a million websites. > > Well, I claim that it is at least OBVIOUSLY EQUIVALENT to what Cantor is > saying. It's true I didn't go to a web site and cut and paste any > "official" text from his original document, but in essence my > understanding > is: > > 1. Cantor shows that any map from N to R cannot be "onto" R. > 2. That my proof above sketches this out: > - Cantor equates real numbers with their digit sequences > (This is why I go from L to LD) > - Cantor shows the existence of a new digit sequence, via > the antidiagonal argument > (That is what I do in step (3). RD is my new digit sequence.) > - Cantor proves that RD is a new digit sequence > - and that consequenctly the corresponding real number > is not in the original map L. > > So I think my proof follows Cantor's pretty closely, although it is my own > notation. Given the amount of effort I put into trying to help you, I'm > disappointed you can't get over some minor notationaly differences to > follow > what I'm saying.
The whole point of what I am doing is to use his original proof in its original form.
Introducing new notation is a crank's way of arguing.
Why not look at my proof, which follows Cantor's proof, instead of changing te notation to produce a proof which you claim is equivalent and arguing about that different proof, which you claim differs only in notation?
> You've also given no indication of in what way you think > my proof differs from Cantors! (Well, that would be OK if my proof was > "nothing like" Cantor's but that's clearly not the case as I've just > explained, so I'm left mystified by your claim...) > >> >> The whole point of my post was to use Cantor's proof exactly, which alows > me >> to demonstrate that Cantor's proof does *not* in of itself prove the >> uncounatbility of the Reals. >> >> Perhaps if you were to tell me where you think the error in my very >> simple >> proof is, then we would be making progress. > > Sure I'll do this. To avoid a further "that's not Cantor's proof" claim > from you, please quote exactly what proof of Cantor's you want me to > comment > on. >
There can be no list of all Reals.
Imagine such a list of purported Reals exists. Compute the anti-diagonal number as follows ... blah blah. This Real is not on the list.
