Why can no one in sci.math understand my simple point?
in [Logic]
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From: Charlie-Boo on 25 Jun 2010 08:26 On Jun 24, 8:49 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > On Jun 15, 2:15 am, "Peter Webb" > > >> No. You cannot form a list of all computable Reals. > > > Of course you can - it's just the list of Turing Machines. > > No, it's not. I asked for a counterexample, to no avail. Don't you think you should substantiate your statement or retract it? Each Turing Machine represents some computable real (all computable reals are included) and you can list those Turing Machines. The Turing Machine represents it as well as any other system of representation. C-B > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, dar ber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Jesse F. Hughes on 25 Jun 2010 10:24 Charlie-Boo <shymathguy(a)gmail.com> writes: > On Jun 24, 8:49 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: >> Charlie-Boo <shymath...(a)gmail.com> writes: >> > On Jun 15, 2:15 am, "Peter Webb" >> >> >> No. You cannot form a list of all computable Reals. >> >> > Of course you can - it's just the list of Turing Machines. >> >> No, it's not. > > I asked for a counterexample, to no avail. Don't you think you should > substantiate your statement or retract it? > > Each Turing Machine represents some computable real (all computable > reals are included) and you can list those Turing Machines. The > Turing Machine represents it as well as any other system of > representation. It is not the case that every TM represents some computable real. Example 1: The TM that never halts and never changes the tape does not represent a computable real. Example 2: The TM that repeatedly changes the value in one cell, never halting, does not represent a computable real. Other than that, of course, your response was mighty insightful. -- Jesse F. Hughes "I just define real numbers to be all those on the number line, as they were defined before Dedekind and Cauchy." -- Ross Finlayson's simple definition.
From: Mike Terry on 25 Jun 2010 14:39 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:f4abb19e-0ec6-4eb1-b5b0-6baafdee37bf(a)x27g2000yqb.googlegroups.com... > On 24 Jun., 21:26, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > > There is a starting list L0 and many following lists. During > construction step n+1 the list is > > An > ... > L0 > > and has an antidiagonal. > > The antidiagonals of all these lists are countable but canot be > written in form of a list (it does not matter if in a separate list or > in a list with spaces or else), because, if written in form of some > list, they would yield an antidiagonal that was not in the list. OK finally I am certain of what you are saying. The "antidiagonals of all these lists" is A0, A1, A2, A3, ... Now let me see how I might construct a list which covers all those anti-diagonals. OK, I've thought of one! My list is: L = (A0, A1, A2, A3, ...) OK, I've done what you said couldn't be done. Let's look at why you thought it couldn't be done - perhaps there's a mistake there? You said that L (as I've defined) has an antidiagonal that's not in the list! I agree that L has an antidiagonal that's not in L. That's normal for antidiagonals, but I don't see why you think that's a problem. IOW we've finally got to the point: - WM has constructed a countable set of antidiagonals A0, A1, A2, A3... Define L = the list (A0, A1, A2, A3,...) - WM claims that AntiDiag (L) is not in L, therefore L does not exist. - MT points out that L does exist and agrees AntiDiag(L) is not in L, but points out this is not a problem. (AntiDiags are never in their input lists, so why would anyone think it would be a problem?) > Hence > all antidiagonals of this process, though countable, cannot be written > in a list. Of course they can - you've not shown any problem yet. Why do you think Antidiag(L) not being in L means that L doesn't exist? > > Regards, WM
From: Virgil on 25 Jun 2010 15:23 In article <f4abb19e-0ec6-4eb1-b5b0-6baafdee37bf(a)x27g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 24 Jun., 21:26, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > > There is a starting list L0 and many following lists. During > construction step n+1 the list is > > An > ... > L0 > > and has an antidiagonal. Not as listed, it doesn't. As listed it is a finite list with different types of elements. If the list were, for example, An,...,A0, L0.0, L0.1, L0.2, ..., THEN it would have an antidiagonal, but only lists of elements all of the same type, such as all infinite binary sequences or all reals, can have antidiagonals, at least according to any standard definition/construction of an antidiagonal. > > The antidiagonals of all these lists are countable but canot be > written in form of a list (it does not matter if in a separate list or > in a list with spaces or else), because, if written in form of some > list, they would yield an antidiagonal that was not in the list. That presumes that one can include all reals in a list. But Onoe of the antidiagonal proofs proves that wrong, and all of WM's weaseling won't invalidate that proof. Besides which { L.0, A0. L0.2, A1, L0.2, A2, ...} is clearly a listing of what WM has just said couldn't be listed. So WM is not only wrong, he is stupidly wrong. Hence > all antidiagonals of this process, though countable, cannot be written > in a list. Except that they have been listed.
From: Virgil on 25 Jun 2010 15:25
In article <adaa005c-c180-4ce0-8d2c-81da912c6b3b(a)w12g2000yqj.googlegroups.com>, Charlie-Boo <shymathguy(a)gmail.com> wrote: > On Jun 24, 8:49�am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > Charlie-Boo <shymath...(a)gmail.com> writes: > > > On Jun 15, 2:15 am, "Peter Webb" > > > > >> No. You cannot form a list of all computable Reals. > > > > > Of course you can - it's just the list of Turing Machines. > > > > No, it's not. > > I asked for a counterexample, to no avail. Don't you think you should > substantiate your statement or retract it? > > Each Turing Machine represents some computable real (all computable > reals are included) and you can list those Turing Machines. The > Turing Machine represents it as well as any other system of > representation. Aren't there Turing machines that don't represent any real at all? > > C-B > > > -- > > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > > > "Wovon man nicht sprechan kann, dar ber muss man schweigen" > > �- Ludwig Wittgenstein, Tractatus Logico-Philosophicus |