Why can no one in sci.math understand my simple point?
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From: Mike Terry on 25 Jun 2010 18:32 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:3ef6da97-a35e-445a-9d62-b4df42474711(a)c10g2000yqi.googlegroups.com... > On 25 Jun., 20:39, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > "WM" <mueck...(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? > > > Please let me know: Does the list conssisting of A0, A1, A2, A3, ... > contain its antidiagonal or not? No, of course not. > > > > 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?) > > That is not a problem. It shows however, that there are countable sets > that cannot be listed. How exactly does it show that there are countable sets that cannot be listed? > > > > > > 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? > > I do not think so. I said, that there are countable sets that cannot > be listed. One is the set of all antidiagonals of my process. So > countability and listability are not the same. OK, but you've still not given any explanation why you think the antidiagonals of your process cannot be listed! (I'm genuinely baffled.) Mike. > > Regards, WM
From: Virgil on 25 Jun 2010 18:47 In article <3ef6da97-a35e-445a-9d62-b4df42474711(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > Please let me know: Does the list conssisting of A0, A1, A2, A3, ... > contain its antidiagonal or not? Not! > > > > 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?) > > That is not a problem. It shows however, that there are countable sets > that cannot be listed. How does it do that? WHICH countable set does WM claim cannot be listed? By designating a set as countable one guarantees it to be listable, at least for standard definitions of countable and listable. I have several times asked WM for the distinction that he claims exists between countable and listable, but he has yet to explain that difference to anyone's satisfaction but his own. > > > > > > 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? > > I do not think so. I said, that there are countable sets that cannot > be listed. One is the set of all antidiagonals of my process. So > countability and listability are not the same. The claim that the set of all antidiagonals to WM's process cannot be listed has been disproved by constructing a rule for listing that set.
From: Peter Webb on 25 Jun 2010 21:39 "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote in message news:87lja3yyv2.fsf(a)phiwumbda.org... 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. _______________________________ And nor is it the case that TMs represent Reals "as well as any other system of representation". Continued fraction representations allow the Real to be calculated to arbitrary accuracy in finite time. TM representations don't.
From: Tim Little on 25 Jun 2010 22:56 On 2010-06-25, Virgil <Virgil(a)home.esc> wrote: > Aren't there Turing machines that don't represent any real at all? it depends upon how you define the representation. You could e.g. define any Turing machine that fails to halt on some input to represent 0. Of course, then you do have some difficulty in proposing an algorithm to compute an antidiagonal. Such is life. - Tim
From: Newberry on 25 Jun 2010 23:37
On Jun 25, 7:24 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(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. If the Turing machine is hacked such that it outputs a digit on any state transition does it not represent a real? The digits of pi, e, sqrt(2) etc. can be generated by an algorithm. We can certainly list such algorithms suitably defined. > > 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.- Hide quoted text - > > - Show quoted text - |