Why can no one in sci.math understand my simple point?
in [Logic]
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From: WM on 26 Jun 2010 17:43 On 26 Jun., 23:14, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > news:511dd6ef-6fe7-4664-bda0-082971c2f8db(a)z10g2000yqb.googlegroups.com... > > > On 26 Jun., 00:32, "Mike Terry" > > > > > Please let me know: Does the list consisting of A0, A1, A2, A3, ... > > > > contain its antidiagonal or not? > > > > No, of course not. > > > The set of all these antidiagonals A0, A1, A2, A3, ... constructed > > according to my construction process and including the absent one is a > > countable set. > > Yes, I've agreed that several times. > > > > > > > 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? > > > See the one above: The results of my construction process. > > No. The above shows that you've constructed a countable sequence of numbers > (A0, A1, A2, A3,...) which CAN be listed, namely consider the list (A0, A1, > A2, A3,...). > > Read carefully! In order to do so, I posed the question: Does the list consisting of A0, A1, A2, A3, ... contain its antidiagonal or not? You said no. Therefore your assertion "which CAN be listed" is plainly wrong. > Of course it is not possible to list a list and its antidiagonal. > > > > > > > OK, but you've still not given any explanation why you think the > > > antidiagonals of your process cannot be listed! (I'm genuinely > baffled.) > > > You said so yourself, few lines above. > > Rubbish. I have never said anything other than that (A0, A1, A2, ...) is a > countable sequence of reals that CAN be listed but not with all its antidiagonals. > > Go on - have one more go, then I'll give up... You'd better give up than make contradictory claims. Regards, WM
From: Jesse F. Hughes on 26 Jun 2010 20:48 Charlie-Boo <shymathguy(a)gmail.com> writes: >> Example 1: The TM that never halts and never changes the tape does not >> represent a computable real. > > It depends on the system of representation. An empty tape typically > represents 0. If the machine does not halt, you can't say that the tape is empty for certain. The usual definitions of computable real require that the machine accept input n, halts and outputs the first n digits, or something like that. > >> Example 2: The TM that repeatedly changes the value in one cell, never >> halting, does not represent a computable real. > > The sequence of values put into a given cell defines the decimal > expansion of a real number. Every machine must distinguish between > scratch values and actual output. Having a certain cell represent > output is also common. > > Google "Turing Machines". So, what real number do you think example 2 computes? And what convention of computable real number do you have in mind? Show me a reference or web page, rather than telling me to google Turing Machines. In other words, stop bluffing. -- Jesse F. Hughes "It's easy folks. Just talk about my approach to your favorite mathematician. If they can't be interested in it, they've demonstrated a lack of mathematical skill." -- James Harris
From: Tim Little on 26 Jun 2010 23:12 On 2010-06-26, Newberry <newberryxy(a)gmail.com> wrote: > If the Turing machine is hacked such that it outputs a digit on any > state transition does it not represent a real? Suppose I have a Turing machine that repeatedly marks and erases a symbol at the start position. What real are you saying that this machine represents? - Tim
From: Virgil on 26 Jun 2010 23:13 In article <382a8f19-157c-4c0a-9ed2-b9c295ce0dd8(a)5g2000yqz.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 26 Jun., 23:14, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > > > news:511dd6ef-6fe7-4664-bda0-082971c2f8db(a)z10g2000yqb.googlegroups.com... > > > > > On 26 Jun., 00:32, "Mike Terry" > > > > > > > Please let me know: Does the list consisting of A0, A1, A2, A3, ... > > > > > contain its antidiagonal or not? > > > > > > No, of course not. > > > > > The set of all these antidiagonals A0, A1, A2, A3, ... constructed > > > according to my construction process and including the absent one is a > > > countable set. > > > > Yes, I've agreed that several times. > > > > > > > > > > > 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? > > > > > See the one above: The results of my construction process. > > > > No. �The above shows that you've constructed a countable sequence of numbers > > (A0, A1, A2, A3,...) which CAN be listed, namely consider the list (A0, A1, > > A2, A3,...). > > > > Read carefully! > > In order to do so, I posed the question: Does the list consisting of > A0, A1, A2, A3, ... contain its antidiagonal or not? No list contains its antidiagonal, nevertheless, any list together with its antidiagonal can be listed, and in a large variety of ways. > > You said no. Therefore your assertion "which CAN be listed" is plainly > wrong. Nonsense. > > > > Of course it is not possible to list a list and its antidiagonal. Whyever not? Given a list {b0, b1, b2, ...} and its antidiagonal, a, how is the llist {a, b0,. b12, b2, ...} NOT a list of the original list plus its antidiagonal? > > > > > > > > > > > > OK, but you've still not given any explanation why you think the > > > > antidiagonals of your process cannot be listed! �(I'm genuinely > > baffled.) > > > > > You said so yourself, few lines above. > > > > Rubbish. �I have never said anything other than that (A0, A1, A2, ...) is a > > countable sequence of reals that CAN be listed > > but not with all its antidiagonals. Since any such list has potentially uncountably many "antidiagonals", of course not. But any list along with up to countably many of its antidiagonals can be collectively listed. > > > > Go on - have one more go, then I'll give up... > > You'd better give up than make contradictory claims. Good advice. Its a shame its author doesn't follow it himself.
From: Tim Little on 26 Jun 2010 23:19
On 2010-06-26, Charlie-Boo <shymathguy(a)gmail.com> wrote: > It is trivial to calculate in binary (using only 0 and 1) and output > any desired string surrounded by 2 whenever we want You use the singular here, by which your statement is correct. Any single finite string can be so produced. What's more, any finite sequence of finite strings can be produced. However, you need an infinite sequence of finite strings to define most real numbers, and there are not enough finite algorithms to produce them all. - Tim |