Why can no one in sci.math understand my simple point?
in [Logic]
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From: Charlie-Boo on 27 Jun 2010 04:35 On Jun 27, 12:08 am, Virgil <Vir...(a)home.esc> wrote: > In article <slrni2dgpo.jrj....(a)soprano.little-possums.net>, > Tim Little <t...(a)little-possums.net> wrote: > > > On 2010-06-26, Charlie-Boo <shymath...(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. > > It is well known that for 'most' reals no such algorithm can exist, the > non-computable transcendentals. That has no bearing on this discussion. (They need not be transcendental.) C-B
From: WM on 27 Jun 2010 05:19 On 27 Jun., 05:13, Virgil <Vir...(a)home.esc> wrote: > > 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. But it is impossible to list the countable set that I described. That is the point! And it is obvious that it cannot be circumvented. Regards, WM
From: Mike Terry on 27 Jun 2010 07:01 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:382a8f19-157c-4c0a-9ed2-b9c295ce0dd8(a)5g2000yqz.googlegroups.com... > 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. The fact that a list does not contain its antidiagonal does not mean the list cannot be listed!. My final word on this: The set you have constructed: {A0, A1, A2,...} is: a) countable b) can be listed, e.g. (A0, A1, A2,...) c) of course the list (A0, A1, A2,...) has an antidiagonal Aw, which is not in {A0, A1, A2,...). (This is obviously irrelevent to (a) and (b)). So you are wrong. (Wrong not in a subtle way, but just in the way of not understanding the basic usage of the mathematical terms involved, like "countable" "list" etc. Regards, Mike. > > > > 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: WM on 27 Jun 2010 07:56 On 27 Jun., 13:01, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > 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. > > The fact that a list does not contain its antidiagonal does not mean the > list cannot be listed!. But that is not the question! Please read carefully. The question is whether there is a countable set that cannot be listed. This set is given by the original list and all its possible antidiagonals. > > My final word on this: > > The set you have constructed: {A0, A1, A2,...} is: > > a) countable > b) can be listed, e.g. (A0, A1, A2,...) > c) of course the list (A0, A1, A2,...) has an antidiagonal Aw, > which is not in {A0, A1, A2,...). (This is obviously > irrelevent to (a) and (b)). And your (a) and (b) ist obviously irrelevant for the present discussion. > > So you are wrong. No. You simply cannot understand the meaning of a process which cannot end (hence the results of which cannot be put in a complete list) unless there is a list containing its antidiagonal. But as I have given the description in clear words, I don't want to repeat it. Probably it would not support your understanding either. You may google under "supertask" to better inform you. Regards, WM
From: Peter Webb on 27 Jun 2010 08:24
"Virgil" <Virgil(a)home.esc> wrote in message news:Virgil-4CE082.01003126062010(a)bignews.usenetmonster.com... > In article <4c259cd4$0$1029$afc38c87(a)news.optusnet.com.au>, > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > >> "Virgil" <Virgil(a)home.esc> wrote in message >> news:Virgil-71EA75.23485925062010(a)bignews.usenetmonster.com... >> > In article >> > <b3413a4e-567b-4dfb-8037-21f14b826ede(a)g1g2000prg.googlegroups.com>, >> > Newberry <newberryxy(a)gmail.com> wrote: >> > >> >> > > No. (3) is not true, as it is based on a false premise (that the >> >> > > computable >> >> > > Reals can be listed). >> > >> > How is countability any different from listability for an infinite set? >> > >> > Does not countability of an infinite set S imply a surjections from N >> > to S? And then does not such a surjection imply a listing? >> >> It implies a listing must exist, but does not provide such a listing. >> >> The computable Reals are countable, but you cannot form them into a list >> of >> all computable Reals (and nothing else) where each item on the list can >> be >> computed. >> >> In order to list a set, it has to be recursively enumerable. Being >> countable >> is not sufficient. > > Both countability and listability appear to be the case if and only if > a listing exists, but neither requires specifying that listing. Is that > not so? No. If we take "listable" to mean we can (ummm) make a list of exactly those elements and no other, then this is not correct. A set can be countable but not listable. AFAIK, "listable" is not a formally defined mathematical term. The formal term in mathematics which is closest to the intuitive idea of being able to explicitly list the members of a set is that it is "recursively enumerable". Not "countable". Being countable is necessary but not sufficient. This is the problem I have with the standard presentation of Cantor's proof - it starts with a "list", and then proves there is an item missing from the list. Proving that no list can be prepared proves only that the Reals are not recursively enumerable, not the stronger condition they are uncountable. I have no problem with the very similar proof of Cantor's that the power set has larger cardinality than the set itself, which suffices to prove the Reals are uncountable. It doesn't talk about "lists". |