Why can no one in sci.math understand my simple point?
in [Logic]
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From: |-|ercules on 18 Jun 2010 21:43 "Virgil" <Virgil(a)home.esc> wrote > In article > <57ebf4f0-686a-435e-aaea-c7696d718bc2(a)k39g2000yqd.googlegroups.com>, > WM <mueckenh(a)rz.fh-augsburg.de> wrote: > >> Pi is constructable and computable and definable, because there is a >> finite rule (in fact there are many) to find each digit desired. But >> as there are only countably many finite rules, there cannot be more >> defined numbers. > > If there are countably many rules then there are uncountably many lists > of rules capable of generating a number. > > > > > >> Therefore matheologicians have created undefinable >> "numbers". > > WM mistakes the issue. > > In pure mathematics, like in games, one has a set of rules to follow. > Differing sets of rules generate differing systems only some of which > are of much mathematical interest. > > The systems of rules we chose to use need not be subject to the > constraints that the system of rules that WM choses to play by are > subject to. > > For example, in FOL+ZFC, a commonly used system of rules which WM doe > not care for, all sorts of things are legitimate that none of WM's > systems of rules will allow. > > WM tries to force everyone to play only by his rules, but most of us > find his system of rules dead boring and of little or no mathematical > interest. > > Fortunately, outside of those classrooms in which his poor students are > compelled to play by his rules, he has no power to impose those rules on > anyone. Unfortunately your last paragraph is speculation and your superinfinity rules are full of contradictions. You show this example as Cantor's method 123 456 789 Diag = 159 AntiDiag = 260 But this method miserably fails on the computable set of reals, THERE IS NO new sequence like 260 in this example. Why don't you rework Cantor's proof to define ALL anti-diagonals instead of 1 particular anti-diagonal? Then the flaw in your method is obvious. CONSTRUCT a new real An AD(n) =/= L(n,n) PROVE that this real is not on the list An AD(n) =/= L(n,n) THEREFORE [ An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) ] -> Superinfinity That is literally Cantor's diagonal proof of higher cardinalities. Cantor's powerset proof is more ridiculous, no box contains the box numbers that don't contain their own box number -> Superinfinity WM is entirely correct. You are a blind lost soul, a sacrificial Virgin. What you call interesting or "paradise" we call ridiculous. Herc
From: K_h on 18 Jun 2010 22:00 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message news:4c1b2def$0$14086$afc38c87(a)news.optusnet.com.au... > > "Tim Little" <tim(a)little-possums.net> wrote in message > news:slrni1m68o.jrj.tim(a)soprano.little-possums.net... >> On 2010-06-18, Peter Webb >> <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >>> Because Cantor's proof requires an explicit listing. >>> This is a very >>> central concept. >> >> Cantor's proof works on any list, explicit or not. >> > > Really? > > How do you apply Cantor's proof to a list constructed as > follows: > > "Define a list L such that the n'th entry on the list > consists of all 1's if the n'th digit of Omega is 1, > otherwise it is > all 0's." > > (Your example). > >> The rest of your misconception snipped. >> >> >> - Tim > > Perhaps if you could point out to me why you believe > Cantor's proof that not all Reals can be listed (as it > appears you do) but you don't believe my proof that not > all computable Reals can be listed. They appear identical > to me. All computable reals can be listed, but there is no finite algorithm for doing so. An "infinite algorithm" could list every computable real. An anti-diagonal, then, could be generated from this list but the algorithm creating the anti-diagonal is implicitly relying on the "infinite algorithm" underlying the list. In that sense the anti-diagonal is not computable. The set of all reals are a different story. Even with an "infinite algorithm" generating a list of reals, there is no way such a list could contain every real. For a proof, do a google search on Cantor's theorem. _
From: Tim Little on 18 Jun 2010 22:12 On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > Bijectable with N and "listable" are not the same. To be "listable" > the set must be countable and recursively enumerable. There is no such requirement for recursive enumerability in Cantor's work. The concept was not even introduced into mathematics until some time after his death. Your insistence that Cantor required lists to be recursively enumerable is bizarre. - Tim
From: Tim Little on 18 Jun 2010 22:14 On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > That, BTW, is my own interpretation of what is happening. It is an incorrect interpretation. > Whether you accept this or not, the simple fact is that Cantor's > proof can be applied to any purported list of all computable Reals > and used to generate a computable Real not on the list Your "simple fact" is simply wrong. Look up the definition of "computable real" and get back to us. - Tim
From: Marshall on 18 Jun 2010 22:17
On Jun 18, 6:09 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-18, Peter Webb <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > The number that is produced is clearly "computable", because we have > > computed it. > > I see you still haven't consulted a definition of "computable number". > No worries, let me know when you have. I suggest it would be more persuasive if you made whatever point you have in mind about the definition of computable number directly. Simply repeating this one-liner makes it seem like you might not have a point. Marshall |