Why can no one in sci.math understand my simple point?
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From: Sylvia Else on 18 Jun 2010 23:57 On 19/06/2010 11:43 AM, |-|ercules wrote: > "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. Computable set of reals? Set of computable reals? Cantor proves that the set of reals is not countable. It's not saying anything about computables. > > Why don't you rework Cantor's proof to define ALL anti-diagonals instead > of 1 particular anti-diagonal? Because there's no need. We hyphothesise that the reals are countable. Since they are countable we can list them, and the list will contain all the reals. Then we show that there's a number that isn't in the list, which contradicts the hypothesis, which must therefore be wrong. Hence the reals aren't countable. It seems straightforward enough. To avoid this proof, you have to dispute reductio-ad-absurdum arguments in general, argue that reals cannot be expressed as infinite strings of digits, or have axioms that exclude the construted anti-diagonal from the set of reals. Sylvia.
From: Sylvia Else on 19 Jun 2010 00:08 On 19/06/2010 1:40 PM, |-|ercules wrote: > "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... >> On 19/06/2010 12:45 PM, |-|ercules wrote: >>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>>> On 19/06/2010 6:50 AM, WM wrote: >>>>> On 18 Jun., 09:37, Sylvia Else<syl...(a)not.here.invalid> wrote: >>>>>> On 18/06/2010 5:31 PM, |-|ercules wrote: >>>>>> >>>>>> >>>>>> >>>>>> >>>>>> >>>>>>> "Sylvia Else"<syl...(a)not.here.invalid> wrote ... >>>>>>>> On 18/06/2010 4:52 PM, |-|ercules wrote: >>>>>>>>> "Sylvia Else"<syl...(a)not.here.invalid> wrote ... >>>>>>>>>> On 18/06/2010 3:03 PM, |-|ercules wrote: >>>>>>>>>>> "Sylvia Else"<syl...(a)not.here.invalid> wrote >>>>>>>>>>>> On 18/06/2010 10:40 AM, Transfer Principle wrote: >>>>>>>>>>>>> On Jun 17, 6:56 am, Sylvia Else<syl...(a)not.here.invalid> >>>>>>>>>>>>> wrote: >>>>>>>>>>>>>> On 15/06/2010 2:13 PM, |-|ercules wrote: >>>>>>>>>>>>>>> the list of computable reals contain every digit of ALL >>>>>>>>>>>>>>> possible >>>>>>>>>>>>>>> infinite sequences (3) >>>>>>>>>>>>>> Obviously not - the diagonal argument shows that it doesn't. >>>>>> >>>>>>>>>>>>> But Herc doesn't accept the diagonal argument. Just because >>>>>>>>>>>>> Else accepts the diagonal argument, it doesn't mean that >>>>>>>>>>>>> Herc is required to accept it. >>>>>> >>>>>>>>>>>>> Sure, Cantor's Theorem is a theorem of ZFC. But Herc said >>>>>>>>>>>>> nothing about working in ZFC. To Herc, ZFC is a "religion" >>>>>>>>>>>>> in which he doesn't believe. >>>>>> >>>>>>>>>>>> Well, if he's not working in ZFC, then he cannot make >>>>>>>>>>>> statements >>>>>>>>>>>> about >>>>>>>>>>>> ZFC, and he should state the axioms of his system. >>>>>> >>>>>>>>>>> Can you prove from axioms that is what I should do? >>>>>> >>>>>>>>>>> If you want to lodge a complaint with The Eiffel Tower that the >>>>>>>>>>> lift is >>>>>>>>>>> broken >>>>>>>>>>> do you build your own skyscraper next to the Eiffel Tower to >>>>>>>>>>> demonstrate >>>>>>>>>>> that fact? >>>>>> >>>>>>>>>> That's hardly a valid analogy. >>>>>> >>>>>>>>>> If you're attempting to show that ZFC is inconsistent, then say >>>>>>>>>> that >>>>>>>>>> you are working within ZFC. >>>>>> >>>>>>>>>> If you're not working withint ZFC, then you're attempting to >>>>>>>>>> show that >>>>>>>>>> some other set of axioms is inconsistent, which they may be, but >>>>>>>>>> the >>>>>>>>>> result is uninteresting, and says nothing about ZFC. >>>>>> >>>>>>>>>> Sylvia. >>>>>> >>>>>>>>> That would be like finding a fault with the plans of The Leaning >>>>>>>>> Tower >>>>>>>>> Of Piza. >>>>>> >>>>>>>>> I might look at ZFC at some point, but while you're presenting >>>>>>>>> Cantor's >>>>>>>>> proof >>>>>>>>> in elementary logic I'll attack that logic. >>>>>> >>>>>>>>> Instead of 'constructing' a particular anti-diagonal, your proof >>>>>>>>> should >>>>>>>>> work equally >>>>>>>>> well by giving the *form* of the anti-diagonal. >>>>>> >>>>>>>>> This is what a general diagonal argument looks like. >>>>>> >>>>>>>>> For any list of reals L. >>>>>> >>>>>>>>> CONSTRUCT a real such that >>>>>>>>> An AD(n) =/= L(n,n) >>>>>> >>>>>>>>> Now to demonstrate this real is not on L, it is obvious that >>>>>>>>> An AD(n) =/= L(n,n) >>>>>> >>>>>>>>> Therefore >>>>>>>>> [ An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) ] proves >>>>>>>>> superinfinity! >>>>>> >>>>>>>>> And THAT is Cantor's proof! >>>>>> >>>>>>>>> Want to see his other proof? That no box contains the box numbers >>>>>>>>> (of >>>>>>>>> boxes) that >>>>>>>>> don't contain their own box number? >>>>>>>>> That ALSO proves superinfinity! >>>>>> >>>>>>>>> Great holy grail of mathematics you have there. >>>>>> >>>>>>>>> Herc >>>>>> >>>>>>>> What are you trying to prove? >>>>>> >>>>>>> There is only one type of infinity. >>>>>> >>>>>> Infinity is a mathematical construct. Before you can even being to >>>>>> discuss it, you have to have a set of axioms. >>>>> >>>>> What was the set that Cantor used? >>>>> Nevertheless he "proved". >>>> >>>> He certainly was using some. For example, the diagonal argument falls >>>> apart if the axioms don't permit the construction of a number by >>>> choosing digits different from those on the diagonal. >>>> >>>> It isn't even clear whether Herc is tying to invalidate Cantor's proof >>>> by finding a mistake in it, or to prove the inverse, which wouldn't >>>> invalidate Cantor's proof, but would only show that the axioms on >>>> which it is based are inconsistent. >>>> >>>> Herc cannot avoid the need to specify the set of axioms. >>>> >>>> Sylvia. >>> >>> How would one dispute axiomatic deductions if that were the case? >> >> What do you mean by "axiomatic deduction"? >>> >>> Are you saying all mathematical facts are either axioms or the result of >>> (X & X->Y) -> Y >>> ? >> >> Mathematics consists of axioms and statements (theorems) that can be >> proved from those axioms. The axioms cannot themselves be proved, nor >> disproved, though they may be shown to be inconsistent with one another. >> >> Sometimes the axioms seem so self-evidently true that people aren't >> even aware that they're there. But they are, if you look. >> >> Sylvia. > > blah blah blah... > > you skipped my question, but don't bother I wasn't arguing anything > just seeing if you knew what you were talking about. Your question was typically vague. They you dived into some notation which might be construed to mean "if X and X implies Y, then Y", which is itself unproveable, and needs to be introduced as an axiom. None of which eliminates your need to specify the axioms in which you're making claims about Cantor's proof. Sylvia.
From: K_h on 19 Jun 2010 00:13 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message news:4c1c3663$0$24370$afc38c87(a)news.optusnet.com.au... > > "K_h" <KHolmes(a)SX729.com> wrote in message > news:RcCdndEvyKIquYHRnZ2dnUVZ_hWdnZ2d(a)giganews.com... >> >> "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. > > Agreed. > >> 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. >> > > No, Cantor's diagonal construction does not prove this, > and nor is any of this machinery part of his proof. > > Cantor's proof applied to computable Reals proves that the > computable Reals cannot be listed. You have contradicted yourself. I wrote "All computable reals can be listed,..." and your reply was "Agreed". Now you write "...the computable Reals cannot be listed". Let's be clear. The computable reals can be listed. Cantor's ideas, applied to such a list, shows that the anti-diagonal of such a list is not computable. That's all. > Cantor's proof applied to all Reals proves all Reals > cannot be listed. That a set cannot be listed does not > prove it is uncountable; computable Reals cannot be > listed, but they are countable. All computable reals CAN be listed. The diagonal argument applied to such a list gives an anti-diagonal number that is not computable. If that anti-diagonal number was computable then the set of all computable reals would be uncountable. But that is not the case. The diagonal argument applied to a list of reals gives another real that is nowhere in the list. So the reals cannot be listed. By definition, a set that cannot be listed is uncountable. The computable reals can be listed so they are countably infinite. The reals cannot be listed so they are uncountably infinite. http://en.wikipedia.org/wiki/Uncountable_set _
From: K_h on 19 Jun 2010 00:42 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message news:4c1c3a11$0$18229$afc38c87(a)news.optusnet.com.au... > > "K_h" <KHolmes(a)SX729.com> wrote in message > news:ALWdnQzHH6Fug4HRnZ2dnUVZ_gKdnZ2d(a)giganews.com... >> >> "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote >> in message >> news:4c1b24ce$0$17177$afc38c87(a)news.optusnet.com.au... >>> >>> "Tim Little" <tim(a)little-possums.net> wrote in message >>> news:slrni1m2q9.jrj.tim(a)soprano.little-possums.net... >>>> On 2010-06-18, Peter Webb >>>> <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >>>>> "Tim Little" <tim(a)little-possums.net> wrote in message >>>>>> Cantor's construction applied directly to lists of >>>>>> computable numbers >>>>>> shows almost nothing. Given any list of computable >>>>>> reals, you can >>>>>> produce an antidiagonal real not on the list. >>>>>> Unfortunately the >>>>>> construction says nothing about whether the >>>>>> antidiagonal is computable >>>>>> or not. >>>>> >>>>> Of course its computable. It is a trivial programming >>>>> exercise to form a >>>>> computable number not on the list, by simply doing >>>>> digit substitutions. >>>> >>>> Only if the list itself is a computable function. >>>> Cantor's proof does >>>> not depend on any such assumption. >>>> >>> >>> If you give me a purported list of all computable >>> numbers, I can explicitly >> >> QUESTION 1: >> But ask yourself: is the list of all computable real >> numbers itself computable? Each real in such a list is >> computable but can you find a computable function that >> says which computable real is mapped to 0, which >> computable real is mapped to 1, which computable real is >> mapped to 2, and so on? >> > > No. > >>> construct a Real not on the list. Of course this number >>> is computable; there is a simple algorithm to compute >>> it. >> >> But ask yourself: since the algorithm you use to compute >> it calls on the "algorithm" I asked you about in QUESTION >> 1, is your algorithm depending on another "algorithm" >> that is non-computable? >> > > The algorithm does not exist. There can be no list of all > computable Reals. No, there is no finite algorithm that can generate the list but that does not mean that such lists don't exist. There are lists that contain all computable reals. >>> No. And in fact Cantor's diagonal proof does not prove >>> the Reals are uncountable, just that they cannot be >>> listed. Which is all he claimed. >> >> No, because Cantor's diagonal argument applies to lists >> that are not computable as well as to lists that are >> computable. > > So Cantor's proof when applied to computable Reals proves > exactly what in your opinion? If I understand your question, all it shows is that the anti-diagonal number is not computable. > So the computable Reals cannot be listed. > Yet they are countable. > So a proof that a set cannot be listed does not prove it > is uncountable. The computable reals can be listed and so they are countable. >>> You give me that list. >> >> Yes, every real in the list is computable. But ask >> yourself: is the function that maps each natural to each >> of these reals itself computable? > > No. > They cannot be listed. Yes they can but not by a finite algorithm. _
From: Tim Little on 19 Jun 2010 01:10
On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > Which of these two statements do you agree with: > > 1. You cannot form a list of computable Reals. > > 2. The computable Reals are countable. Answering the same questions for the fourth time now, 1 is false, 2 is true. > You have failed to explain why Cantor's diagonal proof "proves" the > Reals are uncountable, but the same proof applied to computable > Reals does *not* prove the Computable Reals are uncountable. I have, multiple times now. The antidiagonal of a list of computable reals is not necessarily a computable real. Your insistence that it must be is almost certainly because you don't know what a computable real is. (There are other possibilities, even less flattering to your math skills) In particular, the antidiagonal of a list of all computable reals is never a computable real. - Tim |