Why can no one in sci.math understand my simple point?
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From: Sylvia Else on 18 Jun 2010 23:19 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.
From: Peter Webb on 18 Jun 2010 23:30 "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. Just as there can be no list of all Reals. It is exactly the same proof. However, the computable Reals are countable. Therefore the predicate "cannot be listed" is *not* the same as "are uncountable". Cantor's diagonal proof demonstrated merely te weaker condition that the Reals cannot be listed; it does not prove they are uncountable. >>>> Cantor's proof provides an explicit and simple algorithm to form >>>> such a number, and it is extremely easy to compute. >>> >>> If there exists no algorithm for computing the original list, how do >>> you propose to use Cantor's proof to make an algorithm for the >>> antidiagonal of that list? >> >> Cantor's proof just requires a list of numbers. >> If you give me a purported list of all computable numbers, I can produce >> a computable number not on the list. > > But ask yourself QUESTION 1 > > http://en.wikipedia.org/wiki/Computable_number > >>>>> It takes quite a bit of extra work (not part of Cantor's proof) to >>>>> deduce that there is no computable list of all computable reals. >>>> >>>> Cantor's proof applied to computable numbers proves it immediately and >>>> simply. >>> >>> First you have to define the notion of "computable list", and set up >>> the theorems proving properties about how computable lists and >>> computable numbers relate. >>> >> >> >> No I don't, any more than Cantor had to that for all Reals. >> >> You cannot produce a list of all computable Reals. Just as you can't >> produce a list of all Reals. > > But ask yourself: since there is no computable way to generate a list of > all computable reals does that mean that there is no non-computable ways > of getting such a list? Cantor's proof is based upon a purported complete list to be provided, and then it is proved that the list cannot be complete as it misses some elements which should be on the list. How you form the list is your business. Cantor's proof assumes that you are given the list first. > >> This does not however prove that either is an uncountable set; in fact >> the set of all computable numbers is countable. >> >> >>> >>>>> I agree. It was Herc's deranged ramblings that brought computability >>>>> into this discussion in the first place, when they really are >>>>> completely irrelevant to Cantor's proof. >>>>> >>>> >>> >>>> Well, they are relevant if you try and make the mental jump from >>>> "cannot be listed" to "are therefore uncountable". This simply does >>>> not follow. >>> >>> It follows immediately from the definitions. You appear to have some >>> bizarre definition of "list" that Cantor never used, employing >>> concepts not developed until many decades later. >>> >> >> 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? >> In fact the computable Reals can't be listed either, using exactly the >> same diagonal proof, but they are countable. > > No, computable reals can be listed but there is no computable algorithm > that can generate such a list. But if you were given such a list then you > could create an anti-diagonal number not on the list. That anti-diagonal, > though, is non-computable since it is generated from a non-computable > list. <-- But note, what I just wrote there applies to the algorithm in > QUESTION 1, not the algorithm you use to generate the anti-diagonal. That > is, the anti-diagonal algorithm, in the sense of which digits are changed > on the diagonal (e.g. 9-->0, 1-->2, etc), is obviously computable. Please > read: > > http://en.wikipedia.org/wiki/Computable_number > Again, I ask you, what is it that you think Cantor's proof applied to Computable Reals proves? I might remind you that as the form of proof is identical to that used by Cantor for all Reals, whatever you believe that Cantor's proof applied to computable Reals proves, his proof applied to all Reals must prove the same thing. Its the same proof, after all, except limitting the set to just computable Reals. >>>> The computable Reals cannot be listed either, but they are >>>> definitely countable. >>> >>> The computable reals can be listed and are countable. >>> >>> >>>> A Cantor construction applied to a purported list of all computable >>>> Reals will identify a computable Real not on the list. >>> >>> No, it will not. >>> >> >> Yes, it will. >> >> Give me any list of computable numbers and I will produce a computable >> number not on the list. All I have to do is constrruct the diagonal >> number; this is clearly computable (as the diagonal construction >> "computes" it), and its not on the list. >> >> Go on, try it. If you think you have an explicit list of all computable >> Reals, produce it for me. "Computing" the anti-diagonal is easy. > > Yes it is, but there is no finite way of getting such a list (although > each real in the list is generated by a finite algorithm). 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 diagonal is explicitly constructible given the list. >>> >>> The definition of computable number does not include any >>> presupposition of being supplied with infinite lists of infinite digit >>> sequences. To be computable, a number must be constructible by a >>> finite algorithm *without* being given the list. Don't take my word >>> for it, look up the definition yourself. >>> >> >> We are assuming every number on the list is computable; that is the >> premise. >> >> 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. Yet they are still countable. Reals cannot be listed either. Which Cantor's proof demonstrates. But this does not neccesarily mean they are uncountable. > > _ > >
From: Peter Webb on 18 Jun 2010 23:35 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1o6pf.jrj.tim(a)soprano.little-possums.net... > On 2010-06-18, Virgil <Virgil(a)home.esc> wrote: >> Given the axiom of choice, as in ZFC, any countable set must be, at >> least theoretically, listable, though such a listing need not be >> computable. > > The definition of countability is the existence of a bijection with a > subset of N. Since N can be well-ordered even in ZF, the theorem > (countable -> listable) is easily proven without AC. > > > - Tim Well gee, so you say. I agree that computable reals are countable. But I do not agree this means they can be listed. In fact, I can easily prove they are not. If you give me a purported list of all computable Reals I can use a diagonal argument to form a computable Real not on the list. Clearly there are countable sets that cannot be listed. Cantor proved that the Reals cannot be listed. His diagonal proof does *not* show they are uncountable, any more than doing exactly the same proof with computable Reals "proves" they are uncountable.
From: |-|ercules on 18 Jun 2010 23:40 "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. Herc
From: Newberry on 18 Jun 2010 23:50
On Jun 18, 5:17 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Newberry <newberr...(a)gmail.com> writes: > > On Jun 15, 9:46 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > > wrote: > >> WM says... > > >> >On 15 Jun., 16:32, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> >> The proof does not make use of any property of infinite lists. > >> >> The proof establishes: (If r_n is the list of reals, and > >> >> d is the antidiagonal) > > >> >> forall n, d is not equal to r_n > > >> >As every n is finite, it belongs to a finite initial segment of the > >> >infinite list. > > >> I'm not sure what you are saying. The fact is, we can prove > >> that for every real r_n on the list, d is not equal to r_n. > >> That means that d is not on the list. > > > How do you know that it does not prove that an anti-diagonal does > > exist i.e. that an antidiagonal is a contradiction in terms? > > Because every infinite sequence of digits represents a real number? And > the antidiagonal is one such sequence? If it does not exist then it does not represent anything let alone a number. Now it is clear that it does not exist. Since all the reals are on the list and the anti-diagonal would differ from any of them. This violates the assumption. Hence the anti-diagonal does not exist. > > -- > Jesse F. Hughes > > "Yesterday was Judgment Day. How'd you do?" > -- The Flatlanders- Hide quoted text - > > - Show quoted text - |