Why can no one in sci.math understand my simple point?
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From: WM on 18 Jun 2010 16:50 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". Regards, WM
From: Virgil on 18 Jun 2010 20:42 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.
From: Virgil on 18 Jun 2010 20:57 In article <546be0b1-b176-428c-b7c3-c6794588f082(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 18 Jun., 05:22, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > wrote: > > > Since by definition, "listability" = "countability", Cantor's proof of > > > unlistability proves uncountability. > > > > Really? Where did you get that from? > > From Cantor. > > > > The computable Reals cannot be listed. > > > > Therefore according to you they are uncountable. If one defines countability of a set standardly, it requires existence of a surjection from the set of naturals to the set in question. So proof that there can be no such surjection is clearly sufficient to show that a set is not countable. > No. There is no uncountability. That needs to be understood. There is certainly such a definition, but in order to eliminate uncountability, WM would first have to eliminate infiniteness, which he also fails to do. > Uncomputable numbers are not numbers, because nobody knows what they > are. There are certainly what are called noncomputable numbers according to http://en.wikipedia.org/wiki/Computable_number A real number, a, is said to be computable if it can be approximated by some computable function in the following manner: given any (positive) integer n, the function produces an integer k such that: (k-1)/n <= a <= (k+1)/n Otherwise it is noncomputable.
From: Tim Little on 18 Jun 2010 21:09 On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > Now take the 3rd digit of the 3rd number ... its a "8", so the third digit > of the diagonal number is a "1". > > Continue in this fashion. > > 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. - Tim
From: Tim Little on 18 Jun 2010 21:19
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 |