Why can no one in sci.math understand my simple point?
in [Math]
|
Prev: power of a matrix limit A^n
Next: best way of testing Dirac's new radioactivities additive creation Chapt 14 #163; ATOM TOTALITY
From: Tim Little on 18 Jun 2010 02:49 On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > Wrong. Cantor's diagonal construction explicitly forms a Real not on > that list, and the Cantorm diagonal number is obviously constructibe > if every item in the list is constructible. You are now mixing up "constructible" with "computable", getting yourself even more confused. It is certainly possible to have a list of computable numbers that is not computable itself. For example: Chaitin's Omega is not computable. 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. Do you agree that every entry in the list is computable? Do you think that the diagonal is therefore computable? > If you believe that you have a list of all computable Reals, I can > use a diagonal argument to explicitly construct a Real not on the > list. This number is obviously computable; Cantor provides an > explicit algorithm for constructing it, which can be implemented in > a few lines of code or in a TM. Again, look up the definition of "computable number". Note carefully the absence of the Turing machine being provided with an infinite list of infinite sequences. Come back when you have at least checked a basic definition of the terms you are using. - Tim
From: |-|ercules on 18 Jun 2010 02:52 "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... > On 18/06/2010 3:03 PM, |-|ercules wrote: >> "Sylvia Else" <sylvia(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
From: Tim Little on 18 Jun 2010 02:53 On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > Give me a list of all computable Reals. I specified such a list previously in this thread. It is not a computable list, but you didn't ask for a computable list - just a list of computable reals. They are not the same thing. > I will use Cantor's diagonal argument to explicitly construct a Real > which isn't on the list. This is diagonal number clearly computable In this case it clearly is not computable. > as there is an explicit construction for it. No, there is only a finite algorithm *relative to* being provided with the list initially. There is no algorithm to produce the antidiagonal starting without input, which being computable would require. Come back when you read what the term "computable number" means. - Tim
From: Tim Little on 18 Jun 2010 02:55 On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > I made no premise. Sure you did: you assumed that no list of computable numbers can exist. You also assumed an incorrect definition of "computable". - Tim
From: Tim Little on 18 Jun 2010 02:58
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. The rest of your misconception snipped. - Tim |