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: Sylvia Else on 18 Jun 2010 03:14 On 18/06/2010 4:52 PM, |-|ercules wrote: > "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 What are you trying to prove? Sylvia.
From: |-|ercules on 18 Jun 2010 03:31 "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... > On 18/06/2010 4:52 PM, |-|ercules wrote: >> "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 > > What are you trying to prove? There is only one type of infinity. -oo <--|---|---|---0---|---|---|--> oo Where's yours go? Herc
From: Sylvia Else on 18 Jun 2010 03:37 On 18/06/2010 5:31 PM, |-|ercules wrote: > "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... >> On 18/06/2010 4:52 PM, |-|ercules wrote: >>> "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 >> >> 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. Which set are you using when discussing infinity? Sylvia.
From: Ostap Bender on 18 Jun 2010 03:42 Because you are too brilliant for the rest of Humankind.
From: Peter Webb on 18 Jun 2010 03:47
"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 construct a Real not on the list. Of course this number is computable; there is a simple algorithm to compute it. > >> 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. > > >>> 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. 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. In fact the computable Reals can't be listed either, using exactly the same diagonal proof, but they are countable. > >> 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. > >> 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. I take the nth digit of the nth number on the list. The nth number is given to me as part of the list, and so by definition is computable. I look at the nth digit. That is obviously computable. If it is a "1", change it to a "2", and if it is anything else change it to a "1". That is a obviously computable. This is a trivially easy construction that is clearly computable which will give me a computable number not on the list. > > - Tim |