Why can no one in sci.math understand my simple point?
in [Logic]
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From: Peter Webb on 18 Jun 2010 02:20 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1m1vt.jrj.tim(a)soprano.little-possums.net... > On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> The computable Reals cannot be listed. > > False. > > >> Maybe your definition needs a little work? > > No maybe about it, your definition of "listable" needs work. > > > - Tim Give me a list of all computable Reals. I will use Cantor's diagonal argument to explicitly construct a Real which isn't on the list. This is diagonal number clearly computable, as there is an explicit construction for it. All I have to do is change the nth digit of the nth number on the list, a very easy computation.
From: Sylvia Else on 18 Jun 2010 02:21 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.
From: Peter Webb on 18 Jun 2010 02:26 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1m25q.jrj.tim(a)soprano.little-possums.net... > On 2010-06-18, Virgil <Virgil(a)home.esc> wrote: >> Since you object to there being any bijection from N to any superset >> of S, you must equally be rejecting any surjection from N to S and >> rejecting any injection from S to N, since from any such bijection >> such a surjection is easily derived. > > Yes, Peter is very confused. > No. > >> So in what sense do you claim that the the set S of computable >> numbers is countable? >> >> It is certainly not in any sense that I am aware of. > > Heh, these two sentences would be great to quote out of context. Hey, they are great even in context. The computable numbers definitely are countable, as google will verify. (Or can easily be proved by associating computable Reals with TMs that produce them, and there are only countable TMs). > In > the context of Peter's premise the latter is true, but in ordinary > context of mathematics it is very obviously false. > I made no premise. I simply provided an example of a set which is: (a) Countable, but (b) Cannot be listed explicitly. Cantor's diagonal proof shows that Reals cannot be explicitly listed. This is *not* equivalent to the statement they are uncountable, as the example of the set of all computable numbers shows. These cannot be formed into a list, but are nevertheless still countable.
From: Peter Webb on 18 Jun 2010 02:39 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1m27v.jrj.tim(a)soprano.little-possums.net... > On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> Cantor's proof does *not* demonstrate that Reals are uncountable, it >> just proves there can be no explicit enumeration of them. > > It proves that there is no enumeration of them, explicit or not. I > have no idea where you are getting this strange notion of > "explicitness". > Because Cantor's proof requires an explicit listing. This is a very central concept. I can form a list of sorts of all computable Reals. I can associate every Real with the TM that produces it, and list TMs in order. The trouble is that this is not an explicit list, as you cannot say exactly what number appears at each position in the list. This means Cantor's proof cannot be used, which assumes you explicitly know what number appears at every position in the list. HTH > > - Tim
From: Tim Little on 18 Jun 2010 02:43
On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > If you can construct a list of all computable numbers (which you > can't), then Cantor's diagonal proof will construct a number not on > the list. And that number is definitely computable, because there is > a simple algorithm for producing it. Take the nth digit of the nth > item on the list That requires having the list be computable or provided as input, neither of which is assumed or proven. Rest snipped as it is based on your false premise. - Tim |