Why can no one in sci.math understand my simple point?
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From: |-|ercules on 19 Jun 2010 02:11 "Sylvia Else" <sylvia(a)not.here.invalid> wrote... > On 19/06/2010 11:43 AM, |-|ercules wrote: >> "Virgil" <Virgil(a)home.esc> wrote >>> 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. >> >> Unfortunately your last paragraph is speculation and your superinfinity >> rules >> are full of contradictions. >> >> You show this example as Cantor's method >> >> 123 >> 456 >> 789 >> >> Diag = 159 >> AntiDiag = 260 >> >> But this method miserably fails on the computable set of reals, THERE IS >> NO new sequence like 260 in this example. > > Computable set of reals? Set of computable reals? I mixed up "miserably fails" with "fails miserably" but don't see what your distinction is here. > > Cantor proves that the set of reals is not countable. It's not saying > anything about computables. > OH FFS. >> >> Why don't you rework Cantor's proof to define ALL anti-diagonals instead >> of 1 particular anti-diagonal? > > Because there's no need. We hyphothesise that the reals are countable. > Since they are countable we can list them, and the list will contain all > the reals. Then we show that there's a number that isn't in the list, > which contradicts the hypothesis, which must therefore be wrong. Hence > the reals aren't countable. It seems straightforward enough. Do it anyway and you'll see what a bogus argument you have!! > To avoid this proof, you have to dispute reductio-ad-absurdum arguments > in general, argue that reals cannot be expressed as infinite strings of > digits, or have axioms that exclude the construted anti-diagonal from > the set of reals. > > Sylvia. To support your argument you should at least show that you've formed a new sequence of digits. If you actually read my derivation of herc_cant_3 instead of blindly dismissing it, you'll see it holds, just like all digits of PI appear in order below this line, if interpreted correctly. Herc ___________________ 3 31 314 3141 ....
From: Virgil on 19 Jun 2010 02:12 In article <slrni1ole2.jrj.tim(a)soprano.little-possums.net>, Tim Little <tim(a)little-possums.net> wrote: > On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > >> The relevant point: the *only* input to the Turing machine in the > >> definition is n. The rest of the tape must is blank. Peter > >> apparently believes that a number is still computable even if the > >> Turing machine must be supplied with an infinite amount of input (the > >> list of reals). > > > > No. > > Oh? Then what leads you to believe that the antidiagonal of a (not > necessarily recursive) list of computable reals is computable? > > > > Do you agree that Cantor's diagonal proof when applied to a > > purported list of all computable Reals will produce a computable > > Real not on the list > > No, for the fifth time now. The antidiagonal of a list of all > computable real is not computable. How many more times would you like > me to repeat this simple and mathematically obvious statement? There is, however, some question in my mind about the existence of a list of all and ONLY computable reals. For countability of a set it is certainly sufficient to have a list containing all its members even if that list is allowed to contain other things as well.
From: Peter Webb on 19 Jun 2010 02:12 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1okg1.jrj.tim(a)soprano.little-possums.net... > On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> "Tim Little" <tim(a)little-possums.net> wrote in message >>> Your "simple fact" is simply wrong. Look up the definition of >>> "computable real" and get back to us. >> >> Why? > > Because you clearly don't know how a computable real is defined. > > >> If you believe there is an error in what I have said, identify it. > > I have, many times now. > > >> Or give me at least a hint. As a start, my argument rests on two >> observations: >> >> 1. Cantor's diagonal construction, when applied to a purported list >> of all computable Reals, will produce a computable Real not on the >> list. > > The latter part is incorrect. It produces a real, but not a > computable real. If you believe otherwise, provide a proof that the > antidiagonal real is computable. > Well, there are lots of definitions of a computable Real, I will use Wikipedia's most intuitive definition: "computable reals, are the real numbers that can be computed to within any desired precision by a finite, terminating algorithm." 1. Given a list of computable Reals, we can identify the nth computable Real on the list by simply counting down to the nth entry. 2. Given the nth computable Real on the list, we can count off and identify the nth digit of this Real. 3. With the nth digit of the Real, we can use Cantor's construction to identify the nth digit of the anti-diagonal. 4. As we can specify every digit in the anti-diagonal explicitly and to any desired degre of accuracy, it is therefore computable. 5. But the anti-diagonal number does not appear on the list. 6. Therefore, the list could not have included all computable Reals. Exactly the same as Cantor's proof that the Reals cannot be listed. The only difference is that I have to prove that the anti-diagonal is also computable, but because Cantor's proof explicitly constructs the anti-diagonal it is clearly computable (as long as every item in the list is computable, which is the starting assumption which we prove to be incorrect, just like in Cantor's original proof). HTH Peter Webb
From: |-|ercules on 19 Jun 2010 02:14 "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... > On 19/06/2010 1:40 PM, |-|ercules wrote: >> "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. > > Your question was typically vague. They you dived into some notation > which might be construed to mean "if X and X implies Y, then Y", which > is itself unproveable, and needs to be introduced as an axiom. > > None of which eliminates your need to specify the axioms in which you're > making claims about Cantor's proof. > > Sylvia. HAHAHAHA You never studied theorem provers. You're like Wally Anglesea, one of the thickest morons I've ever come across, but he has in innate ability to regurgitate the arguments of other skeptics, in the right places, imitating intelligence. Herc
From: Peter Webb on 19 Jun 2010 02:19
"Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1ol51.jrj.tim(a)soprano.little-possums.net... > On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> Of course there exists a mapping from N to computable numbers. But >> Cantor's proof requires more than that; it requires the mapping to >> be recursively enumerable such that we can also explicitly list >> them. > > It does not. If you believe otherwise, feel free to show where in > Cantor's proof any requirement for recursive enumerability is stated > or implicitly used. > Its not. > The simple fact is that no such requirement exists. Cantor's diagonal > proof applies to *every* mapping from N to R, recursively enumerable > or not, showing that no such mapping is surjective. No. It requires that any listing be provided in advance, so that the anti-diagonal can be formed. > > Your introduction of recursive enumerability into Cantor's proof is > entirely your own fabrication. Well, I did say it was my interpretation of what was going on. The simple fact is that if insert the word "computable" in front of every reference to "Real" in Cantor's proof that there is no list of all Reals, you produce a proof that thye computable Reals cannot be listed either. yet they are known to be countable. Cantor's proved that the Reals cannot be listed. It is your incorrect assumption that this means they are also uncountable. Some countable sets cannot be listed either, such as the set of Computable Reals. This is because the set of computable Reals is not recursively enumerable, and hence cannot be listed by a terminating process. > > > - Tim |