Why can no one in sci.math understand my simple point?
in [Logic]
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From: Tim Little on 19 Jun 2010 01:10 On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > Which of these two statements do you agree with: > > 1. You cannot form a list of computable Reals. > > 2. The computable Reals are countable. Answering the same questions for the fourth time now, 1 is false, 2 is true. > You have failed to explain why Cantor's diagonal proof "proves" the > Reals are uncountable, but the same proof applied to computable > Reals does *not* prove the Computable Reals are uncountable. I have, multiple times now. The antidiagonal of a list of computable reals is not necessarily a computable real. Your insistence that it must be is almost certainly because you don't know what a computable real is. (There are other possibilities, even less flattering to your math skills) In particular, the antidiagonal of a list of all computable reals is never a computable real. - Tim
From: Tim Little on 19 Jun 2010 01:13 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. Hint: start with a statement of what you are trying to prove, i.e. the definition of a computable real. - Tim
From: Tim Little on 19 Jun 2010 01:24 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. 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. Your introduction of recursive enumerability into Cantor's proof is entirely your own fabrication. - Tim
From: Tim Little on 19 Jun 2010 01:29 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? - Tim
From: Tim Little on 19 Jun 2010 01:39
On 2010-06-19, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > So Cantor's proof when applied to computable Reals proves exactly > what in your opinion? That there is a real not on the list. > I might remind you that as the form of proof is identical to that > used by Cantor for all Reals, whatever you believe that Cantor's > proof applied to computable Reals proves, his proof applied to all > Reals must prove the same thing. It does prove exactly the same thing: in both cases, all such lists omit at least one real. > Its the same proof, after all, except limitting the set to just > computable Reals. No, there is a slight difference: the antidiagonal is a real number. It does not necessarily produce a computable real number. Why, in your opinion, does the construction fail for rational numbers? - Tim |