Why can no one in sci.math understand my simple point?
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From: Tim Little on 19 Jun 2010 23:50 On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Tim Little" <tim(a)little-possums.net> wrote in message >> Where is the phrase "provided in advance" used in the proof? > > Because Cantor's proof requires us to know the nth digit of the nth > item on the list. No, it just requires that the list L (a mathematical object) be of the type that an n'th digit of the n'th item *exists*. > Indeed Cantor's proof starts with a list No, Cantor's proof starts with a conditional assumption that some mathematical object *is* a list. > If you don't what is on the list already, you cannot possibly prove > that some Real is missing. Now it looks like your deficiency might be in knowing what a proof in predicate logic is. You don't need a separate proof for each list, each one based on the contents of that list. You have a single proof that there does not exist a mathematical entity satisfying the definition of a surjective mapping from natural numbers to reals. > My proof is *exactly* the same, except for inserting the word > "computable" in front of Real. And therefore deficient as making that textual substitution leaves a gap between Cantor's proof that antidiag(L) is a real, and your substituted claim that antidiag(L) is computable real. > Give me a list of all computable numbers in the same form as > Cantor's proof requires a purported list of all Reals, ie a list > where the nth digit of the nth term is known. The constructive portion Cantor's proof requires nothing more than a mapping from N to R. In particular, it does not require that the n'th digit of the n'th term in that mapping "is known". That requirement is entirely due to your imagination. > So by definition if the list contains only computable Reals, we know > the nth digit of each term. Careful with your quantifiers! For a list of computable reals, the following is true: for all n, there exists Turing machine T such that T(n) computes the n'th digit of L(n), but the following is false: there exists Turing machine T such that for all n, T(n) computes the n'th digit of L(n). You appear to be claiming the latter. Are you? - Tim
From: Tim Little on 20 Jun 2010 00:09 On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Tim Little" <tim(a)little-possums.net> wrote in message >> If you intend your proof to be "exactly the same as Cantor's, but with >> the only difference being the word "computable" in front of the word >> "Real"", it must start with a conditional introduction. In other >> words, something like: >> >> "Suppose that L is a list of computable Reals. That is, L is a >> function from N to R and for all n in N, there exists a Turing >> Machine T_n such that when provided with k as input, T_n halts and >> outputs the k'th decimal digit of L(n)." > > That's not how Cantor's proof starts. Correct: It didn't feature any mention of computable reals (but as I recall it did feature the relevant properties of the definition of a real number). I am presuming that you want *your* proof to substitute "computable real" for "real", and therefore you need to substitute the definition of computable real number for the definition of real number. That means you also need to make that substitution in the definition of a list of computable real numbers. Are you starting to see now why it makes no sense to just drop "computable" in front of every occurrence of the word "real" in the proof? - Tim
From: Tim Little on 20 Jun 2010 00:19 On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > Of course it is computable. > > Cantor provides an explicit algorithm for computing it. Cantor provides an explicit algorithm for computing it with the list L as input. Look again at the definition of "computable real". Is "the algorithm may take as input an auxiliary list of infinite many real numbers" part of that definition? > I make no mention of finite algorithms. I just do *exactly* the same > construction, but applied to a purported list of all computable > Reals instead of a purported list of all Reals. The definition of your term "computable real" includes references to "finite algorithm". So you do make reference to a finite algorithm, and Cantor does not. > Cantor: Lets form an anti-diagonal using this explicit construction > based upon the nth digit of the nth Real on the list. It is > obviously a Real, and it is obviously missing. Cantor did not make any assumption that it was "obviously" a real. In fact, he went to some trouble to *prove* that it specified a real number. > Peter: Lets form an anti-diagonal using this explicit construction > based upon the nth digit of the nth computable Real on the list. It > is obviously a computable Real, and it is obviously missing. You, on the other hand, make no effort whatsoever to prove that the antidiagonal number is computable. If you actually did, you would run into insuperable difficulties. The devil is in the details in any mathematical proof. You are actively *avoiding* the details and just assuming they work out. That renders your so-called proof invalid, and is exactly why I accept Cantor's and not yours. At this point, and in this respect, it is clear that you are just as much of a crank as JSH and WM. - Tim
From: Tim Little on 20 Jun 2010 00:24 On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Tim Little" <tim(a)little-possums.net> wrote in message >> Only if the list itself is a recursively enumerable function. >> Cantor's proof makes no such assumption. > > Yes it does. It requires that the nth digit of the nth term can be > calculated. No, it merely assumes that such a digit exists. > This is not quite as strong as being re, but it is close. It is not at all close. - Tim
From: Tim Little on 20 Jun 2010 00:27
On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > The anti-diagonal can be computed to any required degree of > accuracy. Only given as auxiliary input the infinite list of reals. The definition of "computable real" provides no such luxury as infinite auxiliary inputs. So you will need to find some other path to your so-called proof. - Tim |