Why can no one in sci.math understand my simple point?
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From: Owen Jacobson on 21 Jun 2010 00:32 On 2010-06-19 22:31:56 -0400, Peter Webb said: > Cantor's original proof requires the list to be provided in advance No, it does not. It's an existence proof showing that: IF a list of reals L(x) (where L : N -> R) exists, THEN a real A_L exists such that for every natural number n, L(n) ≠ A_L. Or, equivalently, showing that: For every function L from N to R, there exists a real r not in the image of L. Being a constructive proof, it's usually *presented* as if it were an algorithm or a procedure, but it's only for ease of comprehension. Unfortunately, that seems to have backfired on you. -o Incidentally, "countable" and "listable" mean the same thing in conventional set theories: an infinite set S is countable if and only if there exists an surjective function f from N to S. Since a "list" of elements of S is most easily formalized as a surjective function from N to S, denying that S is listable is equivalent to denying that it is countable.
From: Peter Webb on 21 Jun 2010 02:08 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1tkli.jrj.tim(a)soprano.little-possums.net... > On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> "Tim Little" <tim(a)little-possums.net> wrote in message >> I do understand that. But I don't actually ask for a "finite algorithm" >> in >> my proof, I only require that I am provided a purported list of all >> comptuable Reals. > > Later in the proof, you claim that the antidiagonal real is > computable. In other words, that there exists a finite algorithm for > computing it. > > >> Yes, in practice, this means that each item is specified by a finite >> algorithm, but so what? I ask for a purported list of all computable >> Reals >> and show there is at least one missing. > > At least one *what* missing? A real? Fine - Cantor's proof shows > that there is a missing real. And if the nth digit of the nth term is computable, then so is the nth term of the anti-diagonal, as there is an explicit construction of the nth digit of the anti-diagonal based upon the nth digit of the nth term. > What's that? You want the antidiagonal > to be a *computable* real? Well, then - you are required to show that > there always exists a finite algorithm to compute it, and that is not > present in Cantor's proof. > Well, I don't actually have to write out all "infinite" decimals in its expansion to prove it is computable. All I really have to be able to do is to compute it to arbitrary accuracy. And Cantor's proof constructs the number to arbitrary accuracy. > So your modified "proof" is either invalid, or you must include > significant amounts of material not present in Cantor's proof. > Well, really only one thing. I have to produce an algorithm to estimates the anti-diagonal to arbitrary accuracy. Fortunately, the digit chnage rule of Cantor uniquely specifies the nth decimal place of the anti-diagonal in a completely finite closed procedure. > > As is well-known to competent mathematicians and computer scientists, > no amount of material will make it valid, as the modified "theorem" is > simply false. It is *not true* that any list of computable reals has > a computable antidiagonal. You give me a list of purported computable Reals, I can compute the anti-diagonal to arbitrary precision. Just like Cantor can compute the value of the missing Real to arbitrary precision. > > > - Tim
From: Peter Webb on 21 Jun 2010 02:14 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1tnji.jrj.tim(a)soprano.little-possums.net... > On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> OK, I have a list of all Reals. Which real is missing? > > Denoting your list by L, antidiag(L) is missing. If you *tell* me L, > I can verify that it is missing. Otherwise you'll have to verify it > yourself. Either way, it is missing. > OK, you tell me your list of computable Reals, and I will verify which one is missing. > >>>> But you have made an excellent point in your example. When Cantor is >>>> confronted with some digit as poorly defined your Chaitan's number >>>> example >>> >>> It's not poorly defined at all; it is precisely defined. >> >> No, its not precisley defined, you can't tell me its value to arbitrary >> accuracy. > > It looks like the concept of mathematical definition is also an area > of incompetence for you. Failing that, there really in no point in > continuing any further. Goodbye. > Funny, rather than address my proof you now seem to just want to be rude. Very crank like behaviour on your part. If this really is goodbye, then cheerio and thanks for the (mathematical) input. > > - Tim
From: Peter Webb on 21 Jun 2010 02:20 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1tmi9.jrj.tim(a)soprano.little-possums.net... > On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> "Tim Little" <tim(a)little-possums.net> wrote in message >>> On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >>>> No, I only need a finite algorithm which will calculate it to n >>>> decimal places with finite input. >>> >>> You have previously stated that you can do this with "about three >>> lines of Java code". >>> >>> Here's your Java method signature: int antidiagonalDigit(BigInteger n). >>> >>> Here are the first twenty digits of the first twenty members of my >>> list: >>> >>> 93525532854532512838 >>> 32127313472944276266 >>> 70595916184994935423 >>> 20733652719572401688 >>> 30472031767118774150 >>> 47190325821263633948 >>> 74236814853458351851 >>> 58521903865615844550 >>> 91701104659863267390 >>> 39510921669610680229 >>> 19656091025330568974 >>> 49591084533660072011 >>> 81683520683673830124 >>> 93720622611168810054 >>> 50284245443806050152 >>> 11702670934156383526 >>> 58534679962278312978 >>> 68478827933494896765 >>> 83579375050000862329 >>> 50241229144880453593 >>> >>> Now let's see your Java code. >> >> Here is my p-code. > > As I suspected, your claim that you could actually do it was pure > puffery, and you probably know as little about programming as you do > about mathematics. > > >> Take the nth digit of the nth item. > > The n'th item of *what*? The list. Of computable Reals. That is supposed to include all Computable Reals. > The list is not a parameter, it is not a > global variable, it is completely out of scope. You fail already. > > Actually, Cantor's proof starts with a purported list of all Reals, and then shows that at least one Real is missing. It assumes that a list exists, then shows a Real is missing. If you didn't supply the list in advance, he wouldn't be able to construct a Real missing from the list. In fact, if you don't say what is on the list, it you can't prove soimething is missing from it. There is no Real which is missing from every list, but every list misses at least one Real. You do believe Cantor's original proof that the Reals cannot be listed, don't you? > - Tim
From: Peter Webb on 21 Jun 2010 02:28
"Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1tm56.jrj.tim(a)soprano.little-possums.net... > On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> The same algorithm does work for all n. Look at the first n terms to >> n places of accuracy. > > Provide computer code, please. Here's the required function > signature: int getDigit(BigInteger n). You get to fill in the > function body. > > >>> For some fixed M, you could embed M digits of the diagonal into the >>> algorithm, and the algorithm would work for all n < M. But that's not >>> enough: it has to work for all n. >> >> I can in fact compute the nth digit of the anti-diagonal for any n >> by hand. > > "By hand" is irrelevant to the definition of a computable real, > especially when your "hand" is connected to "eyes" that have the list > as input. The list is *not* input to the algorithm. > Yeah, it is. Cantor constructs the antidiagonal by taking the list and then computing a missing Real based upon the decimal expansion of every Real on the list. > >> The same objection could be raised to Cantor's proof; after all, he >> accepts a purportedly infinite list as input to his algorithn as >> well. > > Cantor does not claim to be proving existence of any finite algorithm. > You are. > Cantor explicitly calculates the real that is missing be examining the decimal expansion of every item in the list. No more or less than I do. > >> Or are you saying I have to do this without using the infinite list >> itself? > > You may use the infinite list to determine which finite algorithm to > use, but the algorithm must be self-contained and not refer to the > list. Not in Cantor's proof. He explicitly uses the decimal expansion of every Real on the list to construct the anti-diagonal. Of course Cantor has know what's on the list to determine an element which is missing. In exactly the same way as I do. > The *only* input to the algorithm you select will be a natural > number, which denotes n. > > >> I note that Cantor's original proof also requires an infinite list >> of Reals, but you seem to have no problem with that. > > Indeed I don't. Cantor is not claiming that the antidiagonal is a > computable real, and so finite algorithms are irrelevant. > Yet he explicitly constructs the missing Real to arbitrary accuracy. > You *are* claiming that the antidiagonal is a computable real, and so > you must prove the existence of a suitable finite algorithm for the > antidiagonal. > Actually, I only need to be able to specify it to arbitrary accuracy for it to be computable, and I can certainly do that. Or are you trying to claim that while I can calculate the explicit decimal expansion of the missing computable Real, I can't compute the missing computable Real itself? If I can compute every digit in the missing computable number, I have computed the number itself. > > - Tim |