Cantor Confusion
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From: Virgil on 30 Jan 2007 15:45 In article <622tCB$Tw0vFFwT5(a)phoenixsystems.demon.co.uk>, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > Just following up the idea of countability. > > Am I right in thinking that any set of "finitely describable" objects is > necessarily countable? - on the grounds that "finitely describable" > implies that there exists some (multi-dimensional) address space which > uniquely defines an object, and that that address is indexable by a > finite set of natural numbers, and then that that multi-dimensional > address (as with the "standard arrangement" for counting the rationals) > can be ordered such that each address has a unique index in the natural > numbers? > > On that basis I would surmise that e.g. the set of all roots of all > polynomials with rational coefficients are countable - is that correct? Yup!
From: Virgil on 30 Jan 2007 15:47 In article <1170164829.580165.131750(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > The proof has been given. > > > > In which peer-reviewed journal? > > Which proof are you talking about? > > Regards, WM The one you were referring to. Actually any of your "proofs".
From: Virgil on 30 Jan 2007 15:50 In article <1170164725.794981.290930(a)a34g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > Continued > > > > > 2. BECAUSE it is _undefined_ until the "sum of all natural numbers" is > > defined. > > This sum is defined by > I > II > III > ... > WRONG! As usual. A listing of all natural numbers is not the same as the summing of all natural numbers, and WM has, at most, produced a listing.
From: Dave Seaman on 30 Jan 2007 16:17 On 30 Jan 2007 12:14:39 -0800, MoeBlee wrote: > On Jan 30, 4:42 am, Dave Seaman <dsea...(a)no.such.host> wrote: >> On Tue, 30 Jan 2007 01:33:03 -0500, David Marcus wrote: >> > MoeBlee wrote: >> >> On Jan 27, 2:26 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: >> >> >> > The maximum number of bits required to describe any index no. i is >> >> > ceiling(log(i+1,base 2)). For all i>0, i+1> ceiling(log(i+1,base 2)). >> >> > And for i = 0, the real number is 0.000 . So, all terms on the diagonal >> >> > are 0 (and, more of this later, the distance in bits between the 0 on >> >> > the diagonal and the first non zero bit goes as (n - log(n,base2)) for >> >> > bit n. >> >> >> > Following Cantor, we construct an antidiagonal number, different from >> >> > the first row, the second row, etc. Because all the diagonal terms are >> >> > 0, we construct the number .1111.... >> >> >> > This string is certainly not in the hypothetically complete list, but it >> >> > is a real number and = 1.000... , which was excluded anyway from our >> >> > initial list. So we can make no inference that the list is >> >> > incomplete/invalid - we have just generated a number. >> >> >> What you've shown is that applying the diagonal method to your list >> >> only shows a real number not in [0 1). So what? It doesn't contradict >> >> that for any list of real numbers, there is a real number not in the >> >> list. We don't have to prove: For any list of real numbers in a >> >> certain interval, there is a real number in that interval not listed. >> >> Rather, we only have to prove: For any list of real numbers, there is >> >> a real number not in the list. >> > The usual diagonal proof proves that [0,1) or [0,1] (depending on how we >> > do it) is uncountable. So, the fact that the anti-diagonal is in the >> > specified interval is important in the usual proof. The main problem >> > with Andy's argument is that he is using a specific list. >> >> Actually, the main problem with Andy's argument is that he is using a >> defective variation of the diagonal argument. When correctly done, the >> diagonal argument does not produce a number that has a dual >> representation. >> >> >> Your lack of understanding of basic logic in mathematics has lead you >> >> to this irrelevent red herring/strawman about [0 1). >> >> It's not irrelevant. A correct argument applied to that list shows that >> there is a number in [0,1) that is not in the list. One way is to >> represent the numbers in decimal and to carry out the diagonal argument >> in such a way as to produce a number, all of whose digits are either 4 or >> 5. > The irrelevancy is in the non sequitur from giving a list of reals in > an interval such that the anti-diagonal (in a given base) is not in > that interval to the conclusion that that somehow works against a > correct diagonal argument. We only need to show that for any list, > there is a real not in the list. That we can do that for a particular > interval if we wish to do so is gravy as far as the mere question of > the uncountability of the reals is concerned; and that there might be > a way to rig a list of reals in an interval such that the anti- > diagonal (in a given base) is not in that interval does not refute > that for any list there is a real not in the list. That is the sense I > mean. I disagree. The version of the diagonal proof that I have in mind is applied to [0,1), not to R. When correctly presented, the proof shows that there is a number in [0,1) that is not in the given list. The conclusion is that [0,1) is uncountable, and therefore R is uncountable. Andy's mistake was in using an incorrect version of the diagonal proof. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: MoeBlee on 30 Jan 2007 16:46
On Jan 30, 1:17 pm, Dave Seaman <dsea...(a)no.such.host> wrote: > I disagree. The version of the diagonal proof that I have in mind is > applied to [0,1), not to R. When correctly presented, the proof shows > that there is a number in [0,1) that is not in the given list. The > conclusion is that [0,1) is uncountable, and therefore R is uncountable. > > Andy's mistake was in using an incorrect version of the diagonal proof. I understand your point, but I am noting that there is also a mistake in logic even more fundamental. The diagonal proof can be used to apply to certain intervals or to the entire set of real numbers, as the argument is variously formulated. To prove that the set of reals is uncountable, it is only required to show that for any list of reals, there is a real not in the list. That we can get the stronger result that that real not in the list is in a particular interval is fine, but it is not required for merely proving that the set of real numbers is uncountable (unless you set up the diagonal argument in such a way as to require that the real not on the list is in a particular interval). Meanwhile, even IF someone had, say, fooled us into thinking that there is no diagonal method to construct a number in [0 1) but not in the list he gave us, then that would STILL be irrelevent in the sense I am mentioning, since we only have to keep in mind that to show the uncountability of the reals does not require (even if it can be done) showing that for every interval there is a diagonal method to construct a real not in that interval. MoeBlee |