Cantor Confusion
in [Math]
|
From: WM on 12 May 2007 12:15 On 11 Mai, 22:40, Virgil <vir...(a)comcast.net> wrote: > In article <1178878942.895369.182...(a)e65g2000hsc.googlegroups.com>, > > > > > > > I ask: Why must this case be excluded from the proof? > > > > You answer: Because they are excluded. > > > > It was Cantor, or someone of his time, who created that rule, not me. > > > No, it was not Cantor. It was someone who recognized that lim[n-->oo] > > 10^-n = 0 in case of dual representation, i.e., 1.000... = 0.999... > > As Cantor DID recognize dual representations, that is insufficient > reason to be sure it was not Cantor. You are not quite right. It was not Cantor who recognized dual representation. You can convince yourself by reading Cantor's correspondence with Dedekind. On Jne 20,1877, Cantor wrote Dedekind about his proof mapping the area of the square to one of its sides. This letter contained decimal expansion with no regard to dual representation. It was Dedekind who two days later informed Cantor about this lack whereupon Cantor switched to continued fractions. > > > Unfortunately he did not recognize that he same holds for the > > difference between the diagonal number and a list entry in case of > > this limit. The reason is clear. Nobody ever considered the case that > > a real number is changed in the limit lim[n-->oo] except the natural > > "dual representation". > > Whatever does it mean to say that " a real number is changed in the > limit"? It means that the exchange of a digit a_n by another digit b_n changes the value of a number SUM a_n*10^-n the less the larger its index n is. As Cantor's diagonal proof concerns infinitely many numbers we need the limit for n --> oo. In this limit the exchange of a_n by b_n has no effect. Unless the fact was as I described, there is no justification for omitting the replacement of 9 by 0. If the limit n --> oo was not in the play, then 1.000... = 0.999... was never a problem. > > Why is there a problem with lim[n-->oo] 10^-n = 0 in one case but not > > in another one? > > If WM cannot see why on his own, then he has no grasp of the meaning of > Cantor's proofs. If you have a grasp, could you explain it? Regards, WM
From: WM on 12 May 2007 12:38 On 12 Mai, 16:45, William Hughes <wpihug...(a)hotmail.com> wrote: > > > Because you need a path that is not p, and any path that is not p will > > > eventually exit. > > > That is contradicted by the fact that there is always another path p* > > with p. Hence not every path can have exited at a node k(p, n) with a > > finite index n. Why should it be impossible to construct a path from > > these infinitely many nodes, i.e. from the pieces of paths p* =/= p? > > This path p** is not p, because every piece belongs to a foreign path. > This path p** is not p, because every piece belongs to the same > foreign path. (1) Every piece of p belongs to some foreign path. (2) The existence of p proves that these pieces can be put together to one foreign path. You do not oppose to (1). If you oppose (2) and claim that one foreign path does not suffer, then please prove that at least two foreign paths are required. Regards, WM
From: William Hughes on 12 May 2007 12:59 On May 12, 12:38 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 12 Mai, 16:45, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > Because you need a path that is not p, and any path that is not p will > > > > eventually exit. > > > > That is contradicted by the fact that there is always another path p* > > > with p. Hence not every path can have exited at a node k(p, n) with a > > > finite index n. Why should it be impossible to construct a path from > > > these infinitely many nodes, i.e. from the pieces of paths p* =/= p? > > > This path p** is not p, because every piece belongs to a foreign path. > > This path p** is not p, because every piece belongs to the same > > foreign path. > > (1) Every piece of p belongs to some foreign path. > (2) The existence of p proves that these pieces can be put together to > one foreign path. Actually no. p is hardly a foreign path. The existence of p can proves these pieces can be put together to form a path. > > You do not oppose to (1). > If you oppose (2) and claim that one foreign > path does not suffer, then please prove that at least two foreign > paths are required. > The fact that you cannot make a different path, p2, using all the nodes of path p1 follows trivially from the definition of a path. However, since trying to get you to agree to a definition is exceeded in futility only by trying to get you to stick to to one, I decline your offer to shift the burden of proof. - William Hughes
From: WM on 12 May 2007 13:59 On 12 Mai, 18:59, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 12, 12:38 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 12 Mai, 16:45, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > Because you need a path that is not p, and any path that is not p will > > > > > eventually exit. > > > > > That is contradicted by the fact that there is always another path p* > > > > with p. Hence not every path can have exited at a node k(p, n) with a > > > > finite index n. Why should it be impossible to construct a path from > > > > these infinitely many nodes, i.e. from the pieces of paths p* =/= p? > > > > This path p** is not p, because every piece belongs to a foreign path. > > > This path p** is not p, because every piece belongs to the same > > > foreign path. > > > (1) Every piece of p belongs to some foreign path. > > (2) The existence of p proves that these pieces can be put together to > > one foreign path. > > Actually no. p is hardly a foreign path. The existence of p can > proves these > pieces can be put together to form a path. > > > > > You do not oppose to (1). > > If you oppose (2) and claim that one foreign > > path does not suffer, then please prove that at least two foreign > > paths are required. > > The fact that you cannot make a different path, p2, using all > the nodes of path p1 follows trivially from the definition of a path. The fact that you cannot make more than one different paths p2, p3, p4, ... from the nodes of p1 follows equally well from the definition of path. > I decline your offer to shift > the burden of proof. > There is no offer. My proof stands. If you do not agree, you should switch from handwaving to arguing. I can promise you, your pink elephant will not be of any use for that task. Regards, WM
From: MoeBlee on 12 May 2007 14:33
On May 12, 7:34 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > What set can exist without being described by a property? (1) We can't specify such a set with a formula of set theory, otherwise the set wouldn't be a set not defined by a formula of set theory. However, you can look in, for example, Levy's 'Basic Set Theory' to see his mention that there is a proof that there is no definition in ZFC of a particular well ordering of the reals, yet in ZFC, such a well ordering is proven to exist. (2) On most general considerations, we prove IN Z that there are more than a countable number of sets, but we also prove ABOUT Z that there are only countably many formulas. Thus there are sets that are not defined by a formula. (3) Even if we do not specify a particular undefined set, then that does not contradict that the definitions of predicates such as 'x is a set', 'x is a relation', 'x is a function', et. al do not mention anything about there being a formula that defines x. > Then this little know-all replied: That you even think a set theoretic > function must have a FORMULA (!) as a component shows, again, your > lack of understanding even the basics of set theory. > > Therefore I showed him who obviously lacks some knowledge of basics: > Function, as understood in mathematics, is a procedure, a rule, > assigning to any object a from the domain of the function a unique > object b, the value of the function at a. I said from the start that the notion of a function as a rule, domain, and range is found in a lot of math books. That was NEVER at issue. What was at issue is what is the SET THEORETIC DEFINITION of 'function', and it has been showed to you and explained to you over and over and over that the SET THEORETIC DEFINITION is NOT that of a rule, domain and function, as seen, e.g., by the explicit definition that is indeed MARKED "definition" in the very Hrbacek & Jech you cited. The definition is: f is a function <-> (f is a relation & Axyz((<x y>ef & <x z>ef) -> y=z)) and that IS a fomula but it does not itself MENTION a formula nor a domain and range. And that cannot be changed by you just reiterating your mistaken claim over and over again. MoeBlee |