Cantor Confusion
in [Math]
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From: Virgil on 12 May 2007 15:37 In article <1178980831.862369.311960(a)u30g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Mai, 22:03, Virgil <vir...(a)comcast.net> wrote: > > In article <1178871714.942838.90...(a)w5g2000hsg.googlegroups.com>, > > > > > > > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 10 Mai, 22:25, Virgil <vir...(a)comcast.net> wrote: > > > > > > > 1) for every path p', there is a node where p' leaves p > > > > > and > > > > > 2) there is no node where every path p' has left p > > > > > then you must be able to name a node K(n) such that for nodes K(m < > > > > > n) > > > > > at least two paths p' and p'' are required to accompany p. > > > > > > This is true of EVERY path and for each of its nodes in an infinite > > > > tree. > > > > > > For every node K(n) of P, there is one path p' which accompanies it to > > > > the next node and then branches off, and another, p'', which does not > > > > branch off until the node after that, and so on, ad infinitum. > > > > > And the Moon is round and January is the first month in the year. So > > > what? I did expect that. And what you said is true. But it is not an > > > argument. > > > > > Ad infinitum there is a path p' with p, because ad infinitum there is > > > no node where p is single. Therefore, ad infinitum, there is a path > > > which shares with p the node in question and all earlier nodes too. > > > > But not all subsequent nodes unless they are the same path. > > Which node does not belong to another path? Which other path? Which node depends on which path, and cannot be determined until that path is specified. > > > > Ad infinitum there are differing paths p' "with" p, > > In what do they differ? Given any path p and any finite subset of its nodes, there will be UNCOUNTABLY many paths sharing that set of nodes. But given the same path and any infinite subset of its nodes, there will be NO other paths sharing all those nodes. For ANY infinite set of nodes, there is either a single path containing all of them or no path at all containing all of them. > For which pair of nodes are more than one > different paths required? For any pair of nodes both at the same level, no single path can contain both. > > We know that p always has company. I say one path is enough. You say: > > > but there is no > > single path, p', always "with" p unless p' = p > > Please prove your assertion by showing at least two paths which are > required such that p always has company. One of MY definitions of equality of paths, p = p' is that the set of nodes of p equals the set of nodes of p'. From which definition the assertion trivially follows. If WM has a different definition, he should declare it. An alternate, but equivalent definition, of path equality is that at each non-leaf level, n in N, p and p' branch in the same direction, both left or both right.
From: Virgil on 12 May 2007 15:39 In article <1178980936.455550.102150(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Mai, 22:07, Virgil <vir...(a)comcast.net> wrote: > > In article <1178872656.637980.112...(a)n59g2000hsh.googlegroups.com>, > > > > > > > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 10 Mai, 21:30, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1178796129.155631.279...(a)u30g2000hsc.googlegroups.com>, > > > > > > > Yes. But why only see the one side of the medal? For every path p > > > > > > > there is another path *existing in the tree* which is not > > > > > > > different > > > > > > > from p at any node, because p is not single at any node. > > > > > > What the above says is that for every path p there is a path p' in the > > > > same tree which is the same as p at every node but somehow not as a > > > > whole the same as p. > > > > > > But that is false in ZF or NBG trees. > > > > > > Whatever have I said that WM claims would be wrong > > > > > The line preceding your question is wrong. > > > > Then is WM claiming that > > "for every path p there is a path p' in the same tree which is the > > same as p at every node but somehow not as a whole the same as p." > > is NOT false, at least in ZF ands NBG? > > > > > There are not more nodes than we can ask for. > > > > WE can ask for all infinitely many > > How long will it take you? There are no clocks in set theory. > > Or are you infinitely many and everyone has only to ask for one node? > How do you synchronize? No clocks!
From: Virgil on 12 May 2007 15:49 In article <1178986533.351253.70010(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. I did not say Cantor discovered it, but you have just proved that Cantor did indeed RECOGNIZE it. > > > > > 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 Then say it directly instead of trying to obscure it. > 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. YOU may. WE don't. All WE need is test to see whether the diagonal number is different from the nth number in the list, which we have. > In this limit the exchange of a_n by b_n has no effect. So that for WM all real numbers are equal??? > > 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. I have no idea what you are trying to say, but whatever it is, it does not affect the validity of Cantors first diagonal proof ( for binary sequences) nor the variation as applied to decimals. > > > > 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? The diagonal binary string for any list of such binary strings is unlisted, as it differs in at leas one place from every listed string. Similarly for decimals. The proofs in detail are widely available. Read them.
From: Virgil on 12 May 2007 15:59 In article <1178987932.151519.266900(a)q75g2000hsh.googlegroups.com>, WM <mueckenh(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. The existence of p only proves that they can be put together to form p. In order to establish existence of your alleged foreign path, you would have to prove, contrary to fact, that all the pieces of foreign paths which yu claim exist belong to the same foreign path, which they do not. > > 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. In fact, infinitely many foreign are required, but no one of them contains all of p. In a CIBT, the intersection of the node sets of ANY two paths is always finite, containing those nodes in the chain of nodes from the root to the last common node of the two paths. This is equally true in finite trees. So that the set of paths containing any infinite set of nodes is either a singleton set (one path) or the empty set (no paths at all).
From: Virgil on 12 May 2007 16:10
In article <1178992778.495219.71020(a)e51g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. Your proof stands only on its head. A path in any CIBT contains the root node and one child node (either a left child or a right child) for each of its nodes. Two paths are the same (equal) if and only if at every node they branch in the same direction, or, equivalently, if and only if they have exactly the same set of nodes. Two paths are different ( not equal) if anon only if there is some node from which they branch in opposite directionsor, or equivalently, if and only if they have different sets of nodes( the symmetric difference is not empty). > If you do not agree, you should > switch from handwaving to arguing. Our proofs seem to have no effect, so why not handwaving? > I can promise you, your pink > elephant will not be of any use for that task. It is at least as valid as WM's "arguments". |