Cantor Confusion
in [Math]
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From: WM on 12 May 2007 10:34 On 11 Mai, 21:52, Virgil <vir...(a)comcast.net> wrote: > In article <1178870415.724641.38...(a)n59g2000hsh.googlegroups.com>, > > > H&J could have written: "it does not matter that the set of these > > ordered pairs is described by a property" in your opinion? > > They could have, with no loss of correctness, said that. What set can exist without being described by a property? > > > In dealing with sets > > > of ordered pairs in general, one need not describe it as anything but a > > > set of ordered pairs. > > > Did I specify the rule, process, mapping > > You claimed the necessity of one without which no function can be > defined. That is correct. > > > And without having a domain, the function is not defined at all. > > If one can deduce a domain, it is not necessary to specify one. I did not specify one. I said not more than: "you know that a function consists of formula, domain and range, I hope." 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. And now it is definitely enough about this silly topic. Regards, WM
From: WM on 12 May 2007 10:35 On 11 Mai, 21:59, Virgil <vir...(a)comcast.net> wrote: > In article <1178870843.168676.226...(a)e51g2000hsg.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. > > Then WM does not know what an argument looks like. > > > > > 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. > > There is an ambiguity in your argument. > > While it is true that at every node there is more than one path through > that node, it is also true that for any two paths there is a ladt node > that they have in common. We have seen that these two "truths" are incompatible. Regards, WM
From: WM on 12 May 2007 10:40 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? > > Ad infinitum there are differing paths p' "with" p, In what do they differ? For which pair of nodes are more than one different paths required? 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. Regards, WM
From: WM on 12 May 2007 10:42 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? Or are you infinitely many and everyone has only to ask for one node? How do you synchronize? Regards, WM
From: William Hughes on 12 May 2007 10:45
On May 12, 4:41 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 12 Mai, 00:20, William Hughes <wpihug...(a)hotmail.com> wrote: > > > On May 11, 4:41 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > It need not change. Take just the path constructed in my last posting. > > > This path does not exists. As usual, your construction made use > > of an unjustified assumption that a set of paths could be chosen > > in such a way that *all* paths were the same. > > It can. See below. Or say why it cannot, according to your opinion. > > > > > > My question concerned your idea always to use that path p* which takes > > > the next exit > > > In a counterexample I gave I used a path p* that takes the next exit. > > However > > in general I do not claim that this must be true. I do claim that you > > always > > have to choose a path with will eventually exit. > > > >(irony warning). Why did you make this poor choice? > > > 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. Look! Over there! A pink elephant! This path p** is not p, because every piece belongs to the same foreign path. - William Hughes |